J\

HINTS AND ANSWERS;

BEING A

KEY

TO A COLLECTION OF

CAMBRIDGE MATHEMATICAL

EXAMINATION PAPERS,

AS PROPOSED AT

THE SEVERAL COLLEGES.

ISY

J. M. F. WRIGHT, B. A.

AUTHOR OF THE PRIVATE TUTOR, A TRANSLATION OF NEWTON'S PRINCIPIA, SECTIONS I. II. III., &C. &C. &C.

PART I.

CONTAINING EUCLID, ARITHMETIC, AND ALGEBRA.

CAMBRIDGE : PRINTED FOR W. P. GRANT,

AND SOLD DY

WHITTAKER & CO., AND SIMPKIN * MARSHALL, LONDON

1831.

H>

CAMBRIDGE :

PRINTED ET W. METCALFE, ST. MARY'S STREET.

PRE F AC E.

This work comprises not only Hints for the solu- tion of the more difficult parts of questions, but Answers for all, over and above the complete de- velopement of many of the most intricate. The full proof, or investigation, is bestowed upon nearly all the last five Algebraic Problems and Equations in each of those papers given at St. John's College which invariably consists of seven questions, — these papers having been found far more difficult of treat- ment than any proposed at other Colleges. In- teresting questions also have been somewhat fully solved and exemplified ; and the shortest practical methods have constantly been suggested on every available occasion. Notwithstanding so much has been done for the student, the work does not greatly exceed double the bulk of the questions them- selves.

It were almost suoerfluous to direct students in the proper use of these aids. Let no Hint, Answer, or Solution be referred to, before the student's own patient efforts have entirely failed. A re- ference then may be equally advantageous with the

IV PREFACE,

help of a tutor. But less benefit will accrue in proportion to the contrary being practised. To boys at school, whose only aim is to get through their tasks, Keys are detrimental ; but to students at an University, where high honours and rewards await distinguished success at Examinations, the motives being essentially different, they will be rendered to good account — at least, by those who value their own progress in knowledge.

As the subjects of Geometry or Euclid, Arithme- tic and Algebra, form a principal part of the studies for the degree of B.A., as well as for the Examina- tion of Freshmen, it is presumed the volume will not prove unwelcome to those Undergraduates, who are considered the "non-reading" portion of the University.

Another volume, containing " Hints and Answers in Trigonometry, and the Differential Calculus" is in progress, and will speedily be published.

Trinity House, Christ's Piece, Cambridge. Marrli 1, 1831.

BOOKS REFERRED TO.

Euclid's Elements. Simsoris. Wright's Self- Examinations in Euclid. CresweU's Maxima and Minima. Bland's Geometrical Problems. Wright's Pure Arithmetic. Wood's Algebra. Wright's Private Tutor. Barlow's Theory of Numbers.

HINTS AND ANSWERS

IN

EUCLID, ARITHMETIC, & ALGEBRA.

EUCLID.

TRINITY COLLEGE, 1824.

[P. 1.

1. The principles of construction are the three Postulates of Euclid.

2. Euclid, Book i. prop. 4.

3. Take any right-angled a ABC, right-angled at C ; from C draw CM _|_ AB, thus obtaining three right-angled a *. Then, since the a s ACM, ACB, have two angles in each equal, their third angles are equal ; that is, L ACM = Z. B. Similarly, it may be shown that z. BCM = Z. A.

Hence, z. A+ ^- B = L ACM+ z. BCM ** a right angle. Consequently the two angles of a right-angled a which are not the right angle, are equal to one right angle.

Again, take any a whatever abc, and from c draw cm JL ab ; then, by what has been shown,

/_ a+ £. acm = one right z.,

and Z. b 4- Z. bcm — one right Z. ,

.'. Z. a+ z. 6+ jLc — two right z.*.

4. Euclid, i. 35.

5. Let P be the given point in the side AB of the given A ABC. Suppose the thing done, and that the required

line, bisecting the a , meets the side BC in the point Q ; then the a BPQ = | a ABC.

2 HINTS AND ANSWERS [P. 2.

Complete the parallelogram BPQR. This parallelogram ■= 2 a BPQ == a ABC. Whence the Synthetic construc- tion and demonstration are evident.

For upon BP describe any parallelogram equal to the A ABC, (Euclid, i. 44), and let Q be the point where the upper side of the parallelogram enters the side BC. Join PQ. Then v &c. as is evident.

6. Euclid, i. 46, and Wright's Self-Exam, in Euclid, p. 4.

7. Euclid, i. 47. 8. Euclid, ii. 14. 9. Euclid, iii. 2. 10. Wood, Alg. art. 515. 11. Euclid, iv. 11.

12. They are an equilateral a , a square, and a regular hexagon.

13. For problems of this kind, see CreswelVs Maxima and Minima.

14. See Wright's Self-Exam, in Euclid, pp. 73, 74, 75, 76.

15. Euc.v. 12. 16. Euc. vi. 1. 17.Euc.vi.4. 18. Euc.vi.25.

19. To the side C of the rectangle A x C apply a rectangle = B x D ; then the base of A x C has to the base of the whole rectangle the ratio required.

20. For if the straight lines AB, CD cut the given lines AC,BD in A, B; C, D, respectively, and intersect in P; and likewise two other straight lines A'B', CD', parallel to the former, cut the same given lines in A', B', and C, D', respec- tively, and intersect in P' ; then the a s APC, A'P'C are similar, and also DPB, and D'P'B'. .'. &c.

21. There are not data sufficient that the square shall be constructed in position, but only in magnitude; for it is clear the square, when constructed, from these given distances, might all assume any position around the point.

If a, b, be the two shorter distances, and c the longest ; then it may easily be shown that the side of the square is /62 + c2 ± v fo62c2-(2a2-62-c2)^

whence the construction is easy.

P. 3.] IN EUCLID. 3

One solution belongs to that case in which the given point lies without the square ; the other to that in which it lies within the square.

The problem is impossible when

2a2-62-c2 is > 2bc,

i • & + c

or when a is > — - ,

or when -j is > — — ; > ratio of the side of a square to its

b + c *J2 1

diagonal.

This is also evident, from the consideration that two sides of a A must be always greater than the third side.

22. Join the two given points A, B ; bisect AB by the per- pendicular ED, meeting the given line CD in D. Then the centre of the required circle is in ED ; and the distance of its centre O from D is

DE3 ± DE V(DE2. EF2+EF2. EB2-ED2. EF2)

EF making an L with CD equal the given Z. .

23. Let AB be any side of the regular polygon, C being the centre of the circumscribed circle. Bisect the z. A'CB by the straight line CM ; then AM = MB. Again, bisect the Z. MCB by Cn, and draw Cm, Cb' J_ Cn, and make each of them = ^ MB. Draw m'm, b'b parallel to Cn, to meet CM, CB in m, b; join mb ; then, it is easily shown, that mb= MB. In like manner draw ma = AM ; am, mb, will be two sides of the new regular polygon ; and, similarly, all the others may be found.

TRINITY COLLEGE, 1828.

1. Euc. i. 24. 2. Euc. i. 45. 3. Euc. ii. 7. 4. Euc. ii. 11. 5. Euc. iii. 27. 6. Euc. iii. 33. 7. Euc. iv. 5. 8. Euc. iv. 13. 9. Euc. v. 10. 10. Euc. v. 22. 11. Euc. vi. 22. 12. Euc. vi. 33.

13. Let the straight lines AC, BD, cut by AB, make the angles BAC, ABD together less than two right angles. Draw

4 HINTS AND ANSWERS [P. 4.

AE, making the angle BAE + angle ABD equal two right angles ; then, by the assumption, AE and BD cannot meet. Consequently, AC and BD will meet.

14. This is easily proved, in a manner similar to that of Prop. 47. B. i. of Euclid.

15. First, make a square = given rectangle (Euclid, ii. 14). Next, on the given line describe a semi-circle ; at the extre- mity of the line draw a J_ which is = the side of the square ; from the end of this J_ draw a line J_ to it meeting the semi- circle in a point from which the J_ upon the given line will divide it as required.

16. See Bland's Geomet. Problems, p. 231 ; or make use of Euclid, iv. 10.

17. This depends upon Euclid, ii. 12 and 13, and upon B. vi. prop. B.

18. Let the chords AB, ab, be produced to C ; and let O be the centre of the circle ; join OA, OB, Oa, Ob, OC, and produce CO to the circumference in D.

Thenz.DOA= Z.OAC + z. OCA = zOBA-j- z. OCA, == Z.BOC+2Z.OCA, Z.DOA-Z.BOC

2

&c.

Z.OCA;

19. From one extremity A of the base of AB of the a as a centre and a radius = sum of the undetermined sides describe an arc. From the other extremity B draw any line BC to the circumference; join AC, and make z. CBD = BCA, &c.

See also Euclid, vi. 11.

11. (1). Euclid v. 25.

(2). See Wright's Self-Examinations in Euclid, p. 101.

12. (I). Euclid, vi. 3.

(2). The required Locus is a Conic Section, whose focus is C major axis along BC, vertex D, and whose direc- trix passes through B. It is a Parabola, Ellipse, or Hyper- bola, according as CD is —, > , or < DB.

13. (1). Euclid, vi. 19.

(2). Let the first part abC of the a ABC be made by the line ab parallel to AB ; and suppose it be required by the problem that a abC : a ABC : : m2 : n2, m and n being two lines ; then, take aC a fourth proportional to n, m, and AC (Euclid, vi. 12). Hence the rest of the problem is easy.

14. Euc. vi. 31.

For the Lunes of Hippocrates, see Wrights Self-Examina- tions in Euclid, p. 172.

TRINITY COLLEGE, 1828.

1. (1). The Converse of a Proposition is that proposition in which the data and qusesita of the latter are the quaesita and data of the former respectively ; thus, Euclid, i. 6. is the con- verse of Euclid, i. 5.

The 8th, and all demonstrations ex absurdo, are indirect demonstrations ; as to contrary propositions, the term is not to be found in any standard work.

All problems that admit of an infinite number of different solutions, are indeterminate.

P. 9, 10.] IN EUCLID. I 1

Thus, if the problem be To find two squares whose sum shall equal a given square.

Upon the side of the given square, describe a semicircle, and taking any point in the arc of the semicircle, and joining it with the extremities of the diameter, we shall have a solution of the problem. .'. &c.

(2). Definitions 15, 30, contain superfluous conditions, it being sufficient that a circle should have three equal radii ; and that a square should be equilateral, and have one right angle. See Wright? s Self-Examinations in Euclid on those subjects.

(3). Euclid's Definition involves the idea of distance, or the opening of a pair of compasses ; whereas, this, requiring a straight line to be drawn equal to the given straight line as radius, assumes Euclid, i. 2.

2. Euclid, i. 9. See Wright's Self-Exam, in Euclid, p. 15.

3. Euclid, i. 11.

From any point P, without the given line, describe a circle passing through the given extremity A, and cutting the given line in another point Q ; join PQ, and produce it to meet the circumference in R ; join RA : it is the J_ required.

4. Euclid, i. 24. See Self-Examinations in Euclid, p. 21.

5. Euclid, i. 48.

Because otherwise the a and square, although upon the same base, would not be shown to be between the same paral- lels ; and they, therefore, could not be supposed equal.

6. Euclid, ii. 5. 7. Euclid, ii. 14. 8. Euclid, iii. 7. For the greater segment continually decreases, and the less

increases up to that position in which they become equal by Euclid, iii.

9. Euclid, iii. 20.

It is true, and proved verbatim by case 1. The only de-

]2

HINTS AND ANSWERS

[P. 10,11.

duction is, that the angle BEC is greater than two right angles ; or is, what is termed in Fortification, a Salient angle.

10. Euclid, iv. 12. 11. See Wright's Self-Examinations in Euclid, pp. llet seq. 12. Euclid, v. 15. 13. Euclid, v. 17.

14. From similar a s, AaZ, BbZ,

Ka : AZ : : Bb : BZ, Cc : CZ, Dd : DZ, &c. But AZ is > Aa;

.-. BZ is > Bb or BC ; CZ is > Cc, &c, so that DZ will never be exhausted, let the series be continued ever so far.

Also, if lines from b, c, &c. be drawn parallel to AZ, it appears that

AZ : Aa :: AB : Aa-Bb, BC : B6-Cc, CD : Cc-Dd, &c. AB

or, AZ : AB

AB : BC

But BC : BZ

AB-BC, BC : BC-CD, CD : CD-DE, &c.

: BC . CD, : CD : DE, &c.

: AB : AZ, BZ

BC = AB.

AZ

also, CD

BC2 ~ AB'

= AB.

&c.

BZ2 AZ2'

Observe, the sum of them is in Geometric Progression. 15. Euclid, vi. 22. 16. Euclid, vi. 25.

P. 11, 12.] IN EUCLID. 13

17. Take a straight line, a mile long, and to it apply a parallelogram that shall equal the given square (Euclid, i. 45). From either extremity of the base, with distance = a mile, describe a circle cutting the side opposite, and complete the Rhombus.

18. By Euclid, i. 47, diag.2 : (side)2 : : 2 : 1,

.". &c. See also Leslie's Geometry on the subject.

19. From the extremities B, C, of the base, let perpendicu- lars be drawn to meet the bisecting line in b and c ; produce B6, Cc, to meet the sides in B', C, and complete the rectangles AC, and Cb ; then the sum or difference of them will equal the a , &c

20. If B, C, be the equal angles of the base, and A the vertical angle, and P any point in the side AC, draw from P a straight line to AB, produced, so as to cut off a a equal the given a ; we must join PB, and from P draw PQ parallel to PB. Then PAQ will be the a required.

Also PQ is greater than BC. For, since the perpendicular from A upon PQ is less than that upon BC, and the areas of the triangles ABC, APQ, are equal; .'. base PQ is greater than the base BC.

TRINITY COLLEGE, 1829.

1. See Wrights Self-Exam, in Euc. p. 5. 2. Euc. i. 7. 3. Euc. i.26. 4. See p. 13. No. 17. 5. Euc. ii. 13. 6. See Wright's Self-Examinations in Euc. p. 39. 7. Euc. iii. 21. 8. This depends chiefly on Euc. i. 8. 9. Euc. iii. 31. 10. Euc. iv. 5. 11. Euc. iii. 36; and see p. 5. No. 7. 12. Euc. iv. 10. 13. Euclid, v. C. 14. Euclid, v. 16.

15. The perpendicular line is evidently the shortest of the three, for with respect to the others, it subtends an acute angle, whilst they subtend a right angle. It remains, then, to

14 HINTS AND ANSWERS [P. 13.

show, that the line bisecting the base is greater than that which bisects the angle.

From the angle A of the a ABC draw AD bisecting the angle CAB, and meeting the base in D, and AE bisecting the base in E. Then, by Euc. ii. 12, and 13.

CA2 + AB2 = 2CE2 + 2AE2, and (Euclid, vi. B.). CA. AB = CD. DB + AD2, /. 2AE2 = AB2+AC2-2CE2, and 2AD2 = 2AB. AC-2BD. DC, .'. 2AE2-2AD2 =fe (AB-AC)2 + 2(BD. DC-CE2),

See Euclid, ii. 7. But Euclid, ii. 5.

BC. DC + DE2 = CE2,

.-. 2AE2-2AD2 = (AB-AC)2+2DE2,

/. 2AE2 is > 2 AD2 by (AB-AC)2 + 2DE2,

.*. &c.

16. Euclid, vi. 6. 17. Euclid, vi. 19.

18. Bisect the angle ACB of the sector by the line CD, meeting the arc AB in D, and from D draw EF _]_ CD, and meeting CA, CB produced in E, F. Then inscribe a circle in the isoscles a ECF ; this will be the circle required.

19. Euclid, vi. 33.

TRINITY COLLEGE, 1830.

1. Euclid, 15. 2. Euclid i. 42.

3. They are as 18 to 20, or as 9 to 10. Euc. i. 32. cor. i.

4. Euclid, ii. 7. 5. By Euc. ii. 12, and 13.

6. Inscribe in the circle an equilateral a , and from any two of its angles draw tangents. They will meet in the point re- quired ; and its distance from the centre .will thus be deter- mined. The distance = the diameter of the circle.

7. Euclid, iii. 22. 8. Euclid, iii. 35.

P. 14.] IN EUCLID. 15

9. For questions of this kind, see CresweWs Maxima and Minima.

10. Let the _|_ BE cut AC in M, and the J_ CF cut AB in N, the J_s intersecting in P. Then, since in the right- angled as PMC, PNB, the vertically opposite angles are equal; .". the angle NBP = z. MCP. .". AE = AF.

And similarly for the other two pairs of arcs.

11. Euclid, iv. 13. 12. Euclid, v. 8. 13. Euclid, v. 12.

14. Upon AC, CB describe two segments of circles, contain- ing each the same angle. These will intersect not only in C, but in some other point P, which will be the point required.

N. B. — The problem is indeterminate. The locus of P is a conic section.

15. Euclid, vi. 4. 1(5. Euclid, vi. 18.

17. Let AM be drawn J_ base BC of the a ABC; then Euclid, i. 47.

AB2 == AM2 + BM2, and AC2 = AM2 + CM2, .-. AB2-AC2 = BM2-CM2; .'. Euclid, ii. 5. (AB + AC)x(AB-AC) = (BM + CM) x (BM-CM), .-. &c.

18. First, construct a a upon an assumed base, having given the ratio of the sides, and the difference of the angles at the base0 This is easily effected, by first dividing the base in the given ratio, and then making the a as in Question 14 of this Paper, observing that the circles must be such, that the angles of the base shall have the given difference.

This being done, describe a a similar to one found, but having an altitude = the given perpendicular.

16 HINTS AND ANSWERS [P. 15, 16.

BOOK XI. TRINITY COLLEGE, 1824.

1. Ans. Three points which are not in the same straight line.

2. Euclid, xi. 4. 3. Euclid, xi. 6.

4. This is a Definition, and not a Theorem. Euclid, xi. Def. 6.

5. (1). Euclid, xi. 21.

(2). There is no exception. This is also seen in Sphe- rical Geometry.

6. (1). See Legendre's Geometry.

This is the fundamental proposition in Spherical Trigono- metry,

. cos a — cos b cose

viz. cos A = -. — -, — -. .

sin b sin c

(2). No; for any two of them must be greater than the

third. Euclid, xi. 20.

TRINITY COLLEGE, 1825.

1. Euclid, xi. 9. 2. Euclid, xi. 11. 3. Euclid, xi. 17. 4. Euclid, xi. 21.

TRINITY COLLEGE, 1826.

1 . Euclid, xi. Def. 11. 2. Euclid, xi. 6. 3. Euclid, xi. 1 1 . 4. Euclid, xi. 17. 5. Euclid, xi. 21. 6. Euclid, xi. 28.

TRINITY COLLEGE, 1827.

1. See Euc. xi. Definitions 2. Euc. xi. 4. 3. Euc. xi. 11. 4. Euc. xi. 18.

5. Draw a plane at right angles to the straight line AB, then it will be at right angles also to the straight line CD, which is parallel to AB (Euclid, xi. 8). Then the common

P. 17.] IN EUCLID. 17

section of any two planes passing through AB and CD will also be at right angles to the aforesaid plane (Euclid, xi. 19). Hence (Euclid, xi. 6), the common section is parrallel to either AB or CD.

TRINITY COLLEGE, 1828.

1. Euc. xi. 4. 2. Euc. xi. 8. 3. Euc. xi. 14. 4. Euc. xi. 17. 5. Euc. xi. 18.

6. Euclid, xi. 21. A solid angle is measured by that por- tion of the surface of a sphere which subtends the solid angle at its centre.

TRINITY COLLEGE, 1829.

1. Euclid, xi. def. 6. The inclination of two straight lines not lying in the same plane is that acute angle which is formed by drawing from any point in one of them a straight line parallel to the other.

2. Euclid, xi. 5. 3. Euclid, xi. 10. 4, Euclid, xi. 20.

5. The projection of a straight line on a plane is the inter- section of a plane passing through the straight line J_ to the plane of projection.

Hence, the plane which projects the straight line, which is perpendicular to one plane, upon a second plane, is perpendi- cular to both planes, and, therefore, to their intersection.

6. Let AB, CD, be the given lines. Join AC, and com- plete the parallelograms ACDE, ACFB; then complete the parallelogram FCDG ; then complete the parallelogram EDGH, and join BH. The figure will be the parallelopided required.

ST. JOHN'S COLLEGE, 1825.

1. Let C be the centre of the circle, and DF touch the circle in E. Join CD, CE, CF. Then prove the A ECF = z. FCB, and a. ECD = z. DCA, &c.

2. Euclid, v. 12.

D

18

HINTS ,,ND ANSWERS [P. 18.

3. Let ABC be an isoscles a , having AB = AC. Draw BM J. AC; then, Euclid, ii. 13.

BC2 = 2 AC2- 2 AC x AM, = 2 AC x CM, (Euclid, ii.) or BM2+CM2= 2 AM x CM + 2 CM2, (Euclid, ii.) .-. BM2 = 2 AM x CM.

4. For if the polygon consist of 2n sides, it may be divided into n quadrilateral figures, all inscribed in the circle. Whence (Euclid, iii. 22) the proof is easy.

5. Let AB, AC, be the equal sides of the a . From A with radius AP > \ AB but < £ (AB + BC) describe a cir- cle, and from B with radius = 2 AP describe another inter- secting the former in P. Then P is the point required.

6. Join the diagonals, and make use of Euclid, ii. 12 and 13.

7. Bisect each of the sides of the a in P, Q, R ; and join PQ, PR, QR. Then PQR is the a required.

8. First, prove that the as TED, IHF, ICE, IFG, are similar.

9. From similar as, Bp : AB : : BC : AC,

pq : Bp : : BC : AC, qr : pq : : BC : AC,

&c.

„ . AB.BC A , BC BC2 \

• • Bp+pq + qr + Scc..= AQ [I + -^ + ^-2 + ...to oo J

AB.BC ~ AC-BC ' .'• Bp+pq + . . . . : BC : : AB : AC-BC.

10. Euclid, xi. 20.

ST. JOHN'S COLLEGE, 1828.

1. Euclid, v. 13. 2. Euclid, vi. 15.

3. This is easy, after knowing Euclid, vi. 15.

4. Euclid, xi. 20.

5. First, describe any a having each of the angles at the base double of the vertical angle, Euclid, iv. 10.

P. 19.] IN EUCLID. 19

Then, at the centre of the circle, make two angles, which shall equal the supplements of the equal angles of this A , by straight lines drawn to the circumference in P, Q, R. At P, Q, R, draw tangents. These will form the a required.

6. Since the base and area of the a are given, .\ the alti- tude is given. Consequently, place this base so that its extre- mities may lie on adjacent sides of the square. From one extremity draw a perpendicular to the line, and make it = the altitude of the a . From the upper extremity of this J_ draw a line parallel to the base, and meeting a third side of the square, &c. &c.

7. That is the greatest whose base is the side of the square, and whose vertex is in the opposite side of the square, and this = half the square.

8. Let the circles cut in A, B. Join their centres C, D, the line CD cutting the circumferences in E, F. From E draw EG meeting the circle EA in G. Join GF. Inflect FH in the circle AF, making it = GE. Join EH. Then GH is the parallelogram required. Find the limitations.

9. From A, the centre of the circle, draw AC = given straight line, and making an angle with the given radius BA pro- duced, equal to the given angle. From C draw CD parallel to AB, and meeting the circumference in D. Then D is the point required. Find the limitations.

10. If the given lines AB, CD, are parallel, there is no diffi- culty. If not, produce them to meet in E. In EA, EB, take respectively EF, EG in the given ratio. Draw FH, GK J_ AE and EB to meet in L. Join EL. Then EL is the line required.

11. Divide the base AB in E, so that AE : EB may be the given ratio. Then determine a point D in the _j_ CD, such that ED may bisect the angle ADB. See p. 15. No. 14.

12. Let the chord AB be produced both ways from A to C, and from B to D, making AC = BD. From C, draw

20 HINTS AND ANSWERS [P. 20.

CE touching the circle in E, and bisect AB in F. Join EF, and produce it to the circumference in G. Join DG. Then prove that DG = CE, and is therefore a tangent.

To effect this, take O the centre of the circle. Join OC, OF, OD, OE, OG, and use the as COE, DOG.

13. Let BAC be the given vertical angle, then, drawing CMj_AB, the ratio AM : CM is known = r, suppose.

But if m2 be the sum of the squares of the sides containing the vertical angle, and c the unknown side opposite the L C, and a the base ; then, Euc. ii. 13,

a2 = m% _ 2 c x AM. Also, if a2 be the given area of the a ,

2 a2 = c x CM, .'. a2 = ?n2 — 4 r x a2. Take .'. a semicircle on the diameter PQ = m, and inflect a straight line PR = 2 ^r. a. Join QR. QR is the base of the a . On QR describe a segment of a circle containing an

Z. equal = the given angle Z. , and from P draw PS J_ PQ,

2 a2

making it = , and draw ST meeting; the circumference in

a

T. Join PT and QT. The a PQT is that required. All this may be easily put into Geometrical form.

14. See CresweWs Maxima and Minima.

CORPUS CHRISTI COLLEGE, 1825

1. Euc.i. 47. 2. Euc. ii. 10. 3. Euc. ii. 13. 4. See p. 5. No. 7.

5. Let AB be the straight line. Bisect it in C. Draw BDJ_AB, and make it equal to the side of the given square. Join CD, and in AB produced take CP = CD. Then,

since PB x PA + BC2 = PC2. (Euc. ii. 6.) and PC2 = CD2 = BC2 + BC2, .'. PBxPA= BD2. See, also, Bland's Geom. Problems, p. 156.

6. Euc. ii. 10.

7. Let AB be the position of the base ; CD parallel to AB, and distant from it by the given radius ; then CD is the locus

P. 21, 22.] IN EUCLID. 21

of the centre of the inscribed circle. From centre A, at dis- tance = given difference, describe a circle, &c. The locus is a circle.

8. Euc. vi. 4. 9. Euc. vi. 19. 10. Euc. vi. B.

CORPUS CHRISTI COLLEGE, 1826.

I. Euc. i. 21. 2. Euc.i. 44. 3. Euc. i. 48. 4. Euc. ii. 9. 5. Euc. iii. 13. 6. Euc. iii. 35. 7. Euc. iv. 12. 8. Euc. vi. 19. 9. Euc. vi. D.

10. Take CB : AC : : AC : CP, and P will be the point required.

II. Let CAB be the a, having the obtuse angle CAB. About it describe a circle whose centre is O. Join OA, and upon it describe a semicircle intersecting the base BC in D. Join AD. Then AD is a mean proportional between CD, DB. For, join OD, and produce AD to the circumference in E, &c. &c. It is a necessary limitation, that the angle shall be either right or obtuse. Why ?

12. This is Euc. ii. 11 in disguise.

CORPUS CHRISTI COLLEGE, 1827.

1. Euc.i. 5. 2. Euc. i. 35. 3. Euc. i. 27. 4. Euc. ii. 13. 5. Euc. iii. 9. 6. Euc. iii. 21 . 7. Euc. iii. 35. 8. Euc. iv. 10. 9. Euc. v. 25. 10. Euc. vi. 3. 11. Euc. vi. 19. 12. Euc. vi. 31. 13. Euc. ii. 12 and 13.

14. This problem is indeterminate.

From an assumed point D, in the side AB of the a ABC, draw DE parallel to the base BC. Make DF : DE in the given ratio, and from D describe an arc cutting BC in F, and complete the parallelogram whose sides are ED, DF.

15. Let AB be the given base, DC the position of the bi- secting line produced to meet the base in C. Draw BEj_CD, and produce it making EF = EB. Join AF, and produce it to meet CD produced in G, and join GB. Then AGB is the

A required.

22 HINTS AND ANSWERS [P. 23.

16. Let ABCD be the trapezium; bisect the opposite sides AB, CD in E, F. Join EF, and bisect it in P. Join PA, PB, PC, PD. Take any other point F, and join FA, FB, PC, FD, FE, FF.

Then prove that PA2 + PB2 + PC2 + PD2 _ 2AE2+2CF2+4PF2, FA2+FB24-FC2 + FD2 = 2AE2+ 2CF2 + 4PF2+4PF2, then, the excess of the latter four squares above the former is 4 PP'2, which more than proves the prop.

CORPUS CHRISTI COLLEGE, 1830.

1. See p. 11. 2. See Wright's Self-Examinations in Euc. p. 4. 3. See Wright's Self-Examinations in Euc. p. 5. 4. Euc. i. 5. 5. Euc. i. 13. 6. Euc. i. 19. 7. Euc. i. 32. See Wright's Self-Examinations in Euc. p. 25. 8. Euc. i. 40. 9. Euc. ii 1. See Wright's Self-Examinations in Euc. p. 39. 10 Euc. ii. 13. 11. Euc. iii. 20. See Wright's Self-Ex- aminations in Euc. p. 55. 12. Euc. iii. 22. 13. Euc. iii. 35. 14. Euc. iv. 13. • 15. See Wright's Self-Examinations in Euc. p. 75. 16. Euc. v. i. 17. Euc. v. 8. 18. Euc. vi. 4. 19. Euc. vi. 19. 20. Bisect its three sides, &c.

21. First, if the sides opposite A, B, C, be called a, b, c, A being the right angle ; then, prove that

R (a + b + c)_ AD x a

~2 "" 2~~ '

thence that

be ADxDB , AD x DC

a+6+c'r~HAD+DB'?' ~ b+AD + DC

6c2 , c62

thence get r = —, j— — r, r = — y— — x.

• & a (a+b + c) a(a + b + c)

7,2-2 C.21/,2] £2-2 ,

Hence r2 + r'2= \e. \ \ +. ° >= . , ■* * = R2, .'. &c.

a2(«-fo + c)2 (a + b + cy

Observe. The circle inscribed in the a ABC may be hence proved to equal the sum of the other two.

22. Upon the side describe two segments of circles, contain- ing angles = the given Z. and ~ double the given Z.. Inflect a straight line = the sum of the sides, &c

f". 25.] in kuclii). 23

CHRIST'S COLLEGE, 1828.

1. Euc. i. 16. 2. Euc. i. 45. 3. Euc. i. 48. 4. Euc. ii.6. 5. Euc. iii. 17. 6. Euc. iii. 32. 7. Euc. iii. 37. 8. Euc. iv. 4. 9. Euc. iv. 11. 10. Euc. vi. 8. 11. Euc. vi. 15.

MISCELLANEOUS QUESTIONS.

12. The product is .000392.

13. Ans. ^ is > f by &.

14. Ans. 9.9. 11 id

15. Ans. 14/. 5s.

16. His expenditure exceeds his income by 38/. 5s. a year. He is in debt 229/. 10s.

17. v/3 1595641 = 5621,

^/ 175616 = 56.

18. The quotient is 16/. 4.9. 9d, f .

19. The quotient is ^ — 1- . _ + ■„ _ + &c.

^ 2 a 4 a2 8 a3

The product ismnx3 — (np -+- mp -f m)x% + (nq + p1 4- p)x — pq — q,

20. The root is ax— 6 + |.

21. The value is ^\.

22. TheG. C. M. isa-3b.

47— 1

23. The first sum is 272 ; the second is — ^ — , or 5461.

24. See Wood, art. 253.

x

25. (1). -= = 4, and — \, whence x.

(2). x — 9.

(3). x = 0, and \/3, and — \/3.

(4). Clearing of fractions, &c.

,r2-12.r == -32, .-. x = 8 and 4. (5). From the first \/ {x + y) = 3, and — 6, .". x + y — 9 and 36. Whence, by means of the second, the question is easy. (6). x = 7, y = 10, z = 9.

24 HINTS AND ANSWERS [P. 26.

ST. PETER'S COLLEGE, 1825.

I. Euc. i. 4. 2. Euc. i. 9. 3. Euc. ii 14. 4. Euc. iii. 3. 5. Euc. iv. 6. 6. Euc. v. 15. 7. Euc. vi. 25. 8. Euc. xi. 5.

9. To the given base apply a parallelogram equal to double the given a (Euc. i. 44) ; then bisect the given base, and from its middle point draw a JL meeting the side or sides pro- duced of the parallelogram in a point. This point will be the vertex of the required A .

10. With the given radius describe a circle ; then drawing any radius OP, at P, draw a tangent BPA ; at O the centre, make the angle POA = a right z. — \ given z. A. Also, in like manner, on the other side of PO, make the angle POB = right z. — i B, and from A and B draw tangents meeting in C ; then ABC is the a required.

II. From the point A, with a radius equal a side of the given square, describe an arc cutting the given circle in C ; then join BC, and produce it to meet the tangent in P. P is the point required.

When the side of the square is greater than the denominator of the 0 , the problem is impossible.

12. Upon the straight line describe a ^ © , and from either extremity draw a straight line, making an z = | a right z. , and meeting the circumference in P. From P draw PM _[_ to the diameter. M will be the required point of section. There are two such points.

13. If r be the radius of the © , s the side of the square, and o that of the octagon; then, taking the similar right-

angled a s in a J © , whose sides are o, r — ^ ; and 2 r, 0, we

get 2 r : 0 : : 0 : r — ^.

Again, C being the centre of the © , CA its radius, AB the side of the hexagon = C A = r ; then produce AB to P, so

P. 28.] IN EUCLID. 25

that AP = side of the square =s s, and draw PQ touching the © in Q ; then, since

PQ2 = PA x PB = (AP-AB) x PA = (s-r) xs= s2- rs, .'. PQ2 = 02, and PQ = 0.

14. From the given point P draw a tangent to the © in Q, and let R : 1 be the given ratio ; then take x a point in the concave circumference, such that

Px2 : PQ2 : : R-fl : R, and x is the point required.

15. Divide the base BC=«, in the ratio of m : n ; upon a describe a^O , and at D erect the _L DE, meeting the circum- ference in E. Then, with B as centre and distance = BE, describe an arc cutting the base a in A'. From A' draw A'B' parallel to C.

The proof depends upon similar A s being in the duplicate ratio of their homologous sides.

16. From the points where the parallels meet the plane, draw J_s to the plane. The J_s are parallel (Euc. xi. 6). Hence we have two lines which meet parallel to two others which meet. They, therefore, contain equal angles (Euc. xi. 10). Hence, &c.

17. See Wood/louse's Trig., or Hind's.

18. See Woodho use's Trig., or Hind's.

Also, if A = 18°; then 2 sin 18. cos 18 = sin 36= cos 54,

whence sin 18° = ^— j .

4

19. Since the ratio of the sines of the angles A, B, C, is the same as that of the opposite sides a, b,c ; let

a : b : c : : m : n : r,

71 T

and suppose a -j- b 4- c = p ; then b = a — ,c = a. — , rr r m m

(71 T \ 1 H 1 ) = P, whence a, and .*. b, and m m J l

<-.

20. If A be the vertical angle, B, C the others, a, b, c,

E

26 HINTS AND ANSWERS [P. 29.

the opposite sides, and s, s' the segments of a, contiguous to B, C ; then (Euc. i. 47).

c2	—s2 =	(perp.)2	=	P-	s'%
. s2-	-s'* =	e2-62,
and	$+s' :	c + b : :	c-	-b	: A'-

21. From centre C, and radius CA = 1, describe a J © ABD making AB = z. A. Bisect a. BCA by CT meeting the tangent at A in T, join DB, and draw BM J_AC. Then, from similar a s ACT, MDB.

AT2 : AC2 : : BM2 : DM2 : : AM x DM : DM2, : : AM : DM,

or tan2 -= : 1 : : 1 — cos A : 1+cos A.

Also, produce AT to T', making TT= CT, and join CT'; then

tan A + sec A = AT'=tan ACT' = tan{A + i^ - a)},

= tan (j+ 2) = &c-

22. At the first point of observation, let the elevation be 0, the distance from the first to the second d, at which let the ele- vation be 0'; from the second to the third let the distance be d'; and the third elevation suppose d"; then the required height

will be found to be

dd' (Y/4 d')

V:

d' cot 20 + d cot 20" — (d + d') cot d' '

23. Letx— 1, x, x-\- 1 be the sides of the a , 0, ■* — 3 0, 2 0, its angles opposite to these, , x— 1_ sin 0_ sin 0 1

x ~ sin 30 sin0cos204-cos0sin20— cos204-2cos20 x — 1 sin 0 1

and

x+\ sin20 2cos0* Hence x—5, and 0 = cos_1 f , which may be found by the tables.

24. Sin 2" A = 2 sin 2"-1 A. cos 2n~l A = 22. sin 2"~2 A x cos 2"-2 A cos 2n~l A, &c.

For the series, which are both recurring, see Woodhouse or Hind.

P. 30.] IN EUCLID. 27

ST. PETER'S COLLEGE, 1828.

1. Euc. i. 20. 2. Euc.v. 12. 3. Euc. iv. 11. 4. Euc. xi. 11. 5. Euc. iii. 27. 6. Euc. vi.8. 7. Euc. ii. 11. 8. Euc. v. 25 9. Euc. ii. 14. 10. Euc. vi.25. 11. Euc. i. 48. 12. Euc. xi. 6

ST. PETER'S COLLEGE, 1S30. 1. Euc. v. 16. 2. Euc. v. 23. 3. Euc. vi. 5. 4. Euc. vi. 25.

5. Woodhouse's Trigonometry.

6. Let 0 4- <f> be the French angles ; then

10 vno0

d+<i> = — x 75 = -^ = 83°. 333333 &c.

d-<p = 26°. 333333 &c.

.'. 0 - 54°. 83'33"

57° 0 = -^- = 28°. 50'.

Hence, English = T9^ x French are found.

7. See Woodhouse's Trigonometry.

8. Cosec 3 0 =

sin 3 0 sin 0 (4 cos2 0—1)

= cosec 0. (l + 2c)-1, = cosec 0 (1— 2c + 22c24-&c)

n 0 7T— A . 71- + A 2 A 2?r

J. ^ sec — - — sin — -— = cos — cos -^-j

6 6 6 6

2 A 1

= C°S "3" + 2'

... A tt-A . 7T + A „ . A 2 A .A

. . 4 sin -7T. sin — — . sin — ^— = 2sm-~-. cos — — |-sin-~-.

• a . A . A ."

== sin A — sin -^ -f sin -^ = sin A.

10. If a be one of the equal sides, the hypothenuse = a */ 2,

D = TT^79' P = «^2. (1+ ^2), .-. &c. 1+ \/ z

11. If A, B be the objects, and C the given position ; a, b the linear distances, and C the difference of the given angular distances; then, x% = {a — bf-. sec20,

in which tan '2d — y— - vers. C.

[a—by*

28 HINTS AND ANSWERS [P. 31.

12. See Woodhouse's Trigonometry.

A A

13. Sin A. sec -^ = 2 sin -^>

A A _ . A

sin -^ . sec 02 — ^ sm 92 '

• * A A A n . A

.'. sin A. sec -~ . sec ^r-. • • sec 7^ = *> • sin 7^ .

2 22 2K 2"

A A

But, ultimately, when n = 00 , ^ = sin — , .". &c.

Also, since tan 0 = cot 0 — 2 cot 2 0,

1 A 1 A

.'. -? tan -p- = s cot -p — cot A,

1 A 1 A 1 ■ , A o

4 tan 4 = 4 cot 4-2 cot J'

1 , A 1 A 1 A

2» tan 97 = 2""' cot 2» ~~ 2~* 2^'

1 A

.". the sum = -^ cot — — cot A,

and the sum to x is -r- — cot A (by Vanishing Fractions).

QUEEN'S COLLEGE, 1830.

1. Join the two given points, A, B, and bisect AB in C. Draw CD J_ AB, and meeting the given line in D. D is the point required.

2. Prove the proposition ex absurdo, saying, if the alternate angles AGH, GHD (Euclid's fig.) be not equal, make the Z. GHK = z. AGH ; then, (by Playf air's axiom) HK, when produced, will make a a with GH and AB, two of whose angles are equal to two right angles, &c. &c.

3. See Bland's Geometrical Problems, p. 131.

4. See Bland's Geometrical Problems, p. 126.

5. See Bland's Geometrical Problems, p. 143.

6. Let the sides BC, CA, AB, of the a ABC, be bisected in a, b, c, respectively. Join Aa, Bb, Cc. These intersect in P (a well known deduction, for which see Bland),

P. 32.] IN EUCLID. 29

And it may be easily shown, .-. 4 (AB2+AC2 + BC2) = 4 (Aa2 + B62+Cc2) +AB2

+ AC24-BC2. But Aa = $. AP, Bb = •£. BP, Cc = f CP, .-. AB2 + AC2+BC2 == 3 (AP2+BP2 + CP2).

7. Upon the side AB of the given square, describe a semi- 0, and make a square, whose side is CD, equal the given rectangle; from A, draw AE J_ AB, and = CD, and from E, draw EF parallel to AB, meeting the circle in F. Join AF, BF, and AF, BF, are the straight lines required.

8. If AB be the hypothenuse of the right-angled A ABC, take D in AB, so that DB : BC : : AB : 2 AB + BC. For, &c.

9. In the side CB of the a ABC, take CD : CB : : 1 : </2, : : side of a square : to its diagonal. Draw DE parallel AB. Then the a is bisected by DE.

10. The area of the a , and its base, are given in magnitude. Hence, its altitude is given. Upon the given straight line AB, describe, therefore, a segment of a circle containing anz.=the given angle; draw AC _J_ AB and = the altitude, and draw CD parallel to AB, meeting the circumference in D. Join AD, BD ; ABD shall be similar and equal to the A required. Whence the a is easily determined also in position.

11. Bisect the quadrant, draw tangent at point of section, making a a with the radii produced, and inscribe in the a a circle. It is the one required.

12. AB, CD, being the straight lines, take any point E in the line CD, and draw EF parallel to AB. Pass a plane through CD J_ plane DEF, and meeting the line AB in G. Draw GH _j_ CD, and GH is the line required. It is also the shortest distance between the two lines AB and CD.

13. The edges are all equal ; and the altitude =VlX edge.

14. The direction of the cutting plane being given along the side, it will cut the edge opposite the angle through which

30 HINTS AND ANSWERS [P. 33.

it passes, in a given point. Hence, the question is reduced to that of bisecting a a by a line drawn from a given point in one of its sides.

15. The distance required is (see Wright's Self-Examina- tions in Euclid, p. 128),

PF= ^\[a-df+{b-bj^{c-c'y\.

EMMANUEL COLLEGE, 1821.

1. Let iVBCD be the trapezium, AB being parallel to CD. Bisect BC in E, join AE, ED. Then, AED = \ the trape- zium. For, draw BF J_ DC, and EG J_ DC.

Then, trapezium = ABxBF + DCxBF, a ABE + a DEC = AB xEG + DCx EG, &c. &c.

2. By Euc. ii. 12, and 13.

3. Let A be the given point, B that in the circle. Find the centre C, draw BD touching the © in B. Join AB, and on AB describe a segment of a circle, containing an z. equal the l_ DBA. This segment touches the given circle in B, and passes through A.

4. For the angle in a quadrant = a right Z. -f \ a right Z. , &c. &c.

5. See p. 23. No. 22.

6. Seep. 21. No. 11.

7. For a diameter of the circle is the diagonal of the inscribed square.

8. This depends upon Euclid, iii. 36.

9. The centre of the circle is the intersection of the diagonals, and the radius is the perpendicular drawn from the centre upon any one of the sides.

10. Making any isosceles A ABC, with the given vertical L A, and in the side AB produced, take a ABC : scalene A : : AB2 : AB'2, and complete the isosceles a AB'C, &c.

11. Let ABC be the isosceles a having AB = AC. Take

P. 33.] IN EUCLID. 31

any point D in BC, and produce AD to meet the circum- ference in E ; and draw AM JL BC.

Then, AE x AD = AD2+ AD x DE (Euclid, ii.), = AM2+DM2 + BDxDC,

= AMHBM2 (Euclid, ii.), f= AB2.

12. See Wright's Self-Exam, in Euclid, p. 159., art. 329. EMMANUEL COLLEGE, 1825.

1. Euc. i. 4. 2. Euc. i. 44. 3. Euc. ii. 6. 4. Euc. ii. 13. 5. Euc. iii. 15. 6. Euc. iii. 25. 7. Euc. hi. 32. 8. Euc. iv. 8. 9. Euc. iv. 11. 10. Euc. v. 25. 11. Euc. vi. 4. 12. Euc. vi. 19.

1. The figure is equilateral. The angles which two of its sides make with a side of the square are together equal a right angle, and .". the figure is rectangular.

2. In the side AB, let the given point be P ; then, suppos- ing ab to be the side of the parallelogram opposite to AB, a being opposite to A, in ab, take bp = AP, and join Pp ; Yp will divide the parallelogram as required.

3. Let ABC be the a, C being the right /.. Also, CD be the bisecting line, and CE the _j_ meeting the base in D and E. Then z. ECD+ aEDC = right angle. But (if AC be less than CB), a EDC = \ right z. t- B. Whence Z. ECD = | (A— B). Which shows the proposition wrong.

4. Let A be the vertex of the isosceles a , B and C the angles at the base. Then, taking D any point in the base, and AM a J_ upon it, by Euclid, xii. 2, we have AB2 = AD2 + BD2 + 2BDxDM = AD2+BDxDC.

5. This is only a particular case of the Apollonian Tangency, " To describe a O passing through a given point and touch- ing a given straight line and a given circle." (See Self-Exa- minations in Euclid, p. 163.)

6. For the straight line joining the point of contact and the

32 HINTS AND ANSWERS, &c. [P. 35, 36.

common centre is J_ to the tangent ; i.e. _]_ to the straight line placed in the outer circle. It is .". bisected by Euc. iii. 3.

7. Take a straight line AB = the given sum, and from B draw BC _L AB and == the given mean proportional. Then, upon AB describe a \ © , and from C draw CD parallel to AB, and meeting the © in D ; from D draw DE _|_ AB, meeting it in E ; then, AE, EB, are the lines required.

8. Ans. Six.

9. Let AB be the given difference ; on AB, as diagonal, describe a square ACBD ; in AB take AE = difference of AB and AC; also, in AB produced, take AF : AB : : AB : AE, and upon AF, as diagonal, describe a square. This shall be the square required.

10. For the diagonal of the inscribed square = side of the circumscribed square.

EMMANUEL COLLEGE, 1827.

1. Euc. i. 24. 2. Euc. i. 44. 3. Euc. ii. 7. 4. Euc. iii. 13. 5. Euc. iii. 33. 6. Euc. iv. 4. 7. Euc. iv. 11. 8. Euc. v. 25. 9. Euc. vi. 5. 10. Euc. vi. 31.

1. Let ABC be the a ; bisect AB, BC, inD, E, and draw the perpendiculars DP, EF, meeting in F. Draw FG J_ AC, and prove that AG — GC. See Euclid, iv. 5.

2. By Euc. ii. 13, and 14

3. See Wright's Self-Exam, in Euclid, p. 161. No. 337.

4. See Wright's Self-Exam, in Euclid, p. 163. No. 341.

5. Let ABC be the A , and P a given point in the side AB ; bisect the base in D ; join AD, PD. Draw AE parallel to PD, meeting BC in E. Join PE. The a ABC is bisected byPE.

ARITHMETIC AND ALGEBRA

TRINITY COLLEGE, 1825.

[P. 37.]

1. Since 21. 10s. = 2±l. = 4/. and 3s. 6d. = frl.

•1^5 = 2 * 7 = _1 07

* * 40 • 2 5x40 100""' "

2. ^rcs. The factors are of the form 2n, 5m ; for these being multiplied by 5" and 2m will become 10" and 10"1 respectively, which are the denominators of decimals.

3. The work is 54. 6

6. 93

1 4 1 6 4046 28 30

proof by casting out elevens.

3 0 4. 8 7 6 square -feet expressed in the decimal scale.

34 HINTS AND ANSWERS [P. 37.

By duodecimals :

Feet. Inches.

54. 6

6. 9. 312ths

327. 0 40. 10. 6 I. 1. 7. 6

sq.

feet 369. 0. 1. 0, or 41 sq. yards. 0 feet. l12th- 6,44t1'"

See Bonnycastle's Arith.

. „ men killed

4. Number of cannon ce -,

time x number of rounds in a minute*

270 500

l|x| '• lxf

. 27 x 10 _ __

or, 1 : g — : : x : 600,

.«. * = f = 20.

5. The gain, at 10 per cent., upon 25/. is 50s. or 21. 10s.

.'.30 gallons must be sold for 27/. 10s. or

.*. — ^r — - = -^r-s- = 18s. 4oJ. the answer.

6. (a2+«* + ^2) (a2— a^+^2) = («2 + *'2+«^) (a2-f #2— ax) = (a2+:r2)2-a2#2 (P. T7. vol.i. p.70)=a4 + 2a2a;2+tf4-a2tf2 = a4-fa2 x2 + x4.

7. The quotient of a— x) a (is 1-f _ + _ -i_ ... to oo

a a2

Since the quotient is a geometric series whose common

ratio is -, it is arithmetically equal to when a is > x •

a a—x

for such convergent series =

8. Wood, art. 90, and seq.

1 a

i x a—x

a

P. 38.] IN ARITHMETIC AND ALGEBRA. 35

9. The principle of indices in bn denotes a continued pro- duct of the same letter b, and is assumed • hence it is assumed that b2 denotes b x b. If b'2 = a ; since the operation of extracting the square root of b2 is the inverse of b x b or b-, we analogously assume the root of b2 to be (b2)^ = «-.

10. The denominator of (a) is erroneously printed. The fraction should be

x2-\-{a— U) x — ab (x-\-a)(x—b) x—b x2+(a-\-b) x + ab {x -f a) (x -f b) x + b' 03) gives

x /l x j_ ^ \ x /l 2a?a— #"l

"TZ^l ~ J~c {\—xfi" l-x\ (l-;r)2J

a? ,. ' 0 0N #1(1 — #)3+ xz\ x4

= (1=^3- I1"3 * + 3*a> = (t-,). - =* + (IT7F-

(q24-a6 + a62) (a- 6) (q + & + 62) (a— 6) W glVeS«6(a2 + aA-«6+^) 6 (a2+62) '

(S) gives

a+ /(a2-;g2)-a+ /(a2-;r2) = 2 J (a* -a;*)

tf — cP + O? #2

(c) gives

—a"" + 6 a16 6T c3.

11. For the expansion (1 + x)n, see Private Tutor, vol. i. p. 24; for the middle term, &c. see Private Tutor, vol. i. pp. 199 and seq.

12. (a x-x2) 3 = (a x) 3 ^1 - -J

».(»— 1)..^(«— £+2] u*~lt Generally the pth term of (1 + u)n=~ L 2 3...(p— lj~

36 HINTS AND ANSWERS [P. 38.

./'

see Private Tutor, vol. i. p. 198. Let u = — '-, and ?i = —m;

a

then, when^j is even, ( V-1 is negative, and — m( — m— l)x

. . . ( — m— p-\-2) is negative, .'. the p\h term is positive; also, when/? is odd, ( — -V-1 is positive, and so also is — m(— m— 1) x . . . ( — m — p + 2), .'. the joth term is positive. Hence, in

1 — -) ? is

m (m + 1 ) (m + 2) (m+p— 2) /a?y_1

W "

1. 2. 3 .... (p-1)

Let m ■=. — - ; then the /?th term of ( 1 — -J ^ 33,3,3 3 /#\p~'1

J IS

1. 2. 3. 4 (jo— 1) W '

or -1 4.7. 10. ... |l+3(p-2)l ,

'3?-1" 2. 3. 4. ... (^-1) ' ' 1

whence the general term required is evident.

,o /fa2 , & «\ a 6 .

k VTa ~2 — J = X ~ ~~ y common extraction.

By the rule for binomial surds,

V (12-/ -1 - 5) = V {-5 + 12^-1), = / -5 + ^/(25+144) + / -5-^(25+144)

2

/=A+Il + v/^i_3

= 2 + 3 y — I,

which is easily verified by involution.

14. ThPfrfT, farm ,•«*•("- ^ (n~2)- '•(»-? +2) ,-,+,

^ 1. 2. 3 (p-D

15. Wood, art. 90.

16. JfW, art, 377,

P. 38.] IN ARITHMETIC AND ALGEBRA. 37

17. Wood, art. 182.

18. If a be the first term, b the common difference; then 2n terms = \ 2a -f (2n — 1) b\n, and n terms = \2a + [n — I) b\ -; .'. the latter half of 2n terms = \2a +

(2n- \)b\n- |2 a + (n - 1) 6| g = \2a+ [Zn-\)b\?l,

£ 2

which is the sum of 3 n terms of the series;

19. gr— ri = t V $ « V « — iVi a \^«3

2 a a , ci\/a(2 1\ a */ a

= -9-^-6^a = -3-b -g) ="rr-

20. Effect oc cause x time; .".if x be the time required, and c, c, the given causes, a : a{ : : c x t : cl x tv and a : c x £ : : & : (c + c,) x x

But c, = -J— X c 1 a tx

b t c _bt c bt,tl

a c + c, a a,£ rt ?. 4- « J

1 c + -\c ' T J

21. The square being c2=a?—b2, we have Wood, art. 258. V(a + 6) = ^/a ± ^ f ~ &"> V " ^ f ~ ">-

22. If a be the digit, which recurs 2n times, the number is N = a + a. 10 + a. 102 + a. lO2""1

= a £1 + 10 + (11 - I)2 + (11 ~1)3+ ••• (11-1)2-^

= a 1 11 x Q + 1 - 1 + 1 - I to (2m - 2) terras^

= 11a. Q, which is divisible by 11.

This is only a particular case of the general proposition, that if the sum of the digits in the odd places of any number — sum of those in the even places, the number is divisibli by 11. See Wright's Self-Instructions in Arithmetic, p. 48.

38 HINTS AND ANSWERS [P. 39.

23. The operation is 160.61 (12. 86

I Proof by 11

22)60 44

248) 1861

1714

SeejBar/ott>,T.N.p.234.

2546) 14900

12830

2090 remainder.

b b3

24. (a) Since?/ = - x. .*. x3 =x3 = d

c c 3

each of which has 3 roots. See P. T. vol. i. p. 399. (/3) Clearing of fractions.

-v/ (1 + x) - V (1 - -r2) = >/(!- W(l -^2) .-.^(1 + .r)_ V(l- ») = 2V (1 -x*)

.-. 1 +2r + 1 -x-2s/(\ -;t'2)=:4-4x2 or V(l —a;2) = 2^—1

.-. — a?2 = 4#4 - 4^2 . ^4_ ^^ = 0 4

3 -v/3 , V3

.*. .%■ = 0 and x2 = -g, or the roots are 0, 0, -~- and — -g—

(y) Rationalizing the denominator by multiplying by ax 4- 1 + \/ O2 a;2 — 1), and we get

\ax + 1 + V(a23:2 — l)|a__ &2 a? 2~(aa; +1) 2

or {ax + 1) I\/ (a-g +1) + V(ax— 1)P _ ft2 x ax + 1 .'. 2 aa? + 2 ^/ («2 «2 — 1) = &2 ;r ... 4 ft2 ^2 _ 4 = 64 a;2 + 4 a2 .r2 — 4 «62 *2 /.(4 a- /;2) 62 .r2 = 4

ana ,r = ± , -^ r^..

h sf (4 a. — b1)

P. 39.] IN ARITHMETIC AND ALGEBRA. 39

TRINITY COLLEGE, 1826.

1. Proper, improper, decimal, terminating decimal, circu- lating decimal, non-terminating decimal, and continued fractions.

13 4

2. The sum is 15 + ^ -f jk — 15-f-

Q 2 3 5 , . . . 5 7 35 0 3

7 : 4 : : fi : *ourt" ProP°i'tional = - x - = T^ — ^ Tp

4. The root is 17. 2.

1 121 ll2

5. For 12. I = 12 + y^= -y^ = -^ no square.

G. JFoo6/, art. 79.

_3

7. a 2 means the square root of the cube of a, and a ~ 5 means — or 1 divided by the 5th power of a. The original meaning of am is the continued product of a x a x a . . . . m factors, .'. a~b and a 2 are not included in this.

8.

/ 2 ^ /•, *\i fi . i * ia-i)^2 « \

= « (l+j^) nearly

= a + ^ nearly. Hence (145)* = (122 4- 1) * = 12;+ ~ nearly.

9. (1) x = 2

(2) 9 - 6 x = 12 - 8.c + 10 .'. a; = y

(3) a?2 - 2.r = 3, .;. x = 1 ± V 4.= 3 and - 1

(4) x2 -f 4.r + 4 = 4a? + 5 .'. a- = ± 1

(5) ^-2^ + ^il1] 41 21

4*y = wi''-X + y = V -4T=±~2

i ll

and.r — ?/ = —

40 HINTS AND ANSWERS [P. 40.

.". x — 8 and — f and y = f and — 80. (6) Multiply the 2d by 3, and subtract the result from the first, and we get

11 x-25y = - 17, also, adding the 3d to the second, 9 y — 2x = 12. From these equations we get x = 3, y = 2 and 2 = 4.

10. Let # and y be the two parts ; then by the question

x2 + y = y2 -\- x .'. x2 — y2 z= x — y .*. x +- y = 1 the number required.

11. The cost = 'f° + 50 pence, and they make 200 x f = 80 pence,

.'. the loss = 2>\d. \.

12. The cask contains 50 x 9 = 450 bottles, and has

63 o

leaked 3| x 9 = — bottles ;

and .'. the wine to be sold = 450 — — = bottles.

2 2

But the gain is to be 37/. 1 x ]5 = (37. 1) x ^-

Wj\j <£Vj

__ lll/.3y. 2223,

20 " 400 '

.*. the price, per bottle, is

/741 2223N _2^ _ / _ 2223\ 1 17043

V 20 400 / 737 \ 20 / 7370 : 147400

W1 . , 17043 x 12' 17043 x 3 and the price, per dozen, = 1474QQ - 36850 '

_ 51129 14279

_ 36850 ~~ 36850 17 28558 .. _ _, ,3649

= R-l685--S- = R7^8^3685-

13. Let a be the first term, b the common difference, and x the number of terms required ; then, by the question,

P. 40.] IN ARITHMETIC AND ALGEBRA. 41

a + a -f b = 18, a + 2 b = 12

/. 3 a + 3 6 = 30 and a f b = 10 whence a = 8 and b = 2.

/. |2a + (* — l)b\ % = 28,

or (16 + 2ar — 2) f = 28, or .r2 + 7 / = 28

.*. solving the quadratic 7 /49 4- 112

- 7 ± s/ 161 ~~ 2

which shows the question to be imperfectly stated.

4 2

14. The series is geometric, the ratio being — or -, and

., . 4 x 1031 x 3157 .„ 50718 the sum is _ -gj-. , or 166 -— r

15. V(28-6^3)^V/28+^/(2f-6l^)

728- ^(28* -6* x 3)

728 + 24 / 28-24 n/t ,

= V —%— - V — j—=V26- v/2. See Wbarf,art. 258.

(2v/-l) = V(0 + 2,/-l)=v/^ l_v/(bd=^/2+^2. ,/-!■

16. Wood, art. 255.

17. See Private Tutor, vol. i. p. 21.

18. The general or pth term of (c + x)m is

w(w— 1) . . . (re— j? + 2) "-P-H p-i "17^ . . . ( ^ _TT c -r

Here c = a2, x = — b2, n = %, p — 5, .'. the 5th term is

42 HINTS AND ANSWERS [P. 40.

(f-l)Q-S)(j-3) m l-V _ b2Y 1. 2.- 3! 4 v ] K J

3. 1. 1. 3 , JH or' 16 x 2. 3. 4 ° 6

3 68

128" X ^

or> Too- X 75'

19. Barlow's Theory of Numbers, art. 12.

20. When a and b are not prime to each other, and c does not contain their common measure. Barloiv, art. 160.

21. Let x be the number of crowns, y those of the seven- shilling pieces, which, together, = 13 Napoleon's; then

5 x + ly = 13 x 16 = 208. Generally, if a x -J- by — c

and a (7 — ^/> — 1> then the number of positive solutions of a.r -f- by = c is the

greatest integer in -SL — -£, q and p being the least values

which satisfy aq — bp = 1. Barlow, art. 161. Now, 5q — 7 j9 = 1 gives q = 3, p = 2,

• c? CP one f3 2\ 208 . r

' * t " a = 2°8 X 1.7 " 5J = "3T glV6S 6' Hence the number of ways required is 5. The ways of payment are easily found to be

y = 4, 9, 14, 19, 24, 29 x = 36, 29, 22, 15, 8, 1.

22. Wood, art. 380. The common difference of \, %, £3

being \, the series continued is

_i i _a_ i i a i i i i i j_ i i i 2 _i a i ^

••••2i6J2"t"6 2_r6>2J3>6>6 6 J 6 656 6 J 6 6 ' * •

fir 1 JL iL i i J. ft J. ! !

ul ' ' • • ') 6) 3) 2! 3) 6) UJ 6> TJ ~2> ....

.*. the harmonic series is . . . . 1, y, •§-, 2, 3, 6, oo , — 6, — 3, — 2 . . . .

23. Present value for 19 years, beginning now,

19A + 19 x 9r A „, . , QQ/1

= — — ^ r-~ — Wood, art. 394.

1 + 19?-

P. 41.] , IN ARITHMETIC AND ALGEBRA. 43

■ 19x50(1 +|0)_19x50x29

~ I 19 _ 39

1 + 20

7x5o(l+2o)

7 x 50 x 23

P.V. for 7, beginning now, = =■

1+20

_. _r , 50/19x29 7 x 23\

.*. P. V. required = — ^ — j- — — — ^ — 1,

which is easily reduced.

24. Bonnycastle 's Arithmetic, p. 149.

25. Divide 4321 by 7, successively, in the quinary scale, and the remainders will be the digits in the septenary scale. Barlow. Ans. 1460.

26. A number is divisible by 3 when the sum of its digits is ; by 8, when the number denoted by its 3 last digits is;

by 11, when sum of the digits in odd places=sum in the even.

27. When the denominator is of the form 2™ x 5".

28. See Woodhouse's Trig. Log 3 is obtained from log(a+6),=loga+ ±^ + • (J-^ + * (.Lfi + &c.

TRINITY COLLEGE, 1827.

i o 1 10 ,2fQ2 2 47 94

1. 3- = Tand-of9- = - x -g = ^

15 10 94 1575 1400 1128

/.they are as T, y, 35'oras-420> W "420'

or as 1575, 1400, 1128.

2. -fr of a guinea = ^ x 21*. = 8*. 2d.

4s. 5d. 53

Also, * of 2s. 4M ~

■^8.

, T vi **. ^. - 5 60

44 HINTS AND ANSWERS [P. 42.

.". fraction required is — x — = .

4 60 2£ 150

3. Wood. The quotient is 5000.

4. 3*. \d. = i of a pound = £x . 51. = . 16666 . . . and \ = tjV of 3s. 4rf. = . 00208 . . .

.*. the decimal is . 16874 ....

5. Wood. The greatest common measure of 13536 and 23148, is 4.

The algebraic expressions have no greatest common mea- sure except unit. This may be seen by the common method, and also from the consideration that x4— (p— q) x3-\-(jp—q) X q* x—qA = x4—q4—(p—q) x (x2—q2)

= ix~ <l) (jx + q) \x2— (p—q) x + q2l, none of the factors of which is a factor of the other expression.

6.

x3 -\- a3 x3 + a2

(1) - — - — = ■ — - — = x* — ax + a? a + x x + a

(a + bf (a*-b*)*- Q2+62)2_a2+2a6-+62 4 a2 62, a? — b2~ a4—b4 ~ a?—b2 + a4— b4

a4 + 2 a3 b + 6 a262 + 2 a&3 + b4

a

4

£T4

fl2+2a&A/-l-62+«2-2a6y-l-62_2(fl2-62)

(t3) ' a2 + 62 ~ «2 + 63 '

3?/ 2 x (4) By Evolution, the root is ■£- + = 5.

7. JFbod, art. 111., and seq.

62

8. (1) x =

v ' a — c

(2) The denominator of the right hand side of the

equation being (9), (it has been dropt in printing) ;

we have

12% — 9x + 5$y — 9 = 7* — 21 "\

7x - 35y + 56 = 27 x - 117?/ -f

or, 4 * — 45?/ = 3 1

\0x- 41 y = 28 J

P. 41.] IN ARITHMETIC AND ALGEBRA. 45

.'. 20a.- - 225// = I5i 20 a:- 81y = 56J

41 144 y = 41 .*. y = -rrj; whence x.

r3 — 8

.'. dividing by a — 2

1 a? + 2 a? + 4 _ Q

a;2 + 5 .'. x =2 and |. (4) This equation is erroneously printed. It ought to

have been 2x2 — 2 x + 2 s/ (2 x9-— 7x+6) = 5x—6. It thus becomes

2 #2 - 7 a: + 6 -+- 2 -/ (2 a;2 - 7 a + 6) = 0

.'. V (2«2- 7x + 6) l^ (2x*-7x + 6) + 2\ =0

.'. 2x2 — 7 x + 6 = 0, and ^/ (2x2 — 7.e -f 6) + 2 = 0.

7 7

Hence a-2 — ^ x = — 3, and x2 — -z x = — 1

2 2

••^-^^^/(tI-3) ^-i^V^ii-1)

7±1 . 7 ± ^33

or, x — — -. — and x = ^ ,

4 4

or the values ot x are 2. ^, ^ , -?

(5) Square the second.

a-2 -f y2 — 4 4- 2xy .-. a-4 + ?y4 = 16 4- 16a// + 2x2y2 = 272 /. a;2?/2 + 8 o:y = 128 .". x y = 8 and — 16 .*. x 4- y = + 6 and ± 2^/— 15 1 But a- - y = 2 J

/. x = 4, — 2, 1+ V — 15, 1 — y — 15 1

y = 2, - 4, - 1 + V - 15, - 1 - V - 15*

(6) 2x3-2x2 = I — x2 = - (x2- 1)

2 a-2 (a- 1) + x2- 1 =0 .'. {x- 1) (2 a2 4- x f 1) = 0

46 HINTS AND ANSWERS [P. 42.

xV^ , .— l+'-r'j— .7 -i+^z-7

Hence, the roots are 1, j , j

(7) Makings = -^ — , w' = — j — by the usual

method we shall get

w = 4 id — 2,

9 w + 2 *=— IT" •

200 - 13?/

*^ 9

Then make *»' = 0, 1, 2, 3, 4, &c.

.-. w = - 2, 2, 6, 10, 14, &c. y = - 4, 5, 14, 23, 32, &c. and # = 15, 2, &c. so that the only positive integer corresponding values of x and y are # = 15, 2,

2/ = 5, 14.

144 (8) Since x — y = ; .'. a? + 7/ must be some of

the divisors of 144, that is,

x + y = 1, 2, 4, 8, 16, 3, 9, 6, 12, 24, 48, 18, 36, 72, 144, :.x-y = 144, 72, 36, 18, 9, 28, 16, 24, 12, 6, 3, 8, 4, 2, 1. Hence, the positive integer values of x and y are x ±= 12, 15, 13, 20, 37, y = 0, 9, 5, 16, 35.

9. The Arithmetic mean = — ^ — = A suppose,

Geometric = \/ [ab] = G,

Harmonic = T = H,

a 4- 6

A : G : : G : ? = -***. = H.

A a -f b

The Arithmetic mean is the greatest.

10. (I) a : b : : ma : mb : : nc : nd : : qe : qf, .'. ma -. nc : : mb : wrf,

P. 42.] IN ARITHMETIC AND ALGEBRA. 47

ma : ma + nc : : mb : mb + nd, qe : qf : : ma : mb : : ma + nc : mb+nd, qe : ma + nc : : qf : mb + nd, qe : ma-{-??c + qe : : qf : mb + nd + qf, a : b : : qe : qf : : ma + ne + qe : mb -\- nb + qf,

and so on.

(2) a}" : bm : : cm : dm,

fb c\m /b\m

.'. am + dm = am + I — J —am+ (-J . cm,

which is > bm -j- cm on two accounts, because a is > b, and

b .

.'. -. C IS > c. a

(3) Here is a misprint. It ought to be,

" Ifa + 6 : c + d : : c—d : a—b." Then a* -b2 = c2-d2, .-. a?~c2 = b2-d2, or (a+c) (a-c) = (6 + rf) (6-rf), .*. a-f c : b + d : : b—d. : a — c.

11. (a) Here the common difference is 3,

.-. sum = (4+10x2) H = 132.

3

(/3) This series is geometric, its ratio being - ;

arn-a 3 U2/ ~ J _3(35-2*)_ 633 *'• S - r-l ~~ 3 " 24 ~ 16 '

2 (y) Ought to be — 5 — 3 — 1 . . . . ;

then the sum is 16. The value of

.012363636. . . • = jqqq+ 100000+ IooOOOOO + &°' 12 36 /, 1

36 / J_ l_ \

1000 + 100000V + 100 + 104 + /

12 36 1_

1000 + 100000" ._ J_ Wood, art. 224.

100

48 HINTS AND ANSWERS [P. 43.

12 3600 1 12 4 1 3 1

"T" ^f\r^r^r\f\ no innnT innn" l l o^n '

1000 r 100000 99 _ 1000 ^ 1000 11 250 ^ 11 x 250 34 17 17

~ 11x250 ~ llxl25_ 1375"

12. See Private Tutor, vol. i. p. 21.

13. See Private Tutor, vol. i. p. 263.

14. Let 2m + 1, 2n f 1 be the two odd numbers ; then (2 m + 1 )2 - (2 n + l)2 = 4 (w2 — ?f-) 4- 4 (m - n)

— 4 (pi — ;?) (///• 4- n + 1), in which, if m and rc be both odd or both even, m — w is divisible by 2 ; if m and w be one odd and the other even, then m 4- n + 1 is even ; so that (w — ») (w + n + 1) is divisible by 2.

.-. (2 /w 4- l)2 - (2 n + I)2 is divisible by 4 x 2, or 8. Every prime number > 5 is contained in the expression 6??4 1. Consequently the difference of the squares must be of the form

(6»»4 l)2 — (6 n 4- 1/ = 36 (m2 — rc2) 4- 12 0» - «) = 12 (m - ri) {3 (m 4- n) 4- 1|, which, by reasoning like the above, is shown to be divisible by 24.

15. Let N = a 4- br 4- or2 4- dr3 4- ...

— « 4- b (r 4- 1 — 1) + c (r 4- 1 - l)2 + &c. —a— b+c— d + e... + (r + l)\b + (r+l)c— 2 c4&c.| = (r 4- 1) \b-\- (r 4- 1) r — 2 c 4- &c.| by hypothesis, which is divisible by (r 4- 1).

16. Transform the given number (N) from the common scale of notation to that in which the radix of the system is n, and the new digits will be the numbers of the weights to be taken. Ex. Required to find how many of the weights, 1, 5, 52, 53 . . . will weigh accurately 1769 lbs.

P. 43.] IN ARITHMETIC AND ALGEBRA. 49

5)1769

5) 353-4

5)70-3 5) 1 4-0

2-4

.*. in the new scale the number is 24034, /. the weights are 4 of lib. 3 of 51bs. 4 of 1251bs. and 2 of 625 lbs.

17. The only true principle on which this estimate can be made is that of Malcolm, viz., that if the equated time (x) be between tm and tm^y, then the aggregate interest of the sums Py P2 . . . . POT for the intervals x — tv x — t% . . . . x — tm must = the aggregate discount of the sums P,„ + 1, P„t + 2 . . . . for the intervals tm + ] — x, tm +2 — x, Sec.

Let R be the amount of 1/. for a year, compound interest. Then the amount of Px for (x — tx) years is P, R*_fi, ( Wood, art. 397), and similarly for the other sums. Hence the aggregate interest is

P1R*-',-P1-t-P2R--'2-P.2+P3R'-3-P3...P„R-'«-P„,

or,(-|L- + ^+....^)R--(PJ+Pa+P3+...P^,.(a)

Again, the discount of any sum = worth when due — its present worth.

Let w be the present worth of s pounds due n years hence

s at compound interest; then w R" = s, .'. w = ^, .'. thedis-

count of s, at compound interest, =5 — ^- . Hence the above

described aggregate discount is

P P

P , _ mAr\ i P Xro+2 , frr

Rw+i R'»+2

H

50 HINTS AND ANSWERS [P. 43.

°r> Pot + ! + Pot + 2 + Pjn + 3 + • • • •

-(^-4t + &c-)R* {b)

which being equal to the aggregate of interests (a), we have

T> x __ 1 "t~ ■*• 2 ~r . ■ . • X OT + -Tot + 1 ~r • • »

— p p p p

x 1 I __2 , _J«. I x m --T 1 ,

R'» + Bh + • ' ' ' R'- + RWi +

whence x may be found by the Tables of Logarithms.

Ex. Supposing 100/. to be payable 1 year hence, and 105/. 3 years hence ; what is the equated time to pay the whole, allowing compound interest at 5 per cent, per annum.

Here Pl = 100, P„ = 105,

tx = 1, *2 = 3, and R = 1 + 20 = go' 21\ 0 205 41 x 2P

20; " 100 x 2° + 105 x ™ " 40° X 2P+21 X2°3 21 + 1VAJ x 213

41^ x 212 ,_ /2iy

~ 2(P x (21 + 20) =r V20/ . . x == <^j which is evidently true; for the interest of 100/. for 1 year

105

= 51. and the present worth of 105/. due a year hen ee, = = 105 x ff = 100/. and .*. the discount is 51.

Ex. 2. What is the equated time of 400/. clue 2 years hence and 2100 due 8 years hence at 5 per cent, compound interest ?

Here

/21V 2500 25

^ _400 x ™ + 2100 x ?? ~= 4 x » + »

25 x 2P _ 2P

_ 202 (4 x 2P + 206) ~ 43 (21s + 5 x 20s) wnence^maY be found by reference to the tables. Or thus

21\# 1801088541 207 = 1285382464

P. 43.] IN ARITHMETIC AND ALGEBKA. 51

. jc 515706077 ' '20 == 1285382464

2578530385

x — .Miorfif — ' nearly. 321345616 3

18. tf(l+'.*y5)= -^•(3+</5) = •* + •*

= * (V 30 + V 6) /y(-16^)=2aV(0+V-l) = 2a{y/°+A/^+1) +

= 2a(V/|+V/-i) = o^/2(l. + ^-1).

19. Log 90 = log 9 + log 10 = log 9 -t- 1 = log (6 x 15),

= log 6 + log 15= 1. 9542426,

.-.log 9 = .9542426. Hence, log 3 = \ log 9 is known,

.'. log 2 = log 6 — log 3 is known, and .'. log 8 = 3 log 2 is known.

20. The probability that an event will happen

number of favorable cases total number of cases

The number of permutations of 52 cards = 52. 51. ... 2. 1,

and that of 13 cards = 13. 12. ... 2. 1,

„ , , 52. 51. ... Z. 1 .'. total number ot deals = /io io 2~~T)4'

The permutations of 48 cards = 48. 47 2. 1,

and of 12 = 12. 11. .. .2.1,

.*. total number of ways in which 48 cards can be dealt, is

48. 47 2. 1

(12. 11 27iy4'

52 HINTS AND ANSWERS [P. 43.

and each of these arrangements may be combined with the per- mutations of the four honors. Consequently, the number of cases favorable to the event in question is

48.47 2.1

(12.11 2.1)4 X *'°-'-x>

48-47 2.1

, ■ . , (12.11 2.1)4 *

.-. the probability required = ^ ^ ^ >

(13.12 2.1)4

134 x 24

~ 52.51.50.49 x '

133 133

~ 51.50.49' X 17.25.49'

2197 1

= » = FTT nearly' = yg> nearly-

Hence, it is 8 to 1, or 17 to 2, against the event.

TRINITY COLLEGE, 1828.

1. Barlow's Theory of Numbers, pp. 220 and seq.

2. Prop. 1. Book II. Euc.

3. Rule extended. Place the multiplier under the mul- tiplicand, units under units, tens under tens, tenths under tenths, 8fc. Multiply the multiplicand by the units' digit of the multiplier ; then by the tens' digit, placing this latter result one place to the left of the former result ; then by the hundreds' digit, placing the result one place further to the left, and so on. Again, multiply by the tenths' digit, placing the result one place to the right of the multiplicand; then multiply by the hundredths' digit, placing the result one place further to the right, and so on throughout. Then add up the columns, and the resulting sum will be the re- quired product.

p. 44.] in arithmetic and algebra. 53

Example : 9876 . 329 453 . 7689

29628 . 987 493816 . 45 3950531 . 6 6913 . 4303 592 . 57974 79 . 010632 8 . 8886961

4481570 . 9463681. The same as by the common method.

4. The least common multiple of the denominators being 12, the sum is + + = f£ = f = 2*.

5. (W+ i) «? (2 + i + TV) = ¥ ~ H = V = 8.

6. His debts amount to 560. He pays 4s. in the pound. They receive 50/., 30/., 42/.

7. The root is 23. 12. For rule, see Wood.

8.

9.

e2	—	&
	5b
a9	+	X2

X

15 a2 3 a2

1 = — r (<*> — b).

a+b b v J

2x2 2x* 2x6 - l + ~7jT + -ZT + -^5- + &c- t0 °°

a-6 — iC

a"

a4

aD

10. (1) 7 * = 35, .-. * = 5.

(2) 36 - 9 x - 12 = \2x - 48,

.-. x = y =3f. a + 2r

(3) ,r

I

54 HINTS AND ANSWERS [P. 44.

(4) 2x + 2y = 17. 10x—26y= 67 J '

.'. x = 8,

y = i-

(5) a;2 - | = 34,

1-1-35 .-.* = * ±VXrr+34)= — =g = 6and-y.

(6) a?2 — 12 a; = - 40,

.*. x = 6 ± -/— 4 = 6±2V — 1.

11. Let a; be the shillings he had at first. Then, fa; + 10 = what he had after the first sitting,

.'. f (f ar + 10) + 3 = 63, ,\ x = 100s. = 5/.

12. Let a; be the number bought. Then,

— = prime cost per head.

/o7 £\ . (x — 8) f- — J- •p) = what he sold the remainder for = 57/.

.-. 2 a-2 - 16 x = 2280, /. a;2 — 8 a; = 1140, .-. z = 4 ±^/1156 = 4 ± 34, = 38.

13. Wood, art. 162.

14. See p. 47.

15. x : y : : a2 : 62, and a : b : : y' (a + #) : V (a~ y)>

x : y :: a + x : a — y,

x : a :: y : a — 2y,

.'. 2 x : a : : 2 y : a — 2 y.

But x — y : y : : x + y : a — y,

x + y . . a— y — y £~,

P. 45.] IN ARITHMETIC AND ALGEBRA. 55

and a— 2y =

x — y r.2x:a::2y:-^-::X-y:y.

16. [a — cfx = (d - b) y, .-. a- = — Z y x „

17. fToorf, art. 212.

1455 =(2x5 + 296) 15, .-. b

2 9

18. Let the series be 2, 2r, 2 i» 2r3, 2r4,

.*. 2 r4 = 3 2, and r2 = 4, r = ± 2, consequently the required means are

4, 8, 16, or - 4, 8, — 16. 19.

^(4^3 + 7) = v/Zii + y/I^l = 2 + v3.

20. Let a? = number of seven-shilling pieces, y that of half- guineas ; then

14 x + 21 y = 7 x 40, or, 2 a? + 3 y = 40,

•'• * = 20 ^ y - |

Let y = 0, 2, 4, 6, 8, 10, 12, 14, .'. * = 20, 17, 14, 11, 8, 5, 2, - 1. So that the bill may be paid by 20, 0 ; 17,2; 14,4; 11,6;

8,8; 2, 12.

21. Bar lota's Theory of Numbers, p. 261.

22. Cambridge Mathematical Repository, p. 21.; and Barlow's Theory of Numbers, p. 17.

56 HINTS AND ANSWERS [P. 45.

TRINITY COLLEGE, 1829. 1.

TV of a £. = ffs. ; -5^ of a guinea = -f£s., ^ of a crown = | of a

shilling.

/. the values are as f$, f£, r> or as 2800, 2793, 3040.

2.

-Jj of 2s. 6d. = .f§. sixpences, and half-a-guinea = 21 sixpences,

35 5

fraction required =

" 13x21 n 39

Also 105. == i£. = -5, l|d = | x .05 = .00625,

105. l£d. = .50625/.

3.

07 . 13 23 5 115 A1 .... nmo 144000 lonm 2r-T=TX T3= 124- Als° 14-4^.0012= -^-=12000

14. 4 144 12

a 120 ~~ 1200 100 ~"

4 3 -y/8-2 y7_ (3^/8-2-y/7) (^8 + ^7)

-v/8- V7 _ 8-7

= 24 - 2V 56 + 3-/ 56 - 14 = 10 + -/56, = 17. 4833148 . and not 2 . 51 ... . The question is exactly the same in the original paper. Generally let r be the greatest integer in JV N, and sup- pose c the correction ; then

</N = r + c,

.-. N = r3 + 3 r2 c + 3 re2 + c3,

.-. N _ r3 — 3r2c + 3rc2 +C3}

/t r (N— r3) + 2rc3 + c4= 3r3c + 3r2 c2 + 3 re3 + c4,

= 3r3c + c (3 r2 c + 3 rc2-f c3), = 3r3c + c(N— r3) = c(N + 2r3), r(N-r3) 2 re3 + c4 " c ~ N + 2r3 + N + 2r3'

= N + 2? nearly ' •'• ^ N = r + TR^ nearly-

t> a- f 3/1mn in, lft 10(1010-1000)

By this form3v/lQ10=10 + c=10+ 1q10 + 2oqq '

= 10 + TVr = 10. 0332228,

P. 45.] IN ARITHMETIC AND ALGEBRA. 57

which is true much further than required by the question ; for c is evidently less than 10. Consequently the part omitted, viz. 2rc3+cK 20+TV 201

N +2r3 1000x3010" 30100000*

5. Wood, art. 86.

6. By actual division

=xn + xn~x y + xn~2 y2 -f ... xyn~x -f yn,

x-y

as may easily be shown, and the proof is thence derived.

Otherwise.

Suppose it true for ' " — ; that is, let xn~l— yn~l be

x y

divisible by x — y. Then

x—y) xn—yn (xn~l

xn~xy — y"

x—y x — y

.'. if xn~l — yn~l be divisible by x — y, xn — yn is divisible by x — y. But x2—y2 is divisible by # — y, .'. x3 — y3 is, .". x*-y* is, &c. Q. E. D.

Again, suppose, x2n-2—y2'l~'2 divisible by x + y ; then, x + y)x2n — y2n (x2"-2—x2n-2y

/v.2w— 1 i ?*2n—\ ni

— x2n~ly — y2n

__x1n-\y_x1n-2y%

if(x2n-2—y2n-2)

~2n ».2n r2n— 2 ifin— 2

-. * 2_ _- ^-1—a?*1-^ «/ + «*.* ^ .

37+2/ ^ ^ X -\- y

.'. if x2n~2—y2n~2 is divisible by # + y, -p2n — y2n is divisible by x -f y.

58 HINTS AND ANSWERS [P. 46.

But x2 — t/2 is divisible by x -f y, .'. x* — y4 is; .'. x6—y6, and so on.

Q.E.D.

7. See Private Tutor, Alg. Part I. p. 103. et seq.

3 a5-48 ab*) 2 a4- 1 1 a2 62 + 12 62 ( a4— 16 64 ) 2 a4— 1 1 a2 62+ 12 62 (rejecting 3afrom divisor, «2- 4 62) 2 a4— 1 1 a2 62+ 12 62 /rejecting a2 + 4 62 2a4- 8a262 V 2a2-3 62

3a262+12 62

3a262+12 62, .-. G.C.M=a2-4 62.

8. JFood, art. 106. and 109.

9. (1) 20 - 4x + 8 = 5x+ 10,

• ~ _ o

(2) 7 a? - y = 33 1 12 y - * = 19J '

.-. 7 x - y = 33!

12 y -7.r = 133, /. 11 2/= 166, .-.y^ VV = 15 TV .-. 7 x = y + 33 = 105 JT + 33 = 138 A*

^ = ly y -J- yt — 1" TT'

(3) .«2 - f x = 27,

/. a? = | ±V(t9t + 27) = i ± VW

3 ±21 = — a — = 6, or

1 1 j

_9

2 '

(4) 2^— 5^ + 6+ 10V(2.r2- 5 a; + 6) = 39,

.*. \/(2 x*- 5x + 6)+5=± ^/(39 + 25) = ± 8,

/. /(2-r2- 5 a? + 6) = 3 and - 13,

2 a-2— 5 * + 6 =9 and 169,

.". x*— %x — | and »|*,

.*. ». -i = ± /(H + 1) and ± v/(|| + '43)

, , ± \/ 1329 = ± i and \ ,

P, 46.] IN ARITHMETIC AND ALGEBRA. 59

, 5 + V 1329 5 - V 1329 . . a: — o, — -£, —^ and — ^ .

(5) a-4 - 18 a;2— 4 a: + 48 = 0,

.-. a,4 - 14;r2+ 49 = 4a:2 + 4* -f 1,

which are perfect squares.

.-. x2 — 7 = 2x + 1,

.'. .c3 — 2a- = 8, /. ar = 1 ± 3 = 4 or — 2.

„ a;4-18ar2-4ar+48 A , 0

Hence. 5 5 5 = 0 = a:24-2.r — 6,

a;2 — 2 a? — 8

:.^ + l = ±/7.

So that the four values of x are 4, — 2, 1 ± \/7.

Let 6y — 6 = 1 1 z#,

. . y = 1 4- ie 4- — ,

which is an integer when

w = 0, 6, 12, 18, 24, 30

and then y = 1, 12, 23, 34, 45, 56

x = 21, 4, the rest being negative, .'. the only values of x, y are 21, I; and 4, 12.

10. Let x be the number of hours required.

work 1 , ,i

cause cc . oc -p — by the question.

time time ■* *

/. I = J- + _J_ + 1, a: a: 4- 6 a: 4- 15 2xy

1 1.1 2a: 4- 21

2a: #4-6 37 4-15 (^4-6)(a:+15) /. a;2 4- 21a: 4- 90 = 4 a-2 4- 42 a:, .-. x 4- 4 = ± V,

the only answer, the negative value being — 10.

11. In this question the "7? and q together,'''' should have been u p and p together."

60 HINTS AND ANSWERS [P. 47.

c- j lU . n. (n — l) . . . (n— p+l) since p and p together gives — ^ — ^ iL^ — -,

and (n—p) and (n—p) together gives

n(n— I) .... n— (n— p— 1) _ w (»— 1) . . . ( jo-V-1)

1. 2 {n—p) 1. 2 . . . [n—p)

_n. (n—l) .... (ff-t-l)xp (jo— 1) .... 3. 2. 1

1. 2. .... . (p— 1) p

rc (rc — 1) .... (n — j» + 2) (n— p+l)

2. 3 . . (n—p— 1) (n—p) x (rc— p+l) (w— j» + 2).. (rc— 1) rc _rc (w—1). ..(w— jp + 2) (w— p+l) 1. 2. 3 . . . (n— \)n

i. 2 ... . (p-iy^r x i. 2. 3 . . . (w— 1)»

_ n (n—l) .... (n— p + 2) (n— p+l) ~ 1. 2 . . . Qo— l)j» which proves the proposition.

12. See Pri»ote 7toor, Alg. Part I. p. 21., and 199.

13. See Private Tutor, Alg. Part. I. p. 20.

.„ l+2x I 1 1

14 — = = = -

l+n l+x l + 2x—x x ,

1 +2x l+2a? ~ T+2~x

(I + 2x\n / x \-» . x

•'• Ittv) = I1 - rr^i =1+*r+i^

w. 0 + 1) / a? \2 + -TT2 ■ (lT2^) + &C"

. V (7 + 2^10) = V/^ + V/^=V5 + v'2.

Also ^(4 + 3^-20) + y(4— 3 a/-20)P = 8 + 2 V (16 + 180) =8 + 28 = 36, /. ^/(4 + 3 ^-20) + -/(4-3-/-20) = ± 6.

16. Wood, art. 378.

17. N = a0 + ar 10 + a2. 102 + a3. 103 + . . . . 0 = a0 - av 3 + a2 32 - a:i 33 + . . .

.'. N=ar 13 + rt2.(102-32) + «3 (103+33) + ffJ(101-34) + ...

15

P. 47.] IN ARITHMETIC AND ALGEBRA. 61

But 102- 3*3 103-f 33, 104-34, 105 + 35, &c. are all divisi- ble by 10 + 3 or 13.

Hence N is divisible by 13.

18. Let 2 n be the even number. Then (2rc)3-4. (2rc)=:8rc3-8rc=8w02-l)=8. (n—l)n(n+l). But (n— 1) n (ft + 1) is divisible 1. 2. 3, or by 6,

/. (2 rc)3-4. (2 ri) is by 6x8, or by 48.

19- Since 2x4++3yx9 = 35} •••2(-4) + 3.(y-9)=0,

which is of the form,

2a/+3«' = 0, . , _ 3,/

Let y' — 2 iv, or y — 9 -f 2^£ 1 .'. x' = — 3iv, or .r — 4 — 3 wj '

which forms contain all the values of x and y.

For making iv = — 4, — 3, — 2, — 1, 0, 1, we get

x = 16, 13, 10, 7, 4, 1

2/= 1, 3, 5,7,9,11.

The method applies generally ; thus, if

x = a, y = /3 satisfy the equation,

a# + by — c; then it may easily be shown, that

2? = a — blVl y — j3 -f aw J ' Examples. (1) 18 #— 23y=4 is satisfied by #=13, 2/=10. .-. x = 13 + 23^=13, 36, 59, 82, 105, 128 .. . y = 10 + 18m7=10, 28, 46, 64, 82, 100 .. . {2)25 x— 16 y = — 7 is satisfied by a? = 1, 7 = 2, .'. ar = 1 + 16etf == 1, 17, 33, 49, 65, 81 . . . y — 2 + 25 «; = 2, 27, 52, 77, 102, 127 .. .

20 Log 2 = i log 16 = . 30103,

log 30 = log 2 + log 15 = j ; 176091s} = 1-471213,

,\ log 3 = .471213. Hence log 27 = 3 log 3 = 1. 413639. Also log 4 ^=log|^=41og3-log 2- 1 = 1.884852- 1.30103

= .583822.

62 HINTS AND ANSWERS [P. 47.

21. Wood, art. 400, and 402.

22. Let x be the required number of white balls ; then the whole number of balls is x + 11. The number of ways in

which all four can be drawn red == ' ' ' , and the whole

number of ways in which four balls may be drawn is (x + H)_0 + 10) (x_+ 9) Q + 8)

~1T 2. 3. 4

.*. the probability that all four are red 11. 10. 9. 8

(x + 11) (a? + 10) (j? + 9) (x + 8) Again, if of m things, m be of one kind, and these be taken p and p together, the number of these combinations in which p of the iii things are contained, is (see Francoeur 's Pure Math. art. 478.)

$m-rri) C [p-p')\ x {in C p), (m—rn1) (m—trl— 1) . . . \m— rri— (p— p — $\

or " t IT ~r:r (P-P')

m'.(m'—\) . . . (m'-p'- 1) x 1. 2 ... p

Hence the number of ways in which 2 of the 1 1 red balls can be drawn, taking the balls 4 and 4 at a time, is x. (ar — l) 11. 10 _ 55 x (x-l) 1. 2 X 1. 2. 2

.". the probability that of the four balls, two will be red and two white, is

55 x (.r-1) 2. 3. 4

2 ~X O + ll) (.t + 10) 0+9) 0 + 8)

11x60^0— 1) _^_

(«+ll) (a? + 10) 0 + 9) 0+8)' Hence by the question

11. 10. 9. 8 11x60 a; O-l)

(a;+ll)O+10)(a: + 9)O + «) (ar+ 11)0+10)0+9)0 + 8) x*—x = 3 x 4 = 12,

.'. * = i ± V(12 + i)=l|I=4

the number required,

P. 47.] IN ARITHMETIC AND ALGEBRA. G3

TRINITY COLLEGE, 1830.

1. The sum is 2548 U-

2 i"1)

2. -jV of a pound

yj of a guinea = ^k-s.,

* of St. 9*4 =(3f|)i=fi^

.'. they are as |f , ffr, ffc

or as 11 x 48 x 20, 21 x 21 x 48, 91 x 11 x 21, or as 10560, 21168, 21021.

3. I of a groat = f of a shilling = -§- of a sixpence =■ T2T of half-a-crown.

4, If = . 0625 x 21*. = 1. 3125.5. = 1*. 3|rf.

12

3. 7500 4

3. 0000.

5. See Wright's Pure Arith. Decimals. 6.

= 2 a2. (2) The greatest common measure is a2—b2, and the frac- tion in its lowest terms is

3a2+2 62 5a(2a + 36)'

(3) *-{('^+»(^}*

V (.r4 4- 2j?3— 2a?a— 2j-+1) x

64 HINTS AND ANSWERS [P. 48.

= x/ — 1 — -/ — 2.

(5) It = (^J =^ = (^) the best form for numerical operation.

7. The quotient is a6— 2 a12 6* + 4 a24 F - &c.

8. Wood, art. 109, 1 10.

9. (1) x = 2.

(2) * = 2i

(3) x = 12, and - 21.

(4) {x + yY = x4 + 4x3y-\-6xhf + 4[xy:i + 7ji = 97 + 4 a;y (a:2+2/2) + 6 x2 y2, {x + yf = x2 + y2+2xy,

whence x2 y2— .50 xy = — 264. Hence, the equations become x y = 4 4 and 6 l and # + ?/ = 5 J '

From square of the second, take 4#y, which will give x — y — ± */ — 151, and ± 1 1 and x + y = 5 I '

Hence, x = ^ (6 ± y' — 151), and 3 and 2 y = |(5 T V — 151), and 2 and 3.

(5)*=3Q-2/-^Z-3.

Make .*. y — 1 = 0, 5, 10, 15, 20, &c. .". y = 1,6, 11, 16, 21, &c. and x = 29, 21, 13, 5, - 3, &c.

10. Let x be the number of half-pence spent, and y that of the oranges ; then,

- = cost of each,

y

P. 49.] IN ARITHMETIC AND ALGEBRA. 65

.*. by the question,

X x i ,

y+5 y and 5 = - + 1

y-3 y

x = 6r\ half-pence 3^d. T'r,

11. (1) It is arithmetic, and the common difference = 6,

S = (6 + 6 x 12) V = 13 x 39 = 487.

(2) Is geometric, the ratio being — f.

_ arn g _ H-4)5-i _, 7

(3) This is arithmetic, the common difference being — \ . :. S = - (f — 5. f) 3 = 4 — 3 = 1.

Also .'. .0132132. . . = TV x . 132132, &c. Make;z = .132132, &c. .*. 1000 x = 132 + #,

■ • 3? == -9-99- = T1T>

/. .0132132. . . = tKt-

12. Similarly to the question in p. 47.

13. There are eighteen letters, consisting of three a's, b, two i's, three rc's, o, r, two s's, four Z's, m. But of n things, if k be of one kind, k' of another, k" of another, &c. ; then the number of different permutations is (JFrancceufs P. M. art. 483.)

1. 2. 3 (?i — 2) (n - \)-n

1.2.3.. . (k— l)*x 1.2,3... (#-!)# x 1.2.3.... (A"— l)^xk

Hence, the number of different ways required, is

1. 2. 3. 4. 5. 6. 7. 8.9. 10. 11. 12 17. 18.

2.3x2x2.3x2x2.3.4 = 44100 x 8570 x 4896 = 1,850,549,652,000.

14. Private Tutor, Alg. part i. p. 21, and p. 200.

15. Private Tutor, Alg. part i. p. '~3.

K

66 HINTS AND ANSWERS [P. 49.

16. Barloio's Theory of Numbers, p. 317.

17. The reciprocals of the terms of an harmonic series being an arithmetic series, let these reciprocals be

h h+x, i + 2x, $ + 3x, | + 4.r, ± + 5x, then | + 5j- = ~,

* or — -

.*. the series is

2 9 2 2

9 ~ ~ 12

' 1 + 2x 1 + \x 1 + 6jr 1 + 8a;' '

or, 2, Va, 3, 4, 6, 12.

18. n!+(n l)2 - 5 = 2ra2 + 2rc — 4 = 2 (rc2 + n - 2)

= 2. (n - 1) O + 2). Now, •/ n is prime to 3, it must be of the form 3w+\, or 3w -f 2, but n -j- 1 is also prime to 3 ; .'. n can only be of the form 3iv -J- 1.

Hence, rc2 + (rc + 1)2.-5 is of the form 2. 3m>. 3(w> + l,or of 18. w{w+l), which is divisible by 36, for either w or w + 1 must be even.

19. Let n — 2w ; n3+20n 8 to3 + 40 m? w3 + 5w w.(uP + 5)

then,

48 48 6 6

But w = 3 w', 3w'+ 1, or 3 w' + 2,

n3+20n _ 3w'(9w'2 + 5) {3 to' + 1)(9m/2+ 6 m/ + 6) 48 6 °r 6 •

(3 m;' + 2) (9w'*+ 12 w' +9)

or

6

m/ (9 m/2 + 5) (3 m/ + 1) (3m/2+2k/ + 2) 2 °r 2

(3 to' + 2) (3m''2+ 4?^' + 3)

or

2 each of which is divisible by 2, when id is either even or odd.

20. Wood, art. 378.

P. 49.] IN ARITHMETIC AND ALGEBRA. 67

21. Log5=logl0-log2=l-|log,4=l-.30103=.69897.

22. Let x be the time the first college ,ias it, and A be the rent charge. By Wood, art. 402.

1 ~~ R* A

P = -^ r- A, and it — -^ T when x is go ,

K. — 1 rv — 1

i--L i-l

A. Rx RJ

.'. R 2_ j - R_ l A= K _ x A,by the question.

/ 1 \ loo- 9 lo0- 2

Hence, 2. (l - ^= 1, ,. R*=2, and * = ^ =^

log 2 " log 105 - log 100

_ lo§2 ~~ .0211893

.3010300

_ .0211893

= 14 . 25 nearly,

= 14 . 5 years, nearly.

23. Assume ax = A + B x + C x- + D x3 + &c. A, B, C, &c. being independent of x, a2* = A + B (2 x) + C (2 ^)2 + D (2xf + &c. = A2+2AB^+ B2 x* +2BCa;3+&c. 2 AC«2+ 2ADa'3+ &c. .-. A2 = A, 2 AB = 2B, B2 + 2 AC = 4 C, 8 D = 2 BC + 2 AD,

B2 B3

.-. A = 1, B = B, C = ^, D = J^, &c.

B2 B3

.'. a*= 1 + Bx + y-2, x2 + y^-g, x3 &c.

Let a; = -^-, rS

- l 1

■'•«" = 1 + 1 + -j-^- + Y^g + -.-=e, suppose.

68 HINTS AND ANSWERS, &c. [P. 49.

. R ^g a ..15= i- j

log e

which gives the series.

^ r^j 24. The probabilities due to the assertions of A, B, and C,

r&*~ l^-^- are respectively

.'. the probability required is | + | — y = t%-

ALGEBRA.

PART II.

TRINITY COLLEGE, 1819.

1. Wood, art. 266.

[P. 50.]

2. Wood, art. 271. Two typographical errors in this question. In first line of it, for terms read term, and for least term read Imt term.

3. Let the roots be of the form -, a, ar ; then

r

- x a x ar — 8, .'. a — 2, r

which being one of the roots, the other two may thence be

found by the solution of a quadratic. But this may be done

directly ; for

? + 2 + 2r = 7,

	r
• ,	2r2-	■ 5r =		2,
.*.	r2 -	2 '	—	1,
• •	r —	.5	±	h
."•	r =	: 2, or	b
	.-. the	roots are	1.2,	1

4. Wood, art. 277.

70 HINTS AND ANSWERS [P. 50.

5. " Or equal/' it ought to have been added. See Wood, art. 331., and Private Tutor, Alg. part ii. p. 71.

J

6. Since the reciprocals of quantities in harmonic progres- sion are in arithmetic progression, it' !/ = -, the roots of the

JD

equation in y will be in arithmetic progression. That is, the roots of

_L _ 21 + Jii _ 288 = 0.

y3 y2 y ™y3-hf+ w?/-m = a j

Of this equation let u — v, u, u -f v, be the roots,

••• u = i,

• • v — 125

.'. the roots of the equation in y are

j. i i i I _j_

6 12J65 6~Tl25

nr — "— -L !

K>1 12) 6) T;

.'. the roots of the given equation on

12, 6, 4.

7. (1) One root in each is respectively — 3 and 1.

(2) First deprive it of the second term, and then use the rule in Wood.

8. See Private Tutor, Alg. part ii. p. 202. The new equa- tion is

y3 — ^—L-1 y2 4- — ^— y = 0.

pq—rJ pq—r* pq — r

N. B. — The second term should be —px1, and not —qpx2.

9. The equation of limits has one of them ; that is

3 x1 + V° x — 0, .". the equal roots are — ff, .'. the other root is ■£•§- .

P. 50.] IN ALGEBRA. 71

10. The equation is Reciprocal; .'. b = -.

a

AlSoa + J=a+_=p ^>

P"=(« + i)

1\" „ „ o W (ft— 1)

-) =an + n a"~2 -\ \ — — -i a"~4 + . . .

"J la -W

n{n—Y) 1 1 J_

1. 2 a»~4 + l,'a»-»+^'

= ^i + „(^+^)+±^)((^+_L1)+...

= an + bn+n (a"~2+ 6«-2) + "i^"1) (a"-4+ 6"~4)+ . . .

making Sn=an+bn, Sn^=an'°-+bn-2+ . . . q _ n c ^ (ft— 1) Q ft(ft— l)(ft— 2) g „

Ow — 7? ft O^ j ^- Ow_4 j ^ 3 ^n-6' &C'

Similarly S_=p«-»- (ft-2) Sw_4- ("~2) (^~3)Sw_6-&c.

/ft (ft- 2) (ft- 3) ft(ft-l)(ft-2)\g "

+ 1 j^-g- 1. 2. 3 Jbw"« + &C'

r> 1 , r>« ft (ft— 1) ft ,rt , 1N 71 (71 — 3)

But ft(ft-2) A— — '=-(2ft-4-% + l)= ^ 2y

,ft(ft-2)(ft— 3) w(«— 1) (%_ 2) »(« — 2),0 n , ,.

and 1. 2 1. 2. 3 -OT^-9-^1)'

_ ft (ft-2)

-Tr2T3 i2n~8)

_ n (n— 2) (ft— 4) : 173

. C n „ o ft (ft— 3") ~ . ft(ft— 2)(ft— 4)c . 0

.•.Stt=yw-ropw-2+ ^ 2 ;- S„_4 + v 1 M3 £SB_6+&c

Similarly,

s»-4=^"-4-(^-4) ^-6+ (y?~t! (v7) s>

n— 8

(ft-4)(ft-6)(ft-8)

<-. r • (ft — 6) (ft — 9) „

1. . <w

72 HINTS AND ANSWERS [P. 50.

(n_G){»— 8)(rc— 10) c

1.

+ v"~"n: t ; s_12 + &c

_, n (n — 3)

.-. Snz=pn— np'k-2+ -j — -J.pn-4 —

{ \ 2 / 0-4) ^ 1 / 3 1\ pn'G + &c.

n(n— 3) . n(n—4)(n—5) = pn— npn~2 + -^ — ij->n"4 | 0 q i9 + &c-

Otherwise.

Assume the formula true for an~l + bn~l, and also for an'2-\-bn~2, and then multiplying an~} -f &n_1 by a+i, prove it true for an + bn. Then, it being true for a2 -f- ti1 and a3 -\- b3, it is for a4 + 64, .'. for ab-\-b5, and so on.

1 1 . For the proof of Waring' s rule, see Wood, art. 344. The reducing cubic being 8 n3 + 4 q ril -f (85 — 4 r p) « + 4 y * + 4 p2s — r- = 0,

and v p = — 3, ^ == — 12, r — 12, 5 == — 4, /. 8 w3 - 48 rc2 + (- 32 + 144) n + 192 - 144-144 = 0, .*". n3- 6n2+ 14 n — 12 = 0,

of which a root is evidently 2. Substituting this in O2 + p # + w)2 = (jo2 -f 2 rc + y) #2 + [2p n + r) x 4- rc2 4- ,<r, Ave have

(#2 _ 3 9 + 2)2 = a?2 + 0 + 4 — 4 = ar2, /. .r2 — 3x 4- 2 = ± x,

and a;2 — 2 J" =

^ > the tour roots. a:= 1 ± -/-I/

— ^

12. JJW, art.311.

13. A superior limit is 262. Wood, art. 304.

Otherwise.

Since .r4 - 19 x3 4- 1 17 *2 - 261 x + 162 = 0, Make ,r — y 4- p, and substitute; then

P. 50.] IN ALGEBRA. 73

/. 0 = y4 + (4e-19)^+(6e2-57e+117)y2 + (4e3-57e2

+ 234 e — 261) y —

+ e4- 19e3+ 117e2-261e + 162, in which every coefficient being positive when e = 10, 10 is > the greatest root of the equation. Again, to find the Inferior Limit, since all the roots are

positive, (Wood, art. 311.), if we make x = -, the Superior

u r

Limit of the equation in u will give us the Inferior Limit of the equation in x. That is, the Superior Limit of u*-n w3 + H tf - t'A u + rtj = 0 will give an Inferior Limit of the given equation. But u is < | f + 1 < ffr,

• * i'o -> i 8

.T IS >

T7

.*. x is > £f; tnat is> iy is less than the least root. The roots actually are 9, 6, 3, 1.

14. Dividing the equation by a*x2, we have /x2 cr2\ (x a\

afe + «)*+«(£ + fW»+8,

. ^ a — m ± >/(»i2 — 4 n + 8) " a x = 2 = C' suPPose-

whence the root is easy.

15. The new equation being

yn + f yn_1 + | yn-* - 1 yn~* + &c. =o.

But let w = a; x — ^, or x = — 2 u> .'. =p2" m"T j9 2"-1 M'^ + y 2n-2 ?/"-2±r. 2"-3 un~3 q=&c. = 0.

Or, ttn + ^ W*-1 -f f M»-2 _ £ Mn-3 + &q _- 0.

2 4 8

.'. the values of y = the values of w = those of — ^

a 6 c of p Or

2' ~ 2' ~ 2' 2' 2' *G-

74 HINTS AND ANSWERS [P. 51.

TRINITY COLLEGE, 1821.

1. Wood, art. 277.

2. Since ^ — 5 is one root of the equation, — yr— 5 must be another, and the equation is

(x—3) (#-2)0-^-5) (#+\/-5) = 0, or O2— 5a;+6) 02-f5) = 0, or ^— 5 #3+ 11 a;2— 25^ + 30 = 0. Again, since ^ 2 is a root of the next equation, — \/ 2 must also be a root, and the equation is divisible by (x—3) (x—y/2) x (a?+/2),or by (a?— 3) (a2— 2), or by #3— 3#2-2*+6, and the quotient is

x1— Ix +10 = 0, .'. the roots are 2, 3, 5, a/ 3, — a/ 3.

3. Make y — -, and the equation in ?/, whose roots are in

Arithmetic Progression, is

y3- ?/2+ 26?/ — 24 = 0. Let the roots of this be u — v, u, u-\-v; then

u— v + u + u-\-v '= 9 = 3w .'. w = 3,

and (w2— v2) u = 24 .*. 9 — v2 = 8, and z; = I,

.'. the roots required are

i i i

2 5 T> T'

4. See JFood, art. 311.

5. The coefficients of the equation in y,if y = x + e are

es + 2e4 - 50 e3 - 100 e2 + 49 e + 98, 5e4 + 8 e3 - 150 e2 - 200e -f- 49, 10e3+ 12 e2- 150e - 100, + 10e2 + 8e — 50, +. 5e + 2, all of which being positive when e = 8, 8 is a Superior Limit of the roots. Also, it is found that 7 satisfies the equation. .". 7 is the greatest root.

P. 51.] IN ALGEBRA. 75

Again, it is evident there are three negative roots, there being that number of continuations of sign. See Wood, art. 311 . Hence, changing the signs of the roots, the Superior Limit of the new equation

xb- 2x*- 50x3 -f 100 a2 + 49 x -98 = 0 will be the Inferior Limit required.

Here the coefficients of y of the equation in y = x^-e, are e5- 2e4- 50e3-f 100e2 + 49e — 98, 5e4- 8e3- 150 e2 + 200e + 49, 10 e3- 12 e2- 150e + 100, 10 e3— 8c— 50, 5e- 2. These are all positive when e = 8, and the first is 0, when e = 7, .*. — 8 is a limit less than the least root of the given equation, and — 7 is that least root.

6. (1) First deprive the equation of its second term, by making y = x — ^ — x—2, and it becomes

y*-9y— 28 = 0. But by Cardan's rule,

^^[-iVig-d)3}]

Q T

and \ = 3 and - = — 14,

•'• y = 4/ &-14W (196-27)| +4/|l4— /(196-27)|

= 4^7 + ^ 1 = 3 + I = 4,

.'. \a? = 6 one of the roots.

Whence the other two are ^ 3 and — y' 3. (2) Let a, a, b, b, be the equal roots. .*. 2a + 26 = — p, a? 62 = s,

76 HINTS AND ANSWERS [P. 51.

ab = ±*/S.

Hence .;. a = - 1 ± | V" (paHhl6>y/^,

Otherwise.

The given equation and limiting equation have a common measure of the form x2 + P# + Q == 0, which contains the equal roots

4*3+ 3px2-\- 2qx+r) 4#4 + 4px3 + 4qx2 + 4rx + 4s (x+p

4 x* + 3 jo#3 + 2 qx2 + r.r

j9^3+ 2 qx2 + 3rx + 4s' or 4jt?a?34-8<7.z2+ 12?\z-fl6.s 4 px3 + 3jo2 x2 + 2 pqx +pr .'. (8 ?- 3 f) x2+(\2r—2pq)x+\§s—pri consequently the roots are those of the equation,

\2r-2pq pr - 16s

^r Sq-3p2 ~8q-3p2' which are easily found.

The first method is preferable, because of its involving only two of the coefficients p and s. Take the numerical example I x4+ 2 x3— 3x2— 4 a> + 4 = 0. By the first method, a = -i±iV(4 + 32)=r-i±|=lor-2, b = — \ + f = — 2 or 1,

.*. the roots are 1, 1, — 2, — 2.

By the other method, the equation containing the roots, is

2 -48 + 12 -8-64

X + -24-12* --24-12'

or x2 + x = 2

.*. a? = 1 or — 2, as before.

(3) Divide by a2.r2, and the equation becomes x2 , x . a , a2 _ a2 a # x2

X CL

Make 5 + - = u ; then the equation becomes z x

P. 51.] IN ALGEBRA. 77

y? + U — 1 = 0, 1 ^ sfb

u =

2 ~ 2

x a — 1 ± <J 5 or - 4- - =

ax 2

l±-v/5 a:

2 a ''

©-

- 1,

and - = ^ ±y/ rb ,

;.x=^\-l±V5± -/(+ 2^/5- 10)£, and the four roots are

* = | |- 1 4- V5 + V(- 2 ^5 - 10)|,

or=| J- 1 - y/5 + /(2 */5- 10)|,

or= | |- 1 + s/b - V(- 2 -v/5 - 10)f,

or== | |_ l — ^5 — ^(2 V5- 10)|,

all of which are imaginary, and may be reduced to the form A + BV-1.

c 7. Since^S" is a root of the equation,

S + Ra+Qa2+Pa3+ pan-l+an = 0,

.'. S+R + Qa+P«2+ pa^+a^=0,

;. — = -R-Qa-Pa2- —pa^-an'\

a

.'. S is divisible by a,

also, v R,= - + R = -Q«-P«2-. • • .-pa"-*-a*-\

1 a

.-. 5l = _Q_Pa- -j9 a«-3-a"-2,

a

that is, R! is divisible by a,

/. Q - ?l+Q = -Pa-. . . ,-pfl"-4-a"-5, 1 a

78 HINTS AND ANSWERS [P. 51.

.'. Qx is divisible by a, &c. whence the meaning of the proposition is manifest. It is im- perfectly enumerated, however; for it is by no means necessary, although S, Rp Qp &c. may be divisible by a, that the quotients are integers.

8. This symmetrical function of the roots =

a2(63+c3+d3+e3+/3+g3) +fl*(a3+c3+ ) +&c.

= a2(S3-«3) + 62(S3-63) +C3(S3_C3) + &c. — S3 x S2 — S5,

S3 meaning the sum of the squares of the roots, S3 that of the cubes, &c. But, generally, if the equation be (see Wood, art. 352) xn+p xn~x+q #n_2+ &c. = 0, S1+p = 0, S2+^S1+2y = 0, S3+joSa+grS1 + 3r = 0, S4+p S3+q S2 + r Sx+4 s — 0, S6+p S4+q S3+r S2+s St + 5 t = 0, from which five simple equations S,, S2, S3, S4, S5, may be found, and the value of S3 x S2 — S5 may be found. But, eliminating Sv S2, S3, &c. we have

s* = -p (1)

.-. S2=p2-2? (2)

.*. S3 = - p3 + 3 p q-3 r (3)

.*. S4 = p4— $p2q + 4:pr-\-2q* — 4 s . . . . (4)

.". S5 = —p^ + 5pzq—5p2r—5pq2-{-5ps—5qr—5t...(5)

&c.

These are particular cases of Waring's general Theorem.

See Meditationes Alg. Theorem 1.

In this case,

p = — 6, q = 9, r = 10, s = — 45, t = 54,

/. S2 = 18,

S3 = - 408,

S5= -684,

.*. S2 = 18, S3 = - 408, S5 =— 684,

.'. S2 x S3 - S5 = - 8028, the answer.

P." 52.] IN ALGEBRA. 79

9. Taking the equation of limits, as in Wood, art 352, nxn~l— {n— 1) px71'2 + (» — 2) yxn~3— &c.

			xn-	—pxn~	x-\-qxn	"2-&c.
		X-	1 —a	4-	] X-	-b	4-	4- &c. X — c
		1	4-	a X2	+	a2 X3	4-	a3 x~* +	&C.
		1 X	+	b X2	+	b2 X3	+	b3 x* +	&c.
	«	series,
		n X	4-	!5,.	1 X2	+	S2.	1 XJ	^3'x~*	+ &C
Let a: =	1;	then,
s1+s2+s3	+.	# #	. oo	==:	n—	-(».	-l)p+(n	-2)q-	— . . .
	1		rn J_ r*	Xrr-

.00

p— 2?-f 3r— 4s + 5t— &c.

n

~ l—p + q — r + s—t+&.c. ' This is the answer usually given to the present question. But the student must observe, that unless every one of the roots of the equation be a proper fraction, and, consequently, Sj, S2, S3, &c. a converging series, the result cannot obtain arithmetically. For example, in the equation x2 — px 4- 1 =0,

we shall have S, + S0 + . . ..oo = ,- 7 = — 1, which is

1 * 1 — p 4- 1

absurd ; for the roots being of the form a, -, if they are

a real, one of them must be an improper fraction, and, conse- quently, the sum of its successive powers to 00 must be infinite, not considering the sum of the powers of its reciprocal.

In short, the question itself is absurd, except for the equations whose roots are all less than unity.

10. Assume the required equation to be u3 4- P u2 + Q u 4- R = 0. Then-P = aa + aft + bp = a(a + p) + bft =—apl + bp, Q=a2aft + ab a/3 + ah ft2 = a2 7,4- q ?, + qft2, — R = a2 baft2- aft. q qv

80 HINTS AND ANSWERS [P. 52.

and solving the given equations, a, b will be found in terms of p, q ; and a, (8 will be obtained in terms of pv q{ ; which being substituted in the values of P, Q, R, these last will be found, and, consequently, the required equation.

It will be observed, the roots of the required equation are not symmetrical. The question would have been better, if the roots of the required equation had been of the form

aa, a/3, 6/3, ba.

For then, in the assumed equation,

m4+Pm3+Qm2+R« + S = 0,

- P == a. (a-|-/3) + 6 (a+/3) = (a + 6) (a+/3)=m, Q = a? ap + ab afi+ab a?+ab (P+ab a/3+62a/3,

= (a2 + 62) a/3 + (a2+/32) ab + 2qqi=(p*-2q) qx + W-2 7)^ + 2^,

— R = aa. a/3. 6/3 +aa. a/3, ba + aa. 6/3. ba 4- a/3. 6/3. ba

= a2 6 a/32 -1- a26 a2/3 + a62 a2/3 + a62 a/32, = qqiafi+q<Ii ^+qqx 6a + aa1 6/3, = 9'9'i«(«+/3)+aa16(a+/3)=aa1(a+6)(a+/3)=^1aa1 S == aa. a/3. 6/3. 6a = a2 62. a2 /32=a2 a^,

.*. the equation is, in this case of symmetrical roots, ui—ppl u3+(p2ql-^-px2q—2 qqx) u2—ppi qqx u -f q2qx2=0.

11. See Wood, 327.

12. See Wood, art. 332.

13. (1) Wood, art. 278.

14. See Private Tutor, Alg. part. ii.

15. See Private Tutor, Alg. part. ii.

1 6. If the roots be real, there are as many changes of sign from positive to negative, as positive roots, and as many con- tinuations as negative roots; hence, all those consecutive terms which are wanting, being ~j- or — , and indicating

P. 52.] IN ALGEBRA. 81

changes or continuations indifferently, show the imaginary roots. Hence, if the equation be

in which 2 m terms are wanting, then between

± P#p±0. xp~1±0. xp~2± .... ± 0. xJt>--m ± Q^*2"'-1, there are indifferently 2m changes, or 2m continuations, which indicate 2m imaginary roots. .'. if 2m terms are wanting, the equation has 2m imaginary roots. Again, between ± Pxp. ±0. xp~1±0. xp~2 ± . . . . ±0. xP-2m-l±RxP~2m-*, in which 2m +1 terms are wanting. If P and R have the same sign, these are indifferently 2m + 2 changes and continu- ations, and .'. 2m + 2 imaginary roots. But if P and R have different signs, then there are indifferently but 2m changes or continuations, and .'. but 2m imaginary roots.

Examples.

= 0, has two real roots. = 0, has one real root. = 0, has but two imaginary roots. = 0, has 2m + 2, or four imaginary roots. == 0, has four imaginary roots. = 0, has six imaginary roots. + x^ + x + 1, has two imaginary roots indicated by x10 -\- x1 ; and four indicated by xb -J- x.

TRINITY COLLEGE, 1822.

1. See Private Tutor, Alg. part ii. p. 45.

2. Wood, art. 277, and 275.

3. The requadratic is

(x-</3) {x+V'3) (#+y— 5) (x—^—5) = 0, or 02-3) [x2-\-5) = 0, or z* + 2 a?— 15 = 0. Again, since one root of the given equation is of the form <\/a, another must be of the form — a/ a.

M

1.	X2	—	9
2.	X3	—	r
3.	X4	—	s
4.	X*	+	s
5.	xb	—	t
6.	X6	+	IV
7.	xw	+	x"1

82 HINTS AND ANSWERS [P. 53,

Let .". the three roots be

*/a, — V a and 6. Then •v/a. —i/a + b=4, or 6=4, and v'a. — t/°X —6=12, .'. a=3, and the roots are \/ 3, — -y/ 3 and 4.

4. The sum is

a 4- 26 4- 3 c 4- 4d

a + 26 4- 4 c 4- 3d

a + 36 + 2e + 4d

a -f 36 4- 4 c 4- 2d

a 4- 4 6 4- 2e 4- 3d

a 4- 46 4- 3c + 2d 2a + 6 4- 3c 4- 4d 2 a 4- 6 4- 4 c 4- 3 d

2 a 4- 3 6 4- c 4- 4d 2a + 36+4c+ d 2a 4- 4 6 f c 4- 3d 2a + 46+3c+ d 3a 4- 6 4- 2 c 4- 4 d 3a 4- 6 4- 4c 4- 2d

3 a 4- 2 6 4- c -f 4 a7

3a 4- 26 4- 4c 4- a7

3 a 4- 4 6 4- c 4- 2d 3a + 46 4- 2c + d 4a -f 6 4- 2c 4- 3d 4a + 6 4- 3c 4- 2d

4 a 4- 26 4- c 4- 3d 4a 4- 26 4- 3c 4- d 4a 4- 3 6 4- c 4- 2d 4a 4- 36 4- 2c 4- d

60 (a 4- 64- c 4- d) The number of these forms will evidently be that of the per- mutations in four things taken all together. Consequently, 4. 3. 2. 1 = 24 is the number of such forms. Also, a, 2 a, 3 a, 4 a will each occur 3 x 2 x 1 , or six times ; therefore the

P. 53.] IN ALGEBRA. S3

number of a's in the required sum is 6 (a + 2a-f 3a + 4 a) = 60 a, and the sum itself is evidently .".

60 (a + b+c + d). But by the Method of Divisors, it is easily seen, that the roots are 1, — 1, 2, 3. Consequently d=l, and the sum re- quired is

60(1+2 + 3+1) == 420. Again, if a, b, c, d, be the roots of the equation, x* — p x3 + q x2 — r x + s = 0, ab + ac + ad-\-bc + bd-\-cd = q, and a be d = s,

• _L _L _1 J_ i_ I-?

' ' a6 *f ac f ad + be + bd + Yd ~~ *' 1 1 1 1 1 1 q*

' ' a2 63 + a2 c2 + a2 a"2 + 62 c2 + b2d2 + "c2^2 ~ &

*iw

11111 + ttttj + rr-i- + -n-j + -r— > +

ac a6 aa* a6 6c a6 6a1 ab cd ac cd 111111 ac be ac bd ac cd ad be adbd ad cd 1 _1_ 1 -.

+ bcbd + bccd + bdedj

,?2 la \ abc abd acd bed J

-(-

b \abi

+ t -r- 4- -ZT5 +

1

abc abd bed

i (± . J_ , _L.LJj\

c Va6c a6a* ac</ bed J

1 /J_ 1_ 1_ _1_V| f a1 Va6c + «£rf aca" + 6«ay/

/i i i i\ / 1 j_ _L _L>

~ U + b + c + aV Va6c + a&d + ac^ + bedj

\~

abed

£. _ 2.-. 2

s2 ' ' s s s

= \- 2.1. £ + - ( Wood, art. 273.)

q1 — 2 p r + 2 *

*a

84 HINTS AND ANSWERS [P. 53.

In the present example,

p = 5, q = 5, r = — 5, and s = — 6, .". the sum required is 25 + 50 - 12 _ 63 _ 7 144 144 16'

5. Since the reciprocals of roots in Harmonic Progression are in Arithmetic Progression, let 1 1 1 a — b' a a + 6' be the roots of the given equation ; then, ( Wood, art. 273.)

a—b-\-a + a4-b = —• .'. a = — -, and .'. - = — ,

r 3r a q

consequently, x3 — p x1 + q x — r = 0 is divisible by

3r x , the quotient being

xi__ P9- 3r x + y3-3^yr + 9y3 = Q q ' q2

which, consequently, contains the roots j, and — -—?, viz.

the greatest and least. Hence, those roots are easily found ; or they may be obtained independently of the quadratic.

1 l

a? — bl a

whence, b = — — ~ — ...".. — Sr </ q

&c„

6. (1) Let a, b, c, be the roots required ; then, a+b= 13. « + 6 + c=15V !> whence,

a be— 80J

,*. a = 8, or 5, b = 5, or 8. (2) Let 2 a, 36, c, be the roots]; then, 5 a + c = 17 |

2« x 3a + 2oc+3ac = 94J '

IN ALGEBRA.

85

or, 5a-{-c=l7\ .'. I9a2—S5a — 6 a2 + 5rtc = 94 J ' a2— f$a =

-94

_ 9A 1 9 J

P. 53.]

.". a = 2, or -f£,

c = 7, or |4, whence the roots are found.

(3) The equation and its equation of limits having a common measure of the form (x—a)-x (%—b), if this be found, a and b may also be found very easily.

Otherwise. 3a + 26= 13, a?+a2-\-ab + ab + a2 + ab-t-ab±ab + ab + b2=; 67,

or, 3 a 4- 2 b — 13^ whence, a = 3, 3«2+ 6a6 +#J=67J'' 5 = 2.

(4) The general form of triangular numbers being

»<frfel* let the three roots be ^±1), ^±IX^2), 1. 2 -* ■*

^+2)(B+3) TlMn>3.»+9»+8=31i

2 ^

which gives w == 3, or — 6, and the roots are 6, 10, 15, or 15, 10, 6, according as n is made = 3, or — 6.

7. (1) Since 1 is a root {Wood, art. 326.) dividing the equation by a — 1, and the quotient, viz.

x*—(p— 1) x1 + (gr — p 4- 1) .r6 - (r—q +p— 1) #5

+ («— r + q— p+l)x* — (»'— ^+jw-l)^34-('7— p+1)^2 -(p-l)x+l=0, contains the other roots. Divide by x4 ; then,

-(r-^+p-1) (* + -) 4- s — r-f-y — jt? + 1 = 0.

Make x 4 - = m, r

#4

86 HINTS AND ANSWERS [P. 54.

ri i i — „,3 o X2

x3-\- — ,= u3 — 3u,

XA

1

^+-4 = <>2-2)2-2 = tt4-4tt2 + 2,

.*. ti4 - (p — 1) u3+ (q — p— 3) w2 + (2j» + q — ?• — 3) u

+ 2{p-q) = 0.

(2) The roots are 1, -1 \V ~* > "* ~% ~~ >

_] + V-3 2

or 1, —

2 ' -1+ V-3 *

(See Private Tutor, Alg. part ii. p. 399.)

8. The root found by the rule is 4. The rule fails when two of the roots are neither impossible nor equal. ( Wood, art. 331. Private Tutor, Alg. vol. ii. part ii. p. 71.)

9. Since (A + B)5 = A3 + B3 + 3 AB (A +B)

= -r + s£-£-$}*.(A+B)

= _r + ?(A + B), .-. (A + B)3-?(A + B) + r =0. But the given equation is

x3 — q x -f r =0, .'. A + B is a root of the equation. Again, (a A + |3 B)3 = a3 A3 + /33 B3 + 3 «/3 AB (a A + /3 B)

= A3 + B3 + 3AB(aA + i3B),

for a3 — 1 = 0, /33- 1 = 0, and /3 = -.

a

/. (a A + /3B)3= - r + q (a A + /3B), .*. &c. as is evident. And similarly for the other root.

10. Private Tutor, Alg. vol. ii. part ii. p. 121.

P. 54.] IN ALGEBRA. 87

11. (1) Let the required equation be

y2 + P y + Q = o.

Then, P = ^/a + i/b, Q = ^(ab) —Vq, .'. P2 = a + b -f 2 -/(aft) =j» + 2 -/?, .*. t/2 + (i) + 2>/^)y+>v/^ = 0 is the equation required.

(2) xn + q x71'2 + .s a;"-4 + ....== a * (jo a;"-1 + r xn~* + &c).

1

Let x ~ y. a 2 '

.'. t/"fl~^ + g yn-2 a"-^+l + S y»-4 a~?+2 + &C.

a ■ (p y»-» a " + 2 + r i/N~3 a *+ 2 + &c.)

_5 .1 _ n + 2

,rt-l „ 2 1 , „ »,n-3 « 2 T

=jp t/"-1 a 2 ' -f r y""3 a 2 + &c. .'. yn + g ayn-2 + s «2?/"-4-f&c. = p oyB-, + raayB_3+ &c. .*. yn—pay11-1 + qa yn~2— r a2yn~3 -f s a2 ?/"_4 — &c. = 0, in which the coefficients are both rational and entire. (3) Let the required equation be

if-Py2 + Qy-R = 0, then v its arc roots are (seep. 46), a+b 2 ab

— ' ^ai) and 7Tb'

R = gV?*

.'. the required equation is

12. Since a+6-fc =79 ") . , ,

a + b + c'=p'S F l

Also,

ab c = abc' =

r\ . c r

t r • • ~j — ~> r J c r

88 HINTS AND ANSWERS [P. 54.

whence c = r.- — 'K. and c' == r . - — ^7- r — r r — r

13. (I) If y = -; then the Superior Positive Limit of y

will be the Inferior Positive Limit of x. But the equation in

V is

7f + f . if - f y* + y + * = 0, and if w = 7/ 4- e, the coefficients of the equation in w are

4e3+ 2e2— \°.e + 1, 6 e2 + 2 e - 4, 4e + f, which are all positive when e = i, and not so when e = y, .'. i is a wea?' Superior Limit of ?/, or a near Inferior Limit of x.

(2) As in example p. 75, it will be found that 1 is an Inferior Limit of the roots or values of x. Consequently, if all the roots be increased by 1, they will become positive; that is, the terms will be alternately positive and negative. (Wood, art. 265.) Making, in fact, y = x 4- 1, we get, by substitution,

y3_4ya+ V y-i = 0.

14. The given equation being of the form

(x—a) Qp—b) (x—c) . . . . = 0," that of limits, viz.

n xn~x— (n— 1) x"~'2+ . . . . = 0 . . . . (a) is of the form (see Wood, art. 308.)

(x—a) (x—b) . . . . + (x — a) (x—c) ....+....= Q, each term having (n — 1) factors.

Similarly, the limiting equation of (a), viz.

n 0-1) x"-2-(n—l) 0-2) £»--r+- = 0 (p)

is of the form

(x-^-a) (x—b) . . . . 4- (x—a) (x—c) .... + .... = 0, each term consisting of [n — 2) factors.

P. 5.5.] IN ALGEBRA. 89

Lastly, the limiting equation of (/3), viz. n(n-l)(n-2)xn-3-(n-l)(n-2)(n-3)xn-i + 8tc. = 0...{y) is of the form

(x—a) (x—b) . . . . -f (x— a) (x—c). . . . = 0, each term containing n — 3 factors. Consequently, since the given equation has n factors,

n (n— 1) (n — 2) xn~3 — &c. xn—p xn~l + &c.

+ T w l 1.1 , * + &c-

(a:— a) (^—6) (a-— c) (x—a) (x—b) (x—d)

TRINITY COLLEGE, 1823.

1. First find an Inferior Limit of the roots of the given equation, or a Superior Limit of those of the equation

V4— ?/3-19 y2— 11 y + 30 = 0, whose roots are those of the given equation with their signs changed.

Secondly, augment the roots of the proposed equation by this Limit, and the roots will become all positive, and .*. the signs of the terms alternately positive and negative.

But the coefficients of the equation in y, when we make

y — x + e, are e4— e3— 19 e2- lle + 30, 4e3_ 3e8— 38e— 11, 6e2- 3e-19, 4e— 1, which are all positive when e = 5, and not, when e = 4, .".5 is a wear Superior Limit of y, or a near Inferior Limit of x. Make .'.in the original equation

u = x -t- 5, or x = m — 5, and we get

.*. m*-19m3+1 16 m2-224m+1 =0 for the required equation.

N

90 HINTS AND ANSWERS [P. 55.

2. If a be the common root ; then x— a is a common mea- sure of the two equations. By the usual rule, this is

x - 3. .*. 3 is the common root ; whence the other roots are found by reducing the equations to quadratics ; and the roots are

2, 3, 4; -1, 3, 5.

3. Let a be the given root of the equation,

x* — pxn~x -f qxn~2 — &c. = 0, then aM — pa"-1 + qa"-*— &c. = 0, .*. Xn — an—p {Xn~x— am_1) +q(xn-2— a"-2) — &C. = 0.

But each of these terms being divisible by x—a, (see Pri- vate Tutor, Alg. part i. p. 98.) the equation is reducible to the form

which being supposed to have one root, a, may be, in like manner, reduced to an equation of the form

xn-i _ p^ xn-z + ^ xn-z _ &c. = 0,

and so on.

Whence, it is evident, that if any equation have one root, it will have as many as it has dimensions.

This is the meaning of the proposer of the question ; for otherwise, it being easy to give any equation a root, by multi- plying it by x — 1, for instance, the general proposition which so many have endeavoured to establish in a simple manner, would be extremely easy.

4. Wood, art, 271. Or it may be proved better for Ex- aminations, by assuming it true for m dimensions ; then multi- plying the equation x— a, X being the (m + l)th root, and proving it true for m + 1 dimensions. But it is true for two dimensions ; thence it will be true for three dimensions, and so on.

P. 55.] IN ALGEBRA. 91

5. Make y = - ; then the equation in y has the roots -

-r, — ; which equation is be

•'•2/3-p 2/2+ -r y-i = o.

Assume the required equation to be

1112

then P = -3 + ^ + = 1 _ 2£, fFootf, art. 273.

a-* bl cl rl r

o -=_L + _L . _L a% + b'2 + c'2

w a262 a2 c2 JV ^WcT~'

— 2 '

and R = JL. .I. 1 = 1,

a2 b2 c? r2

.'. the equation required is

r2 r r2 r2

6. (1) There being two changes of signs, there are two positive roots, and .'. three negative roots.

(2) This is departing from the usual definition of a root. But see p. 80.

7. Let xm = y ; then

V[i- P*f +qy-r = 0. Take away the second term of this, by making

y =z u -f t. * 3

and then one root of the equation can be found by Cardan's

Method, when two of its roots are either imaginary or equal.

(See Private Tutor, Alg. part ii. p. 71). Hence may be

found, in those cases, the roots of the equation in u ; then

those of the equation in y ; and finally, the equation may be

fully resolved by finding each of the three sets of roots of m

92 HINTS AND ANSWERS [P. 56.

roots each contained in the form xm — y — 0. See Private Tutor, Alg. part ii. p. 399.

8. (1) Wood, art. 304.

(2) The terms being given alternately positive and negative, all the roots are supposed positive ( Wood, art. 265). Hence,

making y = -, the Superior Limit to the roots of the equa-

tion in y is an Inferior Limit to those of the equation in x. But the equation in y is

vn t — vr . ■ • • ± — Vm • • • ± - V% + - ± 1 = o. r u u m u

From this equation a Superior Limit to its roots, is

— + 1, or — + 1, u u

according as u is + or — ( Wood, art. 304) ; that is,

I 1

or

N P

-4-1 -+1 u u

is an Inferior Limit of the given equation, according as u is

positive or negative. Whence the proposition is manifest.

9. Wood, art. 320.

10. (1) First — 1 is a root. Let the other roots be a, -, b, - ; then

a + - + b + v - 1 = 21,

a o

il i a 611,1, 10_

and a.- +ab+ T — a -1 1 — r 4- o. T— b — -t=37

a b a ab a b b

1 u l o,

or a + - + H t = 2-2

a o

lab ._,

aft » a

1 , 1

Let a 4-- =tt,Hr =«J.

M + v = 22i Mr=57f

P. 56.] IN ALGEBRA. 93

tt3	+ 2 uv + v2 = 484 4«y = 228,
.\ Ma	— 2uv +v2= 256,
.'. U	— e? = ± ]6\ .'. w = 19 or 3 + v = 22 1 v = 3 or 19
u
	. 3+*/51 3 whence a = — - — ,- = — 2 a	2
	h 19+^257
	2 '
	,1 19 + -v/257 and6= 2 *
	Otherwise.

Dividing the equation by x 4- 1 , we get

a:4 - 22 x3+ 59 ^2- 22 x + 1 = 0. Dividing this by xa, and arranging

+ ~ - 22 (« + 1) + 59 = 0.

:z*

Make a; 4- — u, x

.'. u?—22u= — 57. Hence u, and .'. the values of #; as is obvious.

(2) Private Tutor, Alg. part ii. p. 25. As to the question holding good when m is any prime num- ber, that is shown not to be the case in the simplest instance. For one root of x3 -f 1 = 0, is — 1, of which the other two

1 ± V— 3 are £ .

11. Private Tutor, Alg. vol. ii. part ii. p. 79.

12. (1) By Wood, art 338, if for x we put 1, 0, — 1, it is found that — 1 satisfies the equation, and is .'. a root without further trouble. Whence the other roots are found by the reduced quadratic to be 3 and 4.

(2) Barlow's Theory of Numbers, p. 32.

94 HINTS AND ANSWERS [P. 56.

13. Wood, art. 510.

14. Wood, art. 352.

15. Let Sp S2, S3, &c. S„, SM+1 denote the sums of

the simple powers, squares, cubes, &c. of the roots a, b, c, &c. of an equation of n dimensions; then, *.*

SOT = am + bm 4-

Sm+l=am+l + bm+i+

if a be the greatest root,

S,+l _ aw+1 + 6w+1+

Sm ' am + 6m +

Suppose b = a. b' , c = a. c', &c. ; then,

SOT+1 aOT+' 1 +y«+i+ c,»+l + ■ . . . Sw a"* ' 1 + b'm + c'm +

1 + 6'm+,-fc'm+1+. . . •

= a

I + b' m + d '"+... .

But •/ a is the greatest root, .". b' , c', &c. are each < 1,

1 _j_ b' "l+1 +

.". the greater m is the more nearly , ,-, - is equal

C

to unity, and the more nearly to a does ™+1 approximate. Example. Required to approximate to the greatest root of

X* — 3x?+2x-\ =0,

S-p = 0, S2-pSi+2q=0, S3-pS2+qS-3r = 0,

S*—P S3 + ? S2_ V Sl = 0>

S6—p S4 + q S3-r S2 = 0, &c. In the present case, p =3, q = 2, r = I, .-. S, = 3, S2 = 5, S3 = 12, S4 = 29, S5 = 68, S6 = 158, S7 = 367, S8 = 853, S9 = 1983, SI0 = 4610, S„ = 10717, Sla = 24914, S13 = 57918.

P. 56.] IN ALGEBRA. 95

Hence, if a be the greatest root of the equation, we have

SQ 5 S, 12 29 68 158

a = S; = 3' °r = S; = T' °r = 12' °r 29' °r "68 '

367 853 1983 4610 10717 24914

or T58 ' °r 367' °r -853 ' °r T983' °r 46l0 ' °r 10717'

57918 or

24914"

To approximate to the least root of an equation in x, make y = -|, and then, as above, approximate to the greatest root of the transformed equation in y.

16. (1) First, make y = —; then, the equation in y is

5 43 19 3 13 , 3 1

y +my~40y~ 120^ f 40y~~ 120 _U'

whose roots are -,

11111

a b c a e But, as in the example of p. 94,

si— P = °> S2—pSl-\-2q = 0,

S3— p S2+q S, — r = 0,

S — — —

' 1 T 60' "2 '" 602 20'

43 Z432 19\ 19 43,13 3 ~= 60 V602 + 60/ 40 60 40' whence the numerical value of S3 may be found. (2) The Function = a2(63 + c3 + d3 + e3) + 62(a3 + c3 + ^3+e3) 4- c3(a3 + 63 + d3 + e3) 4- #(a3 + 63+c3+e3) + e3(a34-63-fc3+d3) = a2 (S3- a3) + 62 (S3- 63) + c2 (S3- c3) + &c. = S3. S2— S5. But, S2, S3, and S5, may be easily found from the equations S-p = 0, S2-^S1 + 29 = 0,

96 HINTS AND ANSWERS [P. 57.

S3-pS^qS-3r = 0,

S5— p S4 + q S3— r S2-K S,— 5 t = 0, p, q, r, s, t, being the coefficients in the general equation, xb—p x* + q x3 — r x2~\-sx—t = 0.

TRINITY COLLEGE, 1824.

1. Private Tutor, Alg. part ii. p. 15.

2. Reducing the roots to a common denominator, they are

a2 + b2 a2 + c2 b2 + c2

Assume

a2	b2 '	a2	c2 '	b2c2
V-	_f-	-2q-	— X2	= X2.	f	-2q- r2	-X2
	r2
		X2
		p2-	-2q r2	, X1 —	I r2'	x\

.'. x*—(p2—2q)x2 + r2y = Q, \ and x3 + p x2 -{- q x -\- r = 0, J from which x being eliminated, the equation in y will be that required. See Private Tutor, Alg. part ii. p. 139.

Otherwise. Assume the required equation to be

y* _ p ^ + Q y _ R = 0,

. D a2 + b2 a2 + c2 b2 + c2 tnen, P - -^- + -^p + ^^:2-

_p2—2q—& p2—2q — b2 p2- 2 q — a2

r1 ~*~ j.2 ~* r2

c2 ~b2 a2

= (fe?fifi£±»±*fc_ £ (0. + ^+ c<).

But the sums of the powers are always known in terms of the coefficients. Wood, art. 352. Consequently, P is known, and similarly, Q and R may be found.

P. 57.]

IN ALGEBRA.

97

3. Let a = X cos 0, b — X sin 6, which is allowable, be- cause \ may be of any magnitude, whilst 0 may be as small as required,

then, (a+b J — l)4 == X4 (cos 0-f J — 1. sine)4

J(

8

cos ^ + </-\ sin_V

and

(«"^-l)^ X4(cos \-yf-\. sin 'I),

•••\/(H*\/-l) + tj(a-b */-!) = 2 A cos |

^ (a+ 6 A/ _ 1) - ^/(a - 6 V - 1) = 2 XV - 1. sin | in which X — ^(a2-f 62), and 0 = cos-1

7 ^(tf + py

4. If all the roots are real, and be called a, b, c, d, &c. ; then the factors are x— a, x—b, x—c, &c. and the equation is decomposable into (x — a)(x — b), (x—c) (x — d), &c. or into x2— (a + b) x + a b, x2—(c + d) x + cd, &c. which are real.

If any of them be imaginary, they must consist of pairs of the form a + ft s/ — 1, o — ft *f — 1, and the factors hence obtained are (x— a— ft\/ — l)(x— a+/3y' — 1), or(x—a)2+ft2, or a?— 2 a x+a2+ft2.

p

5. Since a is a root, a"— 1 = 0, .'. a" = 1, .*. (a")" = 1,

.'. ap — 1 =0, or ap is a root. Let p == n+m; then,

a»+m_ I _ Q5 .*. a", a"'— 1 =0.

But an = 1, .". am — 1, or the roots recur when p is > n.

6. (1) This may be done by the Method of Divisors, or by Cardan's Rule

1	14	1, 2, 7, 14	2
0	- 9	=pl, q=3, +9	3
- 1	- 4	+ 1, +2, +4	4

98 HINTS AND ANSWERS [P. 57.

.". 3 is a root. Whence the other two are11^-^- — * 0 — —

fit

By Cardan's Method.

r=^[-iV{Q-(D3}]

= #8+4/1 = 3.

(2) Let the roots be — 3, -, ar, ar3. Then

— • -• ar. ar3 = s, or a* = 5, . . a = « . Again, let y = — , then the equation becomes

a4 ?/4 +£>a3 7/3 -f 9- a2 ?/2 + ?'a?/ + 5=0, a ' a'iU a J "

»?/3 q . r S „

0r y4 + ££_ + JL yS + y + _ = 0,

or y* +JLy3+ £_y» + _L y + 1 = 0,

S* .ST St

which being a recurring equation, whose roots are

— - , -, r and r3

£..= '21. (JJW, art. 325.)

.*. dividing by t/2, we get

Make y + - = u, if y

.-. y2 + -L = M2 _ 3,

and «2 + 4 u + 4 2 = °- s* s-

»■.+ i+ £, •(».+.=)+ 4 = «■

P. 57.] IN ALGEBRA. 99

Whence u, .'. y + - is known. .". y from solution of a qua-

dratic, and thence the four values of x.

(3) Since xli—an=0, .*. (-\ = l = cos2pr± j — 1. sin 2^,

a; 2» , . 2p

.'.- = cos ~ tt ± >/ — 1 sin j7 *->

a 1~ 1^

/ Pit , , . Ptt\

.'. x=al cos ^r- ± \/— l.sin^- J,

and making p=0, 1, 2, 3, 4, 5, 6, we get

x = a,a(cos^ ± </ — 1. sin ^Lafcos^ ± V — 1. sin-J • . •

/ 5tt , , . 5tt\

a (cos-^- ± \/ — 1. sin -^-j, — a,

or =a,a (cos 30 ±\/ — 1. sin30),a(cos 60 ±\/ — I. sin 30) . . . a (cos 150 ± \/ — 1. sin 150), — a,

or=a,^(^3±A/-l)^(l±^3)..^(-A/3±A/-l),-a.

For the entire Theory of Binomial and Trinomial Equa- tions, see Private Tutor, Alg. part ii. p. 399.

(4) The limiting equation has two of them. Hence

3x5-l0x3+l5x + 8=0\ and x4— 2a?2+l=0 J

have a common measure of the form (x — a)2 = 0 ; and v

#4_2#2-f 1 _ (^2_i^2. /# euqier ^— l)2, or (x+lf is that

common measure : that is, 1 or — 1 is one of the equal roots.

By trial, — 1 is found to be that root.

Dividing the proposed equation by {x -f- l)3, or x3 + 3 x'2

+ 3 x + 1, we get for the reduced quadratic

3 x2- 9^ + 8 = 0,

whose roots are

."•■§• ± i \/l5. -v/ — 1.

7. Since imaginary roots enter equations by pairs of the form a ± /3 -y/ — 1 (see Private Tutor, Alg. part ii. p. 46.) a— (3 s/ — 1 is also a root. .'. the given equation is divisible

by

X^-2aX + a2+ /32.

100 HINTS AND ANSWERS [P. 58.

Dividing and putting the quotient, which is of the form #2 + Px + Q, equal toO, and resolving the resulting equation, we find its roots to be

i |-(p + 2a)±^(pa-40-4ap-8a* + 4/3a;2. But 4s =z (a2 + /32) ^0 + 2a)2-(p2-4?-4aj»-8a2 + 4/32)£ which gives the required transformation.

8. If -I =« and /{(I)*- (§)'} = 6 v-l;

then, by Cardan's Rule,

x= \/(a+b^— 1) + V(«— 6-y/ — 1). But (see third question),

a\bsf — 1 = -y/(a2-f62) (cos 0 + V~ 1. sine),

in which cos 0 :

.\ # = 2(a2+62)^"cosl

But a2 + 62 - - - - + Si = i! + 4 4 + 27 27'

.-. a; = 2. d)i cos |, &c.

9. The reducing cubic is ( Wood, art. 332.)

y3 + 2qy2 + (y2 — 4 s) y — r2 =r 0.

.-. if?/=«, e=>y/a, and/= ^ ~ — ~^~^^

.'. one of the component quadratics, is

r

q +a which being resolved gives

j* + Va. x+ -= ^ =0,

At

q + a

P. 58.] IN ALGEBRA. 101

and the other component quadratic gives

10. Divide the given equation by xm, and the result is, collecting the extreme terms,

*m+ „\+P (*-* + ^n) + q (f~ + ^) + • • • = 0,

and making x + - = 2 cos 0, then .r2+ - — =2 cos 2 0, &c. x x2

.'. the equation becomes

2 ^cos md + p cos (ra— 1) 0 4- q cos (m— 2) 0 + . . .\ == 0, or cos ra 0 + p cos (w— 1)0 + 9- cos (m— 2) 0 + .... = 0,

which equation is resoluble into one of the form cos"1 0 + P cosm_1 0 +....= 0,

an equation of m dimensions.

Otherwise.

Since the roots are in pairs of the form, a, -, the equation

is decomposable into quadratic factors of the form (x — a) x

(x J, or of the form x2 - («H — J x + 1, that is,

x^+px^-i+qx*"^ 4. &c.=(ar2— A x+ 1) (.c2— B# + l)&c. &c. see Wood, art. 325.

11. Make q, — A cos 0, 6 = \ sin 0 ; then

X = V (*+n 9 = cos- y^pyj.

and (a + ^-lf+MV-l)'"^^ (cos 0 + A/ - 1 sin 0)'"

+ X (cos 0 — -y/ — 1 sin 0)"', = \ (cos md + s/ — 1- sin w0) 4- X (cos mfl- y' — 1. sin m 0), = 2 X cos md, — 2 ^/(«2-f62) cos mfl.

12. JFood, art. 338.

1	3
0	10
-1	-35

1, 3, —1, —3

1,2, 5, 10,-1, -2, -5, -10

1,5,7,35,-1,-5, -7, -35

o? O^ Oj — o

2, -J, -1, -5 1, -5,-1,-7

102 HINTS AND ANSWERS [P. 58.

The decreasing progressions in which the common difference is a divisor of 10, are 3, 2, 1; 3, 1, — 1 ; —3, —5, —7; consequently the factors to be tried are

2x 4- 2, 2 # 4- 1,2a; — 5,

5 of which 2 x — 5 succeeds, and .'. ■= is a root giving the re- duced quadratic.

13. Since

?F=3)5 = 1-2 i+a* = 1+2 a + 3 a2+ &a ^ division>

and , = 1 + a + a2 + &c.

1— a

a 4- 2a2 4- 3 a3 4- &c.

' " (1-a)2 1-a

similarly, * --1^= 6 + 262+3 63+&c. &c.

.\ 5l + 2«2 + 3 53+ CO = (1_a)2 + -(j— gy2 + &C.

~(l^+ 1^ + &c-> But, ( Wood, art. 352.)

n + (n-l)p + (n-2)q + ... = _1_ _J_ &c 1 +/> + ?+ ••• 1— a 1— 6

• /^4-(re-l)p4-(re-2)?4-...\2_ 1 1 , fe.

"V l+jp + 0+... ; -(l_a)2+(i_6)2+^c-

+ 2{TT-a)(l-*) + (l_o)(l_c) + &c'}'

Also, •.• n xn~x + (»— 1) p x71-'2 + (re— 2) q x"'3 + &c. = (x—d) (x—ft) . . . (re— 1) factors 4- (x—a) (x — c) . . . (re— 1) factors + &c.

.*. n xn + (re— 1) p xn~l + . . .

= # £(#—«) (x—b) . .-. (re— i) factors 4- &c.f

P. 59.] IN ALGEBRA. 103

Taking the Equation of Limits of this, we get a*«^-i+(n.— I)2 px«-* +...— (x-a) (x—b)...(n—l) factors

+ (x —a) (x — c) . . . ( n — 1 ) factors &c. -1- 2 x \ (x—a) (x—b) . . . (n — 2) factors + (x—a) (x—c)...(n—2) factors

+ &c£ Hence, putting x — I,

n

+ (ft-l)aj> + (n-2)«y+... 1 J_ 1 .

.*. *1 + 2*3+3*3+ .... oo (n + (n—\)p+ &c.y rc2 + (w — l)2 ^ + &c.

V l+p + q{-tkc. / l+_p + y+ &c.

It must be observed, as in the example of p. , that unless each of the roots be less than unit, this result will be absurd. For the left hand member will be oo , whilst that on the

right is finite.

TRINITY COLLEGE, 1826.

1. See Private Tutor, Alg. part ii. p. 15.

2. See Private Tutor, Alg. part ii. p. 46.

3. (1) Private Tutor, Alg. partii. p. 112. for this and many

other transformations of the same kind. Ans. p.

(2) See Private Tutor, ibid.

Make y = x (p—x) = p x — x1.

(3) See Private Tutor, Alg. part ii. p. 272.

4. Wood, art. 314.

5. Let -xn—p xn~l-\-q xn~2 — . . . = 0, be the given equa- tion ; then its Limiting Equation is

n z«-*— {n— \)pxn~2 + (n— 2) qxn~:i — . . . = 0;

104 HINTS AND ANSWERS [P. 59'

also, let a, b, c, . . . ; a, fi, y, ... be their respective roots, nxn~l— Q— l)pxn-2+..._ 1 _J_ _1_ &

xn — pxn~l-\-... ~ x— a x—b x — c

then, ■/ xn—p xn~l+ . . . = (x—a) (x—b) (x—c) . . . n factors andrca^1 — (n— l)pxn~2-\- . . . == n(x — a) (x— fi)...(n— 1 )factors the respective products P, P', are

P =(a—a)(a—b)... x 03— a) Q3— b). .ton x (n— l)factorsinall,

and

F= nn {a— a) [a— /3) ... x {b— a) (6-/3)... to (»— 1) X w factors,

whence it is evident, since n [n— 1) is even, that

P : P';: 1 : n\

6. (I) Private Tutor, Alg. part ii. p. 399.

(2) See p. 70. The roots here are in Harmonic Pro-

gression.

(3) This may be done by Cardan'' s Rule, by first taking

away the second term, &c. ; or, more briefly, thus, :r3-2a2+^-2 = 0, .'. x3 + x-2{x2+l) =0, .-. {x2+l)[x-2) = 0, ,". the roots are 2, ± \f — 1.

(4) See p. 98. An erratum is, for —2 read —r.

7. See Private Tutor, Alg. part ii. p. 363.

8. Since, when y = 0, x = 0, make

y = A x + Bx2+C x3 + &c.

y2 A2:*2 AXJ ,

g- == AB x* + &c.

y* A3*3 .

3 = — + &C'

A2 A3

.-. A = l, B -^=0,C-AB + -g- = 0, .-. A - 1, B = g, C = g - 3 = 273> &c'

y*2 y 3

.-. y = x + ^ +^3 + &c = e*— 1. See WW, art. 346.

P. 60.] IN ALGEBRA. 105

9. Generally, ( Wood, art. 352.) if the equation be

xn-\r p xn~l + q xn-'2 + &c. + L = 0, Sm+p Sm-x+q SOT_2 + ....+ m P = 0. In this case, p == 0, q = 0, r — 0, &c. but L = — 1 . Hence, when m = n, Sn = n, in all other cases SOT = 0.

10. See Wood, art. 509.

TRINITY COLLEGE, 1827.

1. Wood, art. 275.

2. Seep. 90.

3. Let y = m n p x. Then substituting, &c.

.'. y3 — a npy + bm2np2y-\-cm3 n3p2 = 0, is the required equation. Secondly, make y = p u ; then,

.'. m3-2m2- 33 u + 14 = 0, is the required equation.

4. Wood, art. 285.

5. Wood, art. 298, and 302.

6. First, transform the given equation in x to that in y, whose roots are the differences of the roots of the given equa- tion ; then find the Inferior Limit of the roots of the equation in y.

If a, b, c, be the roots of the equation x3—p x2-\-q x— r = 0, then, [a-bf = [a + bf — kab, .'. if u be the character for the roots of the required equation, we have

4r 4?'

u2 = {p— x)2 = x'2 — 2 p x + p2 — — ,

'JC 30

p

106 HINTS AND ANSWERS [P. 60.

}

/. x3 — 2px2 + (p2— u2)x — 4r — 0 But x3 — px2 + qx — r — 0

.'. p x2 — (p2 — u2 — q)x + 3r =0.

Also x3 — 2px2 4- (p2 — u2) x — 4r = 0 \

0 /'

4 x3 — 4 ^#2 ± 4 qx — 4r

.'. 3 a;2 — 2p^ + 4q~ p2+ u2 =-- 0 ~l j»:r2 — (p2—u2—q) x-\- 3r — 0 J '

3p.r2 — 2p2x + 4pq— p3+pu2 = 0\ 3px2 - (3p2-3 m2— 3 ^) ar-f 9r=0.J '

(jo2-3k2-3?) ^ + 4pgr-jo3+p«a— 9r =0, • r _ pu2— p3 + 4.pq — 9r 3 m2 — p2-f 3 <y Substituting this value of x in the equation x3 — px2 + qx — r = 0 an equation of six dimensions will result, whose roots are ± (a—b), ± (a—c) ± [b—c). If the equation be of the form x3+qx — r = 0 ; then the transformed equation is [Private Tutor, Alg. part ii. p. 300.) u6 + 6qu*+9 q2 u2+4: q3+27 r2 = 0. In the present instance, q = — 7, r = — 7,

.'. w3-42m2+441 u— 1372+1323 = 0, of which an Inferior Limit may be found, as in p. 73.

7. Make y = a+b + ab=p — x + - ; whence

xy = px — r2 -f r,

or x2 + (y—p)x — r = 0\ and x3 — px2 + qx — r = 0 J '

whence eliminating x, the equation in y will be that required.

See Private Tutor, Alg. part ii. p. 235.

Otherwise. Assume y3 — Yy2 + Qy — R — 0, thenP = 2 (a + b + c) + ab + ac + bc, = 2p + q, Q = (a + b + ab) (a + c + ac) + (a + b + ab) (b+c + bc) 4- (a-fc-t-oc) (b + c + bc), — a2 + (b + c) a + bc + (a + b) (ac + bc) + (a-\-c) be,

P. 61.] IN ALGEBRA. 107

4- 62 4- (a 4- c) 6 + ac + ab (a 4- c ) -|- a26c 4- a6 (6 4- c) 4- a63c _j_ c2 4- (a 4- 6) c + ab + ac (6 4- c) + a6c2, =a24-634-c24-3(a64-ac4-6c) + a2c4-a6c 4- abc 4- 62c + abc 4- 6c2

4- a26 4- a6c 4- a26c 4- ab2 4- a6c 4- a62c 4- a6c 4- ac2 -f abc2, = p2— 2 a 4- 3 a 4- 3 abc 4- a (ab 4- ac 4- be) 4- 6 (aft 4- ac 4- 6c)

4- c (a64-ac4-6c)4-a6c (a4-64-c), — p24-a4-3r4-jo^4-^, R = (a + b + ab) (a + c + ac) (b + c + bc), and multiplying these so as to group all those terms which are of the same dimension together, we have R = a2 62 c2,

4- abc2 (a + b)+a2bc (b + c) + ab2c (a + c), 4- ac(a4-6) (b + c) + bc (a+b) (a + c) + ab (a + c) (64-c) 4- (a + b)(a + c){b + c).

4- r (ac4-6c4-a64-ac4-a64-6c)

+ ac (ab 4- ac + b2 + be) 4- 6c (a2 4- ac 4- a6 4- 6c)

4- a6 (a6 4- ac +- 6c 4- c2) 4- (a24-ac4-a64-6c) (64-c), = r24-2 qr,

4- a2624-a2c24-62c24-3a6c (a 4- 64-c) 2a6c4- a2(64-c)4-62(a4-c)4-c2(a4-6), = r2 4- 2 ^4-3^4- 2 7-4- a2 (p— a) + b2 (p— 6) 4-c2 (p—c) = r24-2ar4-3jor4-2r4-p(a24-624-c2)-(a34-63+c3). But (Wood, art. 352.)

a24-6*+ca— jp2— 2 y, and a34-634-c3=p3— 3 pq + 3r, .*. R = r2-f 2ar + 3^r4-2 r4-J»3— 2^a— (jo3— 3pa4-3r)

= rs + 2gr + 3jw + 2 r4-j9a— 3r,

= r2-\-pq — r4-3pr4-2ar, which gives the equation re- quired.

8. Wood, art. 319. Also see p. 98.

9. See p. 78.

108 HINTS AND ANSWERS [P. 61.

10. Private Tutor, Alg. vol. ii. part ii. p. 78.

11. Private Tutor, Alg. vol. ii. part. ii. p. 123.

12. Wood, art. 338.

As in Wood, it is found, that 3 and — 5, are roots, but not 5. The other two roots are

3±V-11 2

13. Private Tutor, Alg. part ii. p. 399.

14. Since ex—\ + x + — - + — — + &c. = y,

.-. x = log y, and, it is well known, that log y cannot be expanded in a series according to the powers of y, for, if we assume

x = A + By + Cif + &c. some of the coefficients will become infinite. But, if we make

x = A(2/-l) + B(2/-l)2+C(2/-l)^+&c.

thenT = ^-2(r-i)2+AB(2/-i)3+ &c

x3 \3

&c. .-. A = 1, AB + ~ = 0, B = - |,

p _ X J_ J. &r>

v^ — 2 6 — 3 J O^1-.

.'. * = (3/-1)— i (V_l)2+i (y— 1^ + &c.

TRINITY COLLEGE, 1828.

1. Wood, art. 277.

2. Private Tutor, Alg. part ii. p. 15.

3. (1) Assume y=a2 {b + c)—a2 (—p— a) = — - x3— px, and eliminate x from the equations,

x*+pafi'+qx+r=0'l SeePrivate Tutor, Alg.pt. ii. p. 223. x3+px +?/=0 J > o r r

P. 62.] IN ALGEBRA. 109

Otherwise ; assume

y*-Plf + Qy-R = 0, then P =1 a (ab-\-ac) + b(ab-\-bc) + c(ac + bc), = a (ab + ac + bc) + b(ab + ac + bc)+c (ac -J- be + ab) —3 abc , = q [a-\- b-\-c)-\-3 r = — pq + 3r.

Similarly, (see Private Tutor, Alg. part ii. p. 234.) Q = q3 — 2 pqr + 3 r2, R = r3 — pqr2.

(2) See Private Tutor, Alg. part ii. p. 300.

4. This can only be done by resolving a cubic. Wood, art. 285.

5. Let a, a2, b, be the roots ; then

(1) Since the last term is the product of all the roots with their signs changed, and the product of two of the roots is of the form a x a2, that factor which is a cube will give one of the roots, viz. a. For a3= ±8, .". a = ±2, and on trial we find a — — 2, .'. a2 = 4, and the other root = 0+2-4=- 2.

(2) Seep. .

(3} Divide by 4:r2, and there results x* x 2 4

*» 4 rx 2\

-GO^S + D^'-*

•"•| + ~ = -|±y (^-y+2J=P suppose. Hence we easily get x.

6. (1) WW, art. 327.

(2) See Private Tutor, Alg. vol. ii. part ii. p. 71.

(3) Idem, ibidem, p. 78.

110 HINTS AND ANSWERS [P. 63.

7. Reduce the equation to the form

X3 + Qx + R = 0,

then make x2 = y + <p 4- 4X- Eliminate x, and then assume, so as to determine <p, ^ , &c_ Private Tutor, Alg. part ii. p. 106.

8. See p. 79. Equation is

. mp , rffl5 , n n .\ _ ran rc2

9. Let the equation be xn+p xn~l + . . . — M x71"™— . . . — P#n_" ± &c. = 0. Take the most unfavorable case in supposing all the coeffi- cients from M negative and = P, and find such a value of x as shall always make xn — P (xn~m + . . . x2 + x+ 1) positive, and .'. a fortiori, the left hand member of the equation always positive ; that is, find x such that xn is always

xn-m+\_ J p x »-«+l P

> P. T > =— — -.

X — 1 X — 1 X — 1

,T _, xn-m+l . P xn~m+l P

Now P =— is >

X—l X—l x—l

p ,«.n — m-t-1 j. m— I

assume .'. xn— or > — ■ , — , or , . . Ax— l)"*=or>P;

x— 1 ' (x— l)CT-lV '

that is, make (x — l)m = or > P, or make x = or > 1 -f Pm .*. #n— P(^m~n + ...a:24-a;-f 1) is always positive when u = or

2.

> 1 + P", and, a fortiori, xn -\-pxn~l + . . . — M a;"-"1— &c. =0

x is then always positive ; .'. 1 -J- P"' is a Superior Limit.

10. See p. 105.

TRINITY COLLEGE, 1829.

1. Private Tutor, Alg. part ii. p. 15.

2. See Fra?icoeur,s Pure Mathematics.

3. Seep. 110. Quest. 9.

P. 63..] IN ALGEBRA. Ill

4. Since the equation has equal roots, it has a common measure with the equation of limits. Hence the equal roots are 1, 1, 1, — 3, — 3, and the other root is — 1.

5. By Cardan's Rule, or by the Method of Divisors, one of the roots is — 2 ; the other two are 1 ± </ — 1.

6. Wood, art. 342.

7. Wood, art. 311.

8. Wood, art. 275.

9. Wood, art. 325.

10. Every Recurring Equation is decomposable into qua- dratic factors of the form x2— mx+ 1 = 0. ( Wood, art. 325.)

If a, — be the roots of this reduced recurring equation,

"- 2

.'. - is not fractional ; and similarly for the other pairs of reciprocal roots.

TRINITY COLLEGE, 1830.

1. Private Tutor, Alg. part ii. p. 106.

2. ( 1 ) By the Method of Divisors, one root is 6 ; the other

roots are ± \f — 3. (2) The equation is reciprocal, and one of its roots is — I. Dividing by x -f 1 we get x4 - 22 x> + 59 x2 - 22 x + 1 = 0,

,.(* + I)2 -22(* +1)^-57,

1 m _ , . 19 ± A/357 3±s/5 . x -\ — = 19 or 3, and . . x = ^ , or ^ — ,

X 4i Ai

112 HINTS AND ANSWERS, &c. [P. 64.

3. Private Tutor, Alg. part ii. p. 46.

4. Wood, art. 271.

5. Find whether the Equation and its Limiting Equation have a common measure. ( Wood, art. 319.)

6. Private Tutor, Alg. vol. ii. part ii. p. 78.

7. (1) Wood, art. 311.

(2) There are two negative and three positive roots in the equation. Wood, art. 311.

8. Wood, art. 307, and 310.

9. Private Tutor, Alg. part ii. p. 399.

10. Wood, art. 352.

11. Private Tutor, Alg. part ii. vol. ii. p. 123.

s

ALGEBRA,

PART IV.

OR,

ANALYTICAL GEOMETRY.

TRINITY COLLEGE, 1880.

[P. 64.]

1. Let the given straight line be the axis of x, and a, ft the co-ordinates of the given point, and x, y the co-ordinates of any point in the straight line, and d the given angle.

y ~w ft = [x -^, a) tan 0.

— - tan 0. x -^ a. tan 8.

2. Let y=A#-l-B be the equation required, in which A and B are to be determined ; and, suppose

y'=z ax' 4- b y"=dx°+b', the equations of the given straight lines, in which a, b ; a', b , are known. Also, let 0, d' be the angles which the given lines make with x ; and <j> that made by the line y = Ax -f- B. Then v it bisects the given lines, <p — \ (0 — d') +- 0* = d + B' 2 ' Now, at the point where all these straight lines meet, their

co-ordinates are coincident. ; .".

y — Ax + Bi (A -a) x + (B-6) = 0 y = a x + b \ (a —a!) x + b — b = 0, y = a'x 4- b'J

Q

114 HINTS AND ANSWERS [P. 64.

and eliminating y and x, we get

A [b— b')—B(a— a') =a' b-ab1 .... (1)

A i a'

Also, A = tan^> = tan— ^— ,

6 tan -j + tan -^

1 _ fan _. tan_

But tan 0 = a, tan d' = a'

8 -\ + V(l-a?), 6' -l + v/(l-«'J)

' • tan 2 = ~ a ' tan 2 = 2

and, bence

A_ aV(i+^2)+qyf(l+q/2)— («+<) ___ (2\

A"V(l+'4+V(H«'2H««'-^(1+«V"(1+a'2)-l''' ;

which gives A ; and B is found from the form (1).

A more simple detenu lnatio7i.

Take the origin of co-ordinates at the point of intersection ; suppose the line y = Ax + B to be drawn bisecting the < ; take any point (x, y) in it; draw a J_ through this point, meeting the lines y'~ax' -f b, y" ■= a'x" 4- b' in the points (V ,y), {x" , y") : then it may be easily shown, that

2x — x' + x" .

2y = y' + y" = ax' +ax"

x'2 + (a:r'+ b?=x"*\r(dx"+b'f >

Find x' and x" from the two first, and subtitute in the second, &c. ; then resolve the resulting [quadratic, and it will give the equation required.

3. Let a, b, c be the three sides of the A opposite to the <" A, B, C respectively. Also let a, a ; ft, ft' ; y, y be the eo-ordinates of A, B, C.

P. 64.] IN ANALYTICAL GEOMETRY. 11/)

d2 b*

Then, (area)'2 of a_ — — r— . sin"2 C,

a* b* a2 b> r*+P-c*\*

= -T- (1-cos «Q) S ^-. Jl - ( 2ab ) .

= | ^4 a2 ft8— (a2 + 62- c2)^ . But. a2 = G3— y)9 + (/3'-y')2, &■=: (a-y)2 + K-Vft c2 = (u-/3)2 + («'-/3')2. Hence (a2 + 62-c2)2 = 4 |{y-«) (y-/3)- (y'-a) (y'-/3)p, anda262= (y-a)2. (y-/3)2 + (y-a)2. (y'-/3')2

+ (y-/3)2(y'-«'r2 f (y'-«')'2. (y'-/3')2, .-. (area)2 of a = 4 £(y-«)2. (y'-/3')a

+ 2 (y-a) (y-/3) (y'-a') (y'-ft') + (y-ftf (y'-«')2£,

/. area of a = 2 \(7-a) (y'-(3?) 4- (y-ft) (r'-o')?.

This is the complete developement of the problem pro- posed and solved by Geometricus Emman. Private Tutor, and Cambridge Mathematical Repository, vol/ii. p. 96.

4. Let a, ft; «' , fi ; a", ft", &c. be the co-ordinates of the given points ; x, y those of any point in the required locus. Then v the squares of the distances of the given points from the point x, y are respectively

(■>■-*? f [y-m (*-«')* + (y-P'y, 0-«"}2+ [y-py,

if n be the number of points, and c'1 the constant quantity, we have

x* +2/2- -(«W,+«"+ ••• •)•'•- ~~ W + P + P"+ ••')# + a2 + a'» + a"i 4- ... + /y f ^ + ft"2 4. . . c2

■ • ■ l J. I « 77

whence the locus is a circle.

But if a, I) be the co-ordinates of the centre of the circle, and R its radius ; then

(■c-«f \ (y-W = R2,

116

HINTS AND ANSWERS [P. 65.

/. x*+yi-2ax-2by=R*-{a? + b2) ... (2) Comparing this with (1),

a + a' + a 4-

a = — ! — ±. -EL.

n

b== /3 + /?+j3' + ...

n

»=*+*+.£. "'+"" + ■'+*+*»+

ft ft

= ^ , (a + a, + ..)2+(/3-f/6/ + ..)2-<aM-«/a+..+/32 + /3/H..) » + W2

= - 4- («2 + a/2 + ..+/32 + /3/3+..) (l-^)+2(aa/ + aa,r + .. + /3/3, + i3^ + ..)

which determine the co-ordinates, the centre, and the radius R of the circle.

5. Let o, /3 be the co-ordinates of the centre of the 0 , and a its radius ; then its rectangular equation is

{x-a)* + (y-/3)a = a2 But x=zr cos 6, y—i' sin d; hence

r9— 2 (a cos 0+/3 sin 0) r + a3+/32— «8=0 the equation required.

6. Let the equations of the circles be

(x - af + (y - /3)2 = «3

(V - a')2 + (2/' - fi'f = a'2 .

Or" -a*)2 + (y"-P"? = a"2 J

The conditions, that the two first, may respectively intersect the first and third, and second and third, are that VK«-*')2 + (ft-ft')2l < a+ a', VK«-«")2 + (A-/3")2| < *+ «", ^|)a'-V)2 + (tl'-fi"yi < «'+ of. To find the points of intersection of the first two, make x == x, y' = y ; then

(a? -ay + (y-fif = d\

.'■i r '

.-. ,*2 + */ - 2 a x + a 2 — 2 /3 ?y + /32 = a*)

.r2 + .v2 - 2 a'.r + «/! - 2 (fy + &* = a'2) ' ' ' ^

P. 65.] IN ANALYTICAL GEOMETRY. 117

.-. 2(a'-a) ^+o3-a'2-2(/8-/3/) y+(P-0* — tf»-a'*,

Hence find jt in terms of ?/, substitute in the first of equa- tions [a) ; thence obtain two values of the form y = B ± B . Similarly, substituting for y in terms of x, from the same equation we get two values of x ■=. A ± A', suppose ; then the co-ordinates of the points of intersection of the two first circles will be found.

Having thus obtained the points of intersection of the circles, we can easily obtain the equations of the straight lines passing through every two of them, and shall thence find they have one point common to them.

Otherwise.

The equation (6), being that of a straight line, can belong to no other than that which passes through the two points of intersection of the first two circles. Similarly, we find that

(a._oK +(/,_,)?y, =.^i«iW) . ,c)

and

are the two equations of the chords common to the first and third, and to the second and third, circles respectively.

Now, for brevity in the elimination, we may simplify these equations, by transferring the origin of co-ordinates to the cen- tre of the first circle ; that is, we may suppose

a = 0, 0 = 0.

They may be further simplified, by making the axis of x coincident with the line joining the centres of the two first

118

HINTS AND ANSWERS

[P. 65.

circles ; that is, by making 0=0; then our equations become

a'x

a2 — a'fi + a

(a* -a') X„ + (3" y„ =

"•2

"2

a'2_ a"2_ „'2± „"2 , fl*2

^.(e).

a'2+ a"3+/3*2

2

From the first of these it appears that the common chord of the two first circles is J_ to the line joining their centres ; the same is true of the other chords. To find the point of intersection of the first two chords, make x — xt y —v; then

x

an

da",

a2 _ an + a'a

a2 _ a>2 + a'2 a2 _ a"2+ a»2 + Q»2

Z a

+ /*V

2

(/)-

When ce

«2 (a' _ a"} + a'2 a» _g"2a' + ^2 + ^ a> _a>2a»

a

2/

nd .r =

2 a' 0'

a — a + a _2~n7~

f-(g)

which are the co-ordinates of the point of intersection of the common chords of the first and second circles, and first and third circles.

Again, to find the intersection of the first and third common chords, we must have x = xa and y = yg. But then

a*

a X =

a2 + a

o

a'2 _ a"2 _ a>2 + a»2 + 0>2

2

(a"-a')X+0'y

Adding the first to the second, we get

a"x+0'y =

a* - a"2 + a"2 + 0'2

and nx

(£l

2~

'2

P. 65.] IN ANALYTICAL GEOMETRY. 119

which, being the equations, as we before resolved, must give the same values of x and y. Whence the common chords all intersect in the same point determined by equations (g).

7. See Hamilton's Conies.

8. If a, b, be the semi-axes, and a, b' , the semi-conjugate diameters, and 0 be the inclination of the latter ; then

(Hamilton's Conies, or Francoeur 's Pure Math.)

a'* + if* = a'2 + b2

a b' sin 6 = a b

}■

Now, when 0 — 90°, a'b' = ab,

and then

ah2 ± 2 a b' + b'1 = a* ± 2 a b f ¥,

or, d + b' = a -\- b 1 v , i ;, ,

' r > , whence a = a, and b = o.

and d — b' — a — 6 '

9. JFood, art. 50S.

10. Wood, art. 510; and Hamilton, p. 191. #4 _ a2 xi _ & xi + ai Q2

11. y =

r

3

a2 62 Let x = 0 : then w = — — ,

r-2

a2 -f 62 / «4 + 64~2 a2 £2 + 4 c3 y

rW

2 V 4

Let y = 0, then

, a2 + 62 a2-62 ;o

J"2 = — - — ± — - — = a2 or 62,

.'. x — ± a or ± b, the curve cuts the axis of y in one point only in the point

a2b2'

( a a*&\

120 HINTS AND ANSWERS, &c. [P. 65.

but it cuts the axis of x in the four points

(y - 0, * == ± a), (y = 0, x - ± b). Maker = ± oo ; then y = ± oo ; .'. the curve has two infinite branches, one on each side of the origin of co- ordinates, but both on the same side of the axis of x.

The greatest negative value of y is — j — —J , and this

/a?±x* gives x = ± Y — 2 — "

The singular points, and other properties of the curve, are found by the Differential Calculus.

ARITHEMETIC AND ALGEBRA.

ST. JOHN'S COLLEGE, 1814.

[P. 65.]

1 . Multiply by the least common multiple of the denomina- tors, viz. 84, and there results

18a; + 42— 8. T + 28+ 168 + 63=21 j?— 84; whence

x — 35.

2. The first reduces to

11 ^+14 y= 127 ... (1) Again, from the second

9xy-l2x-9xy-\-U0_ 151— 16a; 3y-4 ' ~ 4*/-l '

12jr— 110 _ 151 -16 x or 3y-4 ~ 4^-1 '

52x + 13?/ = 494, 4ar + y=38 . .' . . (2)

The equations, when reduced, are

\\x + Uy = 127i Ax + y = 38 J >

which give x = 9, y —. 2.

3. This easily reduces to the quadratic

11 12

122

HINTS AND ANSWERS [P. 65.

11 / 361 X~ T0±V TOO' 11 ±19

10

= 3, or — f .

4. ^-ISy-H)^

5 - 3 y b'

__x2-\5y — 108 - -3 ,

.*. a:a-152/ = |A/(^_i52/_ 14) + 108?

.'. a^-lSy-H-lv'lar2- 15y-24) = 94, and solving this quadratic,

• t(** 1* 1/1N 3 /9409 . . VKX 15?/— 14) = — - + 4/ ,

J ' 10 - V 100

3 97 94 47

= io±lo=10'and-To' or ~ T'

Again

/. ^_ 15 y- 14 = 100, or --,

.-. x*-\5y =114, or ^ (1)

y£±5-!!,2:c y

V \3y+ 4/~8V+"3"+2'

<3y ' 4/ 83/ 3^2'

■ £!.j_^1- a?4 4;p2 y2

"3y + 4 =~ 64 ^+ -9-+ -4

a?3 a:2 2

+ %+^ + 3- *y>

x* x* 23 „ 2 v2

••64^-%+72r f3^+ T=0- The square root of this gives

8y~T~2~ 7 ' •;i;A2) /. 3a;2- 16 .ry _12?/2 = 0. Divide this by ^2 ; then

12^+16^=3,

x2 x

P. 66.] IN ARITHMETIC AND ALGEBRA. 123

■••1+1 = ± "/(g+s)

.•.|=1,or-^....(3)

5

6'

By means of this equation and (1) we easily find all the simultaneous values of x and y ; one pair of which is x~ 12, y=2.

5. Let x be the number of gallons required ; then it is

easily found that the clear profit on the ale is 17/. 10s., or 17^/.

x the dutv on the strong beer is -rp;l., that on the small beer is

— vttt; — /. Hence the profit on the mixture is 30 r-^ — .

160 r 160

.'.by the question,

SO — 5Q + 3:g • 1 7 i • • 1 0 • 7 30 -j^- . 172 . . 10 . 7.

Hence, multiplying means and extremes, 210 350 + 21* - 175

whence

x = 250 gallons.

6. Let x be the number of men at first in the garrison, and y that of the days the provisions lasted ;

then, at the end of the sixth day, there were left #—136 men. Also, as the provisions lasted y — 6 days afterwards, 10 men dying daily, at last there were left only

a:_136-(7/-6) 10, or x— 10 y— 76 men. But the stock of provisions at first being capable of sustain- ing the garrison for 8 days, allowing the loss of 6 daily, it con- sisted of

(2 x—6 x 7) 4 daily rations, and, at the end of six days, there had been consumed

(2 x— 6 x 5) 3 rations. Consequently, by the question,

(2*-6 x 7) 4-(2 x-6 x 5) 3=6 x 61,

124 HINTS AND ANSWERS [P. 66.

or 4 x- 84- 3 x + 45= 183, .*. x =222 men at first. Again, the number of deaths in the first six days=136. Consequently 86 men had to subsist y— 6 days upon 366 daily rations, 10 of them dying daily. Hence

|(2x86 -10 x(y-7)f x t? = 366,

.'. (86-5y + 35) (y— 6) = 366, which quadratic being solved gives

2/ = 12,orV, of which the former only belongs to the problem.

x and y being found, the rest is evident.

7. Let x be the ratio of the quantities of gold and alloy in the guinea. y the ratio of the values of equal weights of gold and

alloy ; then, if w = weight of alloy in the guinea, and v = value in shillings of one oz. of alloy, then

to. v + x. wxv. ?/=21s., and by the question I .

ff- wv + ff wx. vy. fff = 21.?. J Hence, dividing the second by the first 24 / . 249 \

which gives

xy = 239 .... (2)

Again, by the question

#2-f2/2=H (* + 2/) + 233T|T ... (3) From (2) and (3) x and y are easily determined.

i

P. 66.] IN ARITHMETIC AND AJLGEBRA. 125

ST. JOHN'S COLLEGE, 1815.

1. First multiply by 12, and then reduce. Then multiply

by 29, and collecting coefficients of x, &c, it will be found

that

x = 72.

2. One term of the proportion" 'of the second of the simul- taneous equations is wanting. The student may make it what he pleases ; then reduce the first equation by multiplying by 12, &c. &c.

3. First multiply by 10 and reduce ; then multiply by 2x— 5, and a quadratic will be found, whose roots are

5 and U-

4. From the first of these simultaneous equations, multiply- ing both numerator and denominator by

we get

(x + yy + 2(x + y) y/{x*-tf~)±x>-y* _9_

(x + y)*-(x2-y*) ~ Sy ^ + ^'

Hence,

x+S(&-y*)=$(cc+y) .... (1).

Dividing by x, we have

/(

y2\_ 1 9 y ~ x~l)~8+8'x

Hence,

y2 18 i - 63

x~ri+lW-~c~~U5' x + 290 ~ ~ 290 '

i ill ^15_

x ~ 290' 290'

3 21

=3 and~ i (2)-

126 HINTS AND ANSWERS [P. 67.

Again, from the second of these simultaneous equations, we get

(x2 + yf— 2x (>2-f y) = 506 — x + y,

.'. x2 + y—x = ±^(506 + x2 + y—x)

.'. x*+y—x + 506 + x/ (x2+y-x + 506) = 506,

.*. -^(a^ + y— ^ + 506) = ± 23.

Hence 024-y— #=23 • • \ /o\

But y —\x, or — |-g-:r J ^ '

Hence all the simultaneous values of 0 and y may be found,

of which one pair is 0=5, y=3.

5. Let x be the number of soldiers in front,

then 40 x = number of men in the mass, and 4 x = number of spectators. Hence the number of men and spectators = 44 x ; and, by the question,

(0— 100) 45= 44 x,

.'. x — 4500 men.

6. Let 2x be the number of persons, and £y the subscrip- tion of the youngest ; then they subscribed respectively

y,y + 5y+\0 .... y + 5(2x-l). Hence,

\2y+5(2x-l)l 0=345*... (1). v Also

I2y + 5(x-l)^=22x (2). J

( •

From (2)

2 y +5 0—5=44,

.'. 2y + 5x = 49 . . . . (3) . £49— 5 04-100— 5| 0=345, or 44# + 5 02=345, x* + V a? = 69,

a? + ¥ = ± Y* •'• :r=5' .'. 2 0 = 10 and y— 121.

P 67.] IN ARITHMETIC AND ALGEBRA. 127

7. The geese went | miles an hour. Waggon £ miles an hour.

From the question, B set oft* 5 -f- f = ^ hours after A.

Let # be the rate at which A and B travelled.

.*. when A was with the geese, the waggon was 2 #-+-§ miles from them, or from the 50th mile-stone. But, when A was at the 50th mile-stone, B was x. lT° miles from the 50th mile- stone.

Now, the distance of waggon from where B meets it, when

A was at the 50th mile-stone, is

2 x + f - (50-31) + f x = \ x - V9 miles,

and the distance of B from that place is

10 x

__ + (50 - 31) - 1 x = | x + 19.

and their rates are as f , and x,

. 8* 2 2 9

• • o -^ X ■ — D X -J- ^- ,

or, x2 -'TV#= V-s3-

„ 300.606 . _col ..

rience, a? = — — — nearly = 9.5SI miles an hour nearly.

.'. hence the answer is

lT° X 9.581, or 31.93 miles from London.

ST. JOHN'S COLLEGE, 1816.

1. Multiply by 6; reduce; multiply by 17, &c.

Answer, x = 51.

2. Multiply by 21, and then by 20 ; then the first becomes

515 x - 496 y = - 2634 (1)

Componendo,

30 : y-2x + 15 :: $ :| - | + T'T,

:: 10 : 3y — 4# + 1.

128 HINTS AND ANSWERS [P. 68-

Hence,

5 x — 4 y = — 6 ^ But 515 x ■ - 496 y = - 2634 J ; Hence, x ==? 18j

y = 24.

3. First,

2 x* — 6 x + 2 x2 — 8 x — 5 x + 20 _ 25 ra _ i'x _|_ 12 3 *

Hence, x2 — yT8 * = — W>

• T 5£ — _)_ / ,361 — -+- 2.1

• • «* 13 — Vl32 — 13)

.'. x = 6, and ff .

4. Clearing the first of fractions,

?/2 — 9 x\f x — 81 = 2x\/x. y -f 9 xa/x, .'. y2 — 2 x\fx. y + .r3 = 81 -f 18 a? ^/# f #3, .". 3/ — aV^ = ± (9 -+- xa/x), .'. y = 9 + 2x\/x, and — 9 (1).

Again, clearing the second of fractions, y -f 3 .ry^- = 9 + y x*/x, or, ?/ (1 — x*/x) = 9 — S^-v/a". • • (2).

Hence the simultaneous values of x and y are a? = 0, y — 9 ;

x = l/hy = -9-

5. Let ar be the rate of sailing with a fair wind,

y the distance between Dover and Calais.

Then, by the question,

2x = y (1).

Again,

time through first half = ^ -+■ x = ^-,

time last half = 2x + (* + 2) = 27^2)'

.'. whole time, the wind changing, ■= Ji- + ->, _N.

0 & 2x 2(-r + 2)

P. 68.] IN ARITHMETIC AND ALGEBRA. 120

Also, time, wind not changing, = - ;

x

.'. by the question,

y . y_ + y . . 7 . 6

.r * 2a; 2(.r + 2) * " '

• 52 = I' I? 4. 7 _X_ /9\

' ' x 2x2 x±2 { '

But y = 2 x Hence, x = 5, and y = 10.

6. Let S be the junction of the sewer and river, and C that of the streets,

D the point in CA where the drain is cut. Also, let CA = x, CB = y, CS = z ; then SA = 6, SB = z- 11, and, from the a s formed by these lines, we have

z — 11 _ sin (y, z) _ 6

y sin (6, z) x' '

z_ sinSDC =6_3

r-4 ' sin DSC " 4~2' ^'"

Again, the cost of the sewer is zx.

cost of drain down both streets = 9 (%+y), .'. by the question,

9 (.r + y) = zx+54 (3).

The equations (1), (2), (3), reduce to x z — 11 x = 6 y 2z - 3x = — 12 9 (x-\-y) — zx — 54

Add third to the first, and

9 (x -f y) — 1 1 .r — 6 y — 54, or 3 y — 2 a? = 54 (4).

x\lso, from first and second,

3 #2 — 34 a;— 12 y = 0 (5)

.-. x* _ 14 ^ = 72,

which gives a? = 18, and — 4.

s

130 HINTS AND ANSWERS [P. 68.

But, as the street leads to the river, x must be positive,

.*. x — 18 chains (6).

From (4),

y = 12 +. 18 =30 chains;

and, from (2),

z — 4. (*— 4) = 21 chains, whence, SB = 2 — 1 1 . = 10 chains.

7. Let a? = the number of the family,

y = the sum in pence saved by the man a month, z = the common difference of their monthly savings, and w = the price of wheat per bushel, then y, y—z, y-2 z, .... y—(X—l) z are the respective savings.

Now the seventh child saved y— (9— 1) z, or y— 8z pence, and the eldest saved y— (3— 1) z, or y— 2 z pence, and the 5th saved y— (7— \)z, or y— 6.? pence, and.'.y + y-z + y-2z.. + y-{x-l)z=i. (y-8z)tv-39

or J2y-(.r-l) «| | = f (y-8s)w?-39 (1).

A1 2y-8a+120

Also id = -JL .-L , or y - 4 a + 60 . . (2) .

Thirdly, the two youngest dying, there are left (x—2) in the family,

.-. [y-2z- 12)(*-2) = \2y-{x~l)z\ %- 5 x 2] _x 12(3) it being assumed that the months are calendar,

tm&(y-2z-2) (x-2)={2y-[x-l) z\^ + 25... (4)

these four equations enable us to find x, y, z, iv. Subtracting the (3) from (4), we get 10 0-2) = 130, .'. x = 15.

P 69.] IN ARITHMETIC AND ALGEBRA. 131

Hence, by substitution, we get

(2y-14s)¥ =£(y-8s)w-39 w=y—4:z + 60 I

and(y-2z-2) x lS = (2y-Uz) x V5+25 J 2/2_30y_12^ + 32z2+1502r-234=0 . and2y-f792 + 51=0 /'

from which, first eliminating y, z and y may be found, and

thence w may be found from equation (2).

ST. JOHN'S COLLEGE, 1817.

1. Multiply by 10. Ans. x = 3.

2. From the first \/{y—x) + V (a— x)= Vy- Hence the second becomes

Vy ' V(a—x) :: 5 : 2,

••• y = V («-*)■

Hence, by substitution,

4a 5

y - «

5 4

3. Clearing it of fractions, we get

= X ■ - i s/X : ~- f,

(2b 24\ 5±7

//25 24\ 5±7

V = * ± v< (se+seh— = 2> and ~ *'

.'. a; = 4, and |. 4. From the second equation,

(£)' -»-($=->•

which gives

2/2 = 2^ ± 2^(^-1),

or ^ - 2 A/(.«2-l) = 2.^-4 ^(V-l)/ ' ' ' ^ }

132 HINTS AND ANSWERS [P. 69.

Hence the first equation gives

y-2</(x+l) y-y^-l) _ 2~c + 2~v "" U?

.-. y 2V(x'+l) + 2/-^-l) = 0, /. 2y -= 2^/(^+1)+ </ (x-1) . . . (Z).

Again,

4^2 = 5 a; + 3 + 4 -/(a?2— 1)\ But 4?/2 = 8a; + 8 -v/(>2-l) J'

.'. 3x + 4 Viz2— 1) = 3, 3(ar— 1) + 4A/(*"2-l) = 0, /. 9#2-18# + 9 = 16 a:2- 16, /. 7 a:2 + 18 a; = 25,

r'2 _1_ * 8 <r — 2 5

a, -f- y ^ y ,

9 / 781 175 \ / 256 16

••• *+ 7= * V U+-49-) = ± V -49" = ± T'

.'. x — 1, and — y\ Hence, y = ^/2, and \/ — y + \/ — T, = a/2, and ^/ - V8 +i \/ - ¥• Two other pairs of simultaneous values of x and y may be found by using the second of equations (1).

5. Let x be the number of bushels,

and y the shillings per bushel he gave for it. Then xy is the cost.

By the question, (the interest for 6 months being -Sf-)

x (y-l) = xy + ^ - 5y,

or xy + 40 a-- 200 y = 0 . . . (1). Again, his expected gain was 3 x,

the interest for the year was -J?s.

.".by the question,

■r(y~2) = -ry + |(-(3*-10),-

ot20x—$y = 200 ... . (2).

P. 69.] IN ARITHMETIC AND ALGKBKA. 133

20 #-200

Hence y =

x

.'. 20 #-200 + 40 #-200. 2° x 2°° ^ 0,

#

2000 or #-10 + 2 #-10. 20 + — = 0,

3#2-210# = - 2000,

. Wrt 2000 6000 .'. #a— 70#= — = g-,

x _ 35 = ± y/ (1225 - ™*) = ± A ^5025,

= ±^8 = ±23.62.

.-. # = 58. 62 bushels, or 11. 38 bushels, 20 #-200 20 (#-10)

V =

x x

20 x 48. 62 486. 2 . _ _ , 57

lbs. 7a.

58. 62 " 29.31 ~~ 977

-10) 10 0 . , 59 5.69 = &- 5d- W

(11.38-10) 10 0 _, 59 Also, y = K- n-aQ = 2s. 5d

6. Let # be the days of the expected passage, y the number of men alive ; then the number of deaths = (# + 7 x 3—30) x 3,

= 3 # - 27, .•-. y = 175 -(3 #-27),

or 3# + y = 202 (1).

Again, if to be the water allowed each man per diem,

175 x x x w ■=. whole store of water. Also,

175 x 30 x w — water drunk during the first 30 days,

and to £172 + 169 + 166 (# + 21-30) terms^ = water

drunk during the rest of the voyage ; which, being an arithmetic- series whose common difference is — 3, &c. is

w £2xl72-(#-10)3^ ^-;

134 HINTS AND ANSWERS [P. 70.

.*. by the question,

iv (374 - 3 x) ^- + 175 X30w= 175 x wy

or (374 -3 a;) (ar-9)+ 10500 = 350 a;.. .(2). This equation reduced, becomes

/. a? - 17 x = 2378, whence a; = 58, and — 41, the former of which only applies to the question. Hence y = 202 - 3 x = 202 - 174 = 28.

7. Let x be the gallons at first in the hold, and y the gallons per hour of influx,

z the gallons pumped out by A in an hour ; then | 2 x | or | 2 B

Now, if t be the time of A emptying the hold ; then ty = influx during that time,

j ty + x

and . . — = t.

z

X

:. t.—

z-y . water thrown out by B = xf «,

by A- (l3 *---£_) \ z—yj

that b 1

and the whole water pumped was x + 13 $ y,

.'. by the question,

x z / x \

z — y \ z—y) Again,

(« + 4 *) x (3 + U) = « + (3 + *) y (2) j

also,

■y

(3 + |) = (V__£_JS+100 (3)l

P. 70.] IN ARITHMETIC AND ALGEBRA.

These equations reduced, become

135

■LP

3

V «

#;

z-y x + V 2/

a- -f

y

I" 5 ~

12 *

100,

2 -2/

80 (z—y)2— 6 a? (z—y) — ^ z = 0

30 («-y) + 25 z-8* = 0

12 a z - (1 15 z 4- 1200) (z-y) = 0

To simplify these, make z — y = u; then 80 w2 — 6 .r w — .r 2 = 0 30 u + 25 z - 8 x — 0 12 28— (115*+1200)«=:0,

From the second,

8 .r — 25 2

I

J

?£

30

Hence, substituting in the first,

± (8z-25zT-8x2-*5xz-xz = 0}

which is reducible to

46 x* — 355 x z + 625 s2 = 0

Again, substituting for u in the third, 23 s + 240

(4)

12

x .

6

(8 x -25%) = 0,

575 s2 — 112 x z + 6000 s - 1920 x = 0 From (4), dividing by .r2, we have

625 JL _ 355* + 46 = 0, xa x

(5)

2 X

71

125 V.r

46 625' 71_2 250*

21 92 , 50 = .and —

250 250' 250

71 / /712 46 \

250~= ± V V2502 ~ 625/

_46

6

46

1

, and „ 125' 5

x

125 46

:, or 5 ~

21 250

(C)

136 HINTS AND ANSWERS [P. 70.

Substituting the latter value in (5),

575 s2 - 112 x 5 s2 + 6000 z - 1920 x52 = 0, /. 115 s2- 112 s2 + 1200 z - 1920 2=0, or 3 s2 "= 720 s, s2 = 240 s, ,\ s = 0, or 240, of which the latter, viz. 240 gallons, alone answers the question.

Hence, x = 5 % = 1200 gallons, and •.• V 2 ~ x + V y> ;. V« = 5g + V5 2/> /. 11 s = Ss + 6y, /. 3z = 6 y,

and y = ^ = 120 gallons

that is, x, y and z are respectively

1200, 120, 240 gallons. Again, if we use x — VV5 z5 substituting in (5), we get

x, — - U, 01 — g^j- ,

neither of which answer the conditions of the question.

ST. JOHN'S COLLEGE, 1818.

1 . First multiply by 24 ; collect like terms of the result ; then multiply by 7, &c. Ans. x = 7.

2. Multiply the first by 40 ; collect like terms of the result ; then multiply by 21, and collect like terms.

Again, multiply the second by 18 ; collect like terms of the result. Then eliminate x from the two new equations, and the resulting equation in ?/, will give

29839S y ~ ~ "39757 ' whence x is easily found.

P. 70.] IN ARITHMETIC AND ALGEBRA. 137

3. This equation is

^ -f 41 af T = 97af"* + xT,

Z_ — 2 5.

.-. a?3 — 56 x T -|- .r6, Multiplying by a;3, we get,

#3 _ X2 — 56,

and solving the quadratic,

x* - \ = ± ^/2|5 = ± V,

.-. x* — 8, or — 7, .*. x = 4, or >4/ 49.

272 f

4. From the first,

sy_80 ?/2-272 = - 7/2y/ (a

. 272\ // , 272\ Qn

and solving the quadratic, \/ {f~ ^Pj + i = ± /(j +' 80) = ±i V 321,

a? -

272 , , ,_ 32J_

4 '

. (1).

y2

161 + y' 321

Again, from the second,

36 Q_ #

#2 5 = 35 -,

y2 y

.'. x2y2 — 35xy = 36, and solving the quadratic,

35 /352 + 144 35 ±37 QA , f9.

xy= 2- ± y' -J = — 2 — = 36, or — 1 . . (2).

But from (1),

*f - 272 = 161 -/321 ,», making xy = 36,

138 HINTS AND ANSWERS [P. 70.

V* - 1024 x ? = ^1 161 ± V32\

y _ lv*<* x 161 _ ^321 64()0 2

?/ = ± i 1/(322 ± 2 /321),

/322 - 32C

— b

f /322 -f 320 /322 - 320i

V— T ±V 2—

= ±\ (a/321 ± 1),

36 36 x 5 9 .__,, 1(3).

• * = 7 - ± V321 ± 1 = ±I6" (^321 + 1]J Again, making .»z/ = — 1, and substituting in (1), we get

1 - 272 = ffl V 321 tf,

■2 271 x 2

• ■ ^ — 161 t -v/321' " '

271 161 ± V321

n x 0 xi,

~ 6400" 2

= ±J^w(v/321 ± L> v~l>

_l _ 160 1

and # = = h-

y ^/271 (v/321 ± 1) / — 1

160 (y^321 T 1) y/-\ ^271 * 320 x — 1

(^321 + 1). v^-1-

2^/271

5. Let # be the age of A, January 1, 1799; then the dona- tions amounted to

x + (x -f 1) + (# + 2) ... to 8 terms — (2 x -f 7) x 4 groats, and by the question,

(2 ar + 7) 4 = 7 x 20 x 3 + 18 x 3 f 2 = 476,

which gives x = 56,

.'. the year in which he was born = 1799 — 56 = 1743,

and A's age at his death = 56 J- 7 = 63.

6. Let # be the price of a share in the canal speculation. The profits of the canal speculation = 5x + 5951. ,

P. 71. "I IN ARITHMETIC AND ALGEBRA. 139

.," • r. , 5x + 595 #-+119, .". the gain or each = — =-= = — - — /.

.'. each ventured on the steam boats — - 173= — - — l-

o o

Hence the total advance = — ^ /.

u a ♦ * i i 8^-3200 , 8x+952

Hence the total loss = ^ + ]- + 368,

16 a?— 1144 3

.'. A's loss = - — k = 419 by the question,

/. x = 7001., .'. the price of a share in the steam boat speculation is x— 400

3

100/.

7. Let x be the width of each of the two large warehouse? built by A and B,

y small

The contents of A's warehouses are x3 and y3. Hence, by the question,

x3+y*-xy2-yx* = 73728 (1).

Again,

x1 — y'2 — base of C's, and, by the question,

x* -y2 = 2688 - 8 ^(x* - y2). . . . (2).

From the second, solved as a quadratic, s/itf-y1) = -4 ±52,

= 48, and — 56, .'. x2— y* = 482, and 562 = 2304 and 3136. . (3).

From the equation (1), we have

x3-\-y3—xy (x + y) = 73728, or (a2— xy+y*) (x+y)—xy O+y) = 73728, .'. {x-yf (ar+y) = 73728, that is, (a;2-y2) (*— y) = 73728 (4).

140 HINTS AND ANSWERS [P. 71.

Dividing this by the equation (3),

73728 , 73728 X~y = 230T and J3T31P

= 32 and -^ (5).

Dividing equation (3) by equation (5),

2304 . 3136 x 49

. *+*=-&-"* -rraar-

*o i 492 to . 2401

= 72 and y^ = 72 and -— -.

• , 240 K . . x+y — 72and-^-j

^ 49 J

.-. ^ = 52, y = 20, . 138385 906913

als°' X = -1764-' y = -176^' Also, the width of C's = ^{x2 — y2) =c 48 in the one ease, and is easily also found for the other values of x, y.

ST. JOHN'S COLLEGE, 1823.

1. Clear it of fractions, by multiplying it by 12, &c.

Ans. x = 6.

2. Eliminate z by means of the first and second ; then eli- minate z from the second and third. From the two new equa- tions eliminate y, and there will result an equation in x only. This gives * — 1 ; whence y = 2, and 2 = 4.

Otherwise.

Multiply the first by <j>, and the second by § , and add the third to the results; the sum is

(> + 24>' f 3) ,r+(20 + 3</>'+l) y+m +*'+2)s = 170

f 12 </>' + 13.

P. 71.] IN ARITHMETIC AND ALGEBRA. 141

To eliminate y and z, assume

2 <p + 3 41. + 1 = 0 1

3 y + f + 2 = CM ' Hence, <p = — 4>

and 0' = f. Consequently, (— f + |- + 3) x = — V + y + : 13,

• • f & — T y * + %C — 1 .

Similarly, y and z may be found, independently of the value of x ; or, if two of the equations be reduced, they may thence be found.

3. Divide the equation by xm, and there results

m - n 2l m - n

a?b2 x~>^ — 4 (abyxw'i — (a— b)2, m~n 4 m - n (a—bf

.'. X mn y— j — X2mn =

^(ab) q?V*

Solving this as a quadratic,

?i-M 2 / r 4 (a— 6)21

\/{ab) -V \ab+ a2b2 / '

2 a + 6_2-v/(a6)±(«-f b)

*/(ab) ~ ab ab

la + b + 2^/(ab)l ab

= ±(v«-v^)2

a6

-f win —

a2b2 '

4. From the first equation,

x4— 2x2y2 + yA — 1+2 xy + x2y2>

.-. x2-y2 = ± (a;y+l) (!)•

Then, taking the positive sign, we have

x2-\^y2 + xy, \ (2)

or, (#-1)0+1) =*y(x+-yy

142 HINTS AND ANSWERS [P. 72.

Again, from the second of the proposed equations,

(2t/2+1) 0+1) = x3 + y:i = (x2-x y + 7/2) (#+y), .'. dividing this by (2), we have

2;y2+l _ x2— xy + y* x—l y

But from equations (2),

x2—xy = l+y\ .'. equation (3) becomes

2y2+l = 2^+1 x — l y

Whence 2t/2+1=0,

•'• y= ± V — i-

Substituting for y in (2), we get

•r2 + #-/ — | = i- Whence,

# = ±iAAf±±^-±=±K^f+A/-i)l (5)

and y = ± ^ - |, /

which, taking upper signs together and lower signs together, give two pairs of simultaneous imaginary values of x and y. Others mav also be found.

(4).

5. Let x be the miles an hour he walked backwards ;

then 4 x are forwards.

Let, also, d be the given distance, and t the given time ; then, by the question, d 3d . d , s 3d ^. rtv

4-^ + T^==4"^(*'~^+T"^+)'

.-.i + * = -L- + 3

a; 4# ^ — i 4 r + 2'

_1 = _JL 3

4x 5x — 1 4a: +2'

which being cleared of fractions gives x = 1 .

6. Let x£ be the debt,

y the price of a bond at the beginning of the year ; then, since there must have been as many bonds taken as hun-

P. 72.] IN ARITHMETIC AND ALGEBRA. 143

dreds in the loan, to produce the same interest, the value of* those bonds was at first

X 100 X y'
id, by the question,
X f 40y\ _ xy 100 V 100/ 100 uu	-250,
• ' 500 ** = 650'
xy = 325 x 500 = 162500 .	. (1).
Again, by the question,
100 x30- 100 300-250-
or 2 - m ~ 550	(2).

But,	by (]),	100	= 1625,
		X "2	= 1625 -	- 550,
		.". X	= 2150/.	the debt.
Its interest =	2"0 X	2150 =	107/. 10*.
Hen	^e. n =	162500 16250 3250

' ;/ 2150 215 43 '

.'. the bonds were at 75 ff at the beginning of the year, and at the end, 75 1$ — f x 75 f| = f x 75 $$ = 45 ±f

7. Let # be the number of prisoners at first, and ?/ shillings what he expected for his mill.

Hence, the expense of each man's provisions is « + (a + ar) -f (a + ar -f ar2) -f- &c. to ?^ terms = A suppose. Again, that of the fresh convicts is c (a+ ar) -f 2 c (a + ar + ar2) + . . . to (w — 1 ) terms = B suppose. Now, the estimated expense was n x x «, the estimated labor was nx x p a,

.'. 7i a (p—Y) x = expected gain.

144

HINTS AND ANSWERS [P. 72.

Hence, by the question,

nxx pa + c x pa + 2c x pa + . . . [n— I) terms — x A — B

— na (p — 1) x,

_ B— ffac^l-t-2-j- . . .{n — 1) terms ^

"*.~ na-A ■ ' * (1'*

B\\tA=na + ar + (ar + ar2) + ar-\-ar2 + ar:i + ..ar + ar'i + ..arn~1

A — «a •*. ■ — =« + (a + «^)-l-(« + «?'-f ar2) + ...(a + ar -\- ...arn-*)

= A— (a+ar + .. .ar"-1), 1

= A-a - r

.'. A—na = r A — ar

-r

r»— 1

«a — a?-.

r-1'

r" — 1

• A- "-1

■'— A-^-fe--) «■

Also, — r= (l+r) + (l+r + r2)-f ....(l+r + r2 + ....rn_1),

= w_I+7.|l4-(l+r)+(l+r+ra) + ..(l4-r+...r»-2)^

= »— 1 +r + r < — — (l+r+ »,B_1) jt

rB rM— li

lac r— 1 J

r"~ 1 B^n-l+r-r. ;— J-

' ac 1 — r

1 — r»-i w — 1 — r2.

*. B = acx r (3).

1 — r

n2

Also, 1+2+ . . . [n—l) terms = — . Consequently, substituting in (1), we finally get n

.„ — 1 — r2. — £-—. (1— r)

c 1— /• 2 y

.r = -X

r I —

n

P. 72.] IN ARITHMETIC AND ALGEBRA. 145

ST. JOHN'S COLLEGE, 1824.

1. Multiply by 21, and reduce. Then multiply by 100.

Ans. x = 8.

« ™ , 1 1 1

2. Make - =u,-— v.- = w,

x y z

then the equations are 3u—±v+ w — ^\ i.u + ±v + 2w = V ► .

From these it is easily found, that

u = 2, v = 3, ^' == 4,

3. First multiply by 50 ; then

1350-75 -y/ x -50 x . 148\Ar— IPs tfx

•Var + 8 "~ + 4^-7

1342—80 -v/.r— 50 ar _ 592^/^—40^^^ •"'"" 5^ + 8 4#-7

61\ — 40</x—25x 296^/3?— 20 ay j;

5^x+8 "4^—7

2088 , 4697

This gives x - j^g Vr = |379'

a. . ,. , 1044±v/7567099 solving this, y/ x = fs7Q ~" '

/1044 ± v/7567099V3 '• r" V 1379 J '

which is easily reduced to decimals.

4. From the first,

(^vi)--(^ivi)=i-vi

146 HINTS AND ANSWERS [P. 73.

.". completing the square, and extracting the root,

■•V|-1/| = -l/5 !••■<"

Each of these equations is simultaneous with the second of the proposed equations, viz.

9l/" + 3 Z'-7 = (21 ^2..- 1) x V^ + o4-J3) v y V x * 2v x^2sf{xyy '

Multiplying (1) and (3) by \/{xy), we get ys/x

x — y

\

/2

and9, + 3y = (2\^2x-\).%t + I

2 2>

■ v = *^2 ' " \/2 + *Jx

Also, 18a? + 6.y = 21 y A/2,— y + 1,

.-. (7y—2ty2x)y = 1-18,,

.:^r^(e-2W2x) = 1-18,,

.'. 7ar>/2-42arvr« = >y/2- 18^/2. ,= ^,-18 .?V,, 25,^/2— 24,,/, = -v/-*+<\/2,

.-. sfx (24,-^/2,-1) = 24,^2-2^,-^2,

= >/2 (24,-^/2,-1), .*. U/,-A/2) (24,-v/2,-l) = 0, .*. x = 2, and 24,-^/2,-1 = 0, 1 1

,

T7y V*

12^/2' v 24'

. solving the quadratic,

P 73.] IN ARITHMETIC AND ALGEBRA.

V* = 24T72 ^ W2+2l)

147

V 1±7

24a/2 ~3a/2 * = W and 4

== ^- and -

4^2*

Hence, wheha?= 2, y

2a/2

1,

V2+/2

when.r = — ,?/= :_(a/2 + - — -).

18 ,y 9 a/ 2 * V 3a/ 2/

9a/: when# = _ ,y

— ,/9

V2 +

3a/

I

1 3 a/2 J_ 9a/2X 7 ~21'

1 4 a/2 J_

4/27 "16a/2X 9 ~36'

32^ 16a/2 which are three pairs of simultaneous values of x and y Again, take also the simultaneous equations

Multiplying by \/{xy), we get

x—y = x y/(xy) —y\/% (4)

and (7 - 21 a/2:*;) y = I — 18 # J (5)

/. 7 (1-3 a/2:*;) */ = (1-3 a/2*-) (1 + 3 a/2 x),

.". 1—3 a/2 a; = 0, or x = j^. Substituting this value for x in equation (4), we have

3

i-i8y = 4i/|-3^

15^+3^2-^/^1' •'•7/ + 4572- Vy = T5>

148

HINTS AND ANSWERS

[P. 73.

y + 45^2- v^	\$0y/2) '" 15	1 90* x 2'
	1081
	902 x 2*
■*• Vy =	-^^1081 90 ^/2 '
.-. i/ =	(l+^v/1081)2 16200
From the equations
7y =	= 1+3 ^2. x -i
\/*y-^/l	" a;	>,

other simultaneous values of x and z/ may be found, very nearly as above.

5. Let a; be the number of bags intended to be purchased.

Then, - x 8 £ = the sum transmitted. 5

and, by the question,

8x

-v- -r (a?— 18) = cost of each bag. o

Also,

(3 -r - 5 1) x I x J^re = (g*-5D I+104

S— -91) «rw=(l— ffl)8+soT'

(4 ^ - 63)

8a;

(4 a;— 63) 8 + 621,

a;- 18

.-. (4ar-63) 8x18 = 621 a?- 18 x 621, .-. (4 a;— 63) 16 = 69 .r- 1242, 5x ■ = -.63 x 16+1242 = 234, .*, x = 46 f bags.

6. Let x be miles an hour at which A travels.

y B . . . .

and z the miles C is distant from D.

P. 73.]

IN ARITHMETIC AND ALGEBRA.

149

Then, the time B travels, before they meet the first time, is

20

V

hours

the time of A reaching the place of rencontre, is

2-20

x

and, by the question,

z - 20 20 „ = — + 3-

y

X

(1).

Asrain, A reaches D in - hours, and B reaches C in - hours ; ° x y

'. by the question,

z

. = £+3-1 = - +2 x y y

(2).

Lastly, whilst A was performing \ z miles, B had time to go | + 28 + 3 y miles ;

.*. since distance oc rate,

| z : |s + 28 + 3^ :: x : y, :. \y% — \ xz +2% x ±?> xy .... The equations (1), (2), (3), are reducible to yz — 20 y — 20x + Sxy yz— xz = 2 xy 6 yz— xz = 196^ + 21 xy

,-■

(3).

(4). (5). (6).

From (4) % = 20- + 3^-l-20; substituting this in (5) and(6),

y

(y - X) (20- f 3 jf+20^ = 2 xy 1 (7).

and (6y-#) ^20* +3a- + 2o) = l'96ay+*l*yj (8). Multiplying out, these become, when reduced,

.ry + 20^-20 ~-3x°- = 0

t7

(9).

and 3ry- 120^ + 20 — + 3 x* -f 96 .r - 0

150

HINTS AND ANSWERS [P. 73.

Adding (9) and (10) together, we get

4 xy -100 y + 96 .r = 0, or xy— 25y + 24x = 0,

25?/

.'. .r = * (11)

24+y I11'

which, being substituted in (9), gives

25 y' . 20^ 2Q*25V 3x25V 24+y + * y(24 + S/)2 (24 + j/)2-U'

• fy 4 4*252 3X125y

••24 + 2,^ (24+y)2 (24+2/)2 ~ u>

.;. (9*7 + 96) (24+?/) = 2500 + 375?/,

.'. 216?/ + 2304 + 9 2/2 + 962/ = 2500 + 375 y,

9^-63* =§gj°} = 196,

a 7 196

7 ^ /l"49 _L_ 196\ 7

36 28 g

Hence, by equation (11),

Also, « = 20x-+3ar+2O,

2/

= 20 x | + 21 + 20 = 56 miles.

7. Let x be the pounds the first creditor can claim, y the common difference, and z the number of creditors ; then the whole debt is ( Wood, art. 212),

\2x + {z-\)ylZl.

for which the effects being y shillings in the pound, we have

P. 74.] IN ARITHMETIC AND ALGEBRA. 151

Again, the last creditor would receive \x -\-(z — 1 ) y I —d.

Hence, by the question, £r + (*-l)yi^j = h (x + r.+y) - x+ |. . . . (1).

Again, the last creditor having failed of his claim,

I2x+(z-2)y}*^

is now the total debt; v debt oc tt—^, — %,

dividend

/. |2*+(*-i)Sfj||:|2* + (.-a)y|?=i :: I : *

or, 2%z+z (z—\)y : 2xz— 2x-\-{z— 1) (%—z)y : : 3y-f 8 : 3y,

Hence,

3 (?—x) y2+ 6 *y = 3 (s2-3 as+2) ?/2 + 16 xz- 16 *

+8 (**-3;*+2) y,

.*. (6z-6)y* + 6xy-\6xz+l6x-8(z2-3z + 2)y = 0,

or, 3(z—l)y*+3xy— 8xz + 8 #-4(z2-3s + 2) y = 0.(2):

Lastly,

$* + (— l)y|£ : (* + y) S^l :: 45 : 28,

.-. 28y |ar+(«- l)y | = 15 (x+y) (3y+8) (3).

Divide (3) by (1), and the result is

560 = 3Q0*+y(3y+8)

2x-\-y

.'. 112x+56y = 9.o/ + 24.c4-9?/2 + 24y, .'. 9y2— 32y + 9.ry— 88* = 0, and hence,

__ 9y* - 32y

88 - 9y (4)-

Again, eliminate z from (1) and (2), thence obtaining an equation in x and y. Substitute in this the value of x from (4), and the resulting equation in y will give y = 8. Whence by (4) it will be found that x = 20 ; and then by (3) we get

% = 6.

152

HINTS AND ANSWERS

[P. 74.

ST. JOHN'S COLLEGE, 1825.

1. Multiply by 36; then

_, 468.r-72 8 # + 34— -f^— on + 12 #

17 #-32 whence a:

= 21 #—# — 16,

4.

2. From the first (#-f,v)

i * + ?/ xy

_8_ 63 _4_ 63'

from the second (x—y)

*' • xy

.'. by division,

x+yy = 2

.'. # + ?/ = 8*. (x—y),

whence. = J!±I . y

8*— 1

(1)

and substituting in either of the proposed equations for x, the resulting equation in y will give y ; and x will thence be ob- tained by means of equation (1).

Another Method.

Since the equations are homogeneous, make y = ux ; then the equations become

A

and

(# + m#)3 _ _8_] . (1 + m) #2 w 63

(# — W #) 3

M#

_8_ 63

#w

4

63

{i-uy

ux

63

^v

&c. as is obvious.

P. 74.] IN ARITHMETIC AND ALGEBRA. 153

p + q

3. Divide by x2Pq, and there results

1 a2-b2 2pi *p*\ rt

Let ~ — - = m ; then

2p?

•' + x - * 5C^

«24-62 <r.2>ra O '*•'"* — 1

(l2—b2

.'. solving the quadratic,

T ~ a^ZTp ± y (a2-62)2

_a?+ b2 ±2ab _ (a ± b)2 — a2_.fr a2—b2 '

a + b , a — b

= r and r,

a—b a-\-b

2pq 2pq

a—b) \a + b.

x also s= 0.

4. From the first,

3*+y-2 = *v(^— .3^ + 8.).

Squaring and reducing, we get

y2 + (2 #a + 6 #_4) y = ^— «2 + 12 x - 4-

Completing the square of the left-hand member we get, after

deduction of the right,

9 y2+(2x2+ 6x-4)y + (#2+3 x -2)2= jX*+6x*+4x2.

Extracting the root

y + x2 + 3x-2= ±{ox2 +2x)>

.'.y=±x2-x+2, (1).

andy= -%x2-5x + 2/ (2).

154

HINTS AND ANSWERS

[P. 74.

Again, from the second of the proposed equations, we have

f2LL2,+3 = (^-|y(2+y),

... x + y-4x2 + 6 a- = (*** — 3y)V [x + y), ... ^ + y_(4^_3y) ^'(ar+y) = 4a:2-6;r, But f a;2 - 3 -/ y = f x2 - f a:2 + 3 a: -6, = 3 x — 6, .-. x+ y-(3x+6)^ (x+?y) = 4x2-6x, ... ^ (ar + y) = 4 x - 3 ± ^ (V5 & - 19 a; + 9), = *ar-3±(4*-3), = 4 a; — 6 and — jr, ... r +y = 16 a;2 - 48 a: + 36 and a-2, ,\y = 16a-2- 49 a: + 3fri . . . also y = a;2 — x J . . .

From (1) and (3),

V x2 - 48 a- + 34 = 0,

48 /482 68\

•■•* ^^{±s/\^~^ir

48 ±14 n ,34 = -3-— =2 and ^

Hence y = % x? — x + 2, — 2 and ullP .*. two pairs of simultaneous values of a: and y are

„ O « Q/l

(3). (4).

a' = 2i , 34 = 2} anda^

y= 1446 961

(5).

Again, from ( 1) and (4), we get

x2 — x = \ x2 — x + 2,

and a; = ± 2, .-. y = 4 - 2 = 2 and = 4 + 2 = 6, .*. two more pairs of simultaneous values are

*=.2\*r= -2\ . . . . (6).

Again, from equations (2) and (3),

- 4- a-2 - 5x + 2 = 16 x2 - 49 x + 36,

P. 74.] IN ARITHMETIC AND ALGEBRA. 155

V x2 — 44 x = — 34,

	• * ** 2 7 ■*• "" 2"T5
	44 _ A /44'2 - 68 x 27 '• x 27- V 272
	44 ± 10 0 ^34 _ 2? _ ^ and 27,
	.-. ?/ = — # x°- — 5x + 2,
	= _10— 10+2= — 18,
and=-	5 342 34 0 5022 558
2 272 27 ' ~ ~ 27 ~~ 3

- 186.

Whence two more pairs of simultaneous values of x and y, are x = 2 l _34 '-j

y = _ is / and * ~27 [ (7).

y=- 186 ^ Two other sets of simultaneous values may be found, in like manner, from (2) and (4). See Hind's Algebra, Second Edition, p. 502, 56.

5. Let x£. be his annual expenditure,

y£. be the yearly value of his produce ; then, by the equation,

y-x- |=30 (1).

In the next year, the rent — 50 — f x 50 = 40,

tclXcS - -T^Tj

produce = y +^ J = -J', /.by the question, % - *- -4= 40 + 20 -f 5 = 65 . . . . (2).

These first become

Gy - 1 x = 180i . . . . (4). \6y - 13a; = 780 J .... (5). Multiply first by 8, and second by 3, and subtract ; then 17 x = 900,

• ■ 00 ^^ Y"7" "* ~~ *-''w T*7" J

/. y - lff° + 30 = 91 ff/.,

156 HINTS AND ANSWERS [P. 74.

6. Let B ride x miles an hour,

and 2 z be the miles distant from London to Newmarket.

Hence, if t be the hours before A and B meet, 10* + xt = z, . . . . (1). also C overtakes B in t + 4 hours, .-. y (t + 4) = z + x (t + 4) (2).

Again, if t' be the hours before A and B meet, on their return,

10^ + xt' = 5z (3).

and 10 (Y + 1) + y {t + 1) - 6 z (4).

Lastly, 10 (* + 1) = 2 2 + z ^ # * .... (5). From these five equations, x, y, z, t, and t,' may be found as follows :

From the (I) and (3),

t=* t'= 5z

M6).

10 + x, 10 + x'

.'. by substitution in (2), (4), (5), and reduction, 4 #* -f 40 a? — 4 o?# + 2 x z — 40 ?/ — zy + 10 s = 0 lOx + #*/ — 6xz + 10 2/ + 5 yz — 10 a + 100 = 0

and 2 x = 15, .*. x = 7 J miles an hour, and substituting the values of ?/, s, £, *' may easily be found.

7. Let 4x = quantity of fluid in P, at first ; then 21 x = Q . . . .

Also, let 4y be what A pumped out of P ; then 4x— 4y = what was left in P, \ 21 x + 4y = quantity now in Q J Also, after B had pumped,

x—y = what was left in P, ~i

21a: + 4y + 3a?— 3y=24ar-f y=whatwastheninQJ

Again, let w = ~r— ^ — r-r- m P, at first, Qv or spirits '

v == ditto in Q, at first,

P. 75.]

IN ARITHMETIC AND ALGEBRA.

157

X

, Q. of wine -fQ. of spirits. „

then •.■ u + 1 = 7K — n — • •■ mr= ^ — tt—

Q. ot spirits Q,.ot spirits

and t' + l = ditto

.'. Q of spirits in P, at first — ditto in Q, ==

and hence Q of wine in P, at first = ditto in Q, ==

in Q 4%

21

x

u+l

21g

v+l

4ux

U+l

2\vx

vTT

>

Q. of spirits' (t>),

(o).

Hence, also, after B had pumped,

spirits in Q=^ + ~^

* . . ,< 2\vx (3x + y)u

and wine in(j= r + ~-

v + l u+l

(d),

the parts added by pumping being found by Rule of Three statements.

Again, if s denote the strength of wine, then, by the ques- tion, 3 s is that of spirits; and then, also, by the question, f2lx 3x+y\ Q f ^l^ (3x + y) ?n 12

V^TT + ~u~+l ) + U+l + u+l Is -13

(2\x 2\vx \

r +3*+— rr x s )'

\v-fl w-fl /

which, when reduced, gives

21a; v + 3 ,„ . u + 3 ,

T3-. ^T+<3*+>'>-S+T=a •'(1)-

Again, supposing B had pumped 3x—3y fluid from Q into P, after A had pumped 4 y into Q, then

...„ 4a; 4?/ 3 a; —

■PnhmPi=r7l"^T + ^ *x4. V,7XT h^zrJ

= 0*— y) i — t +

y 3 a: - 3 y ( 2]_x 4 y w+1~m+1 + 21a; + 4y U+l~ w + 4 3

+

3£+ 4?/

2lx + 4y \v+l u+l, , wine in P = 4 a;— 4 y + 3 x— 3 y— spirits in P

= (*-y) ( 7 - — -j - ^-^ ^ + — j-J)

158 HINTS AND ANSWERS [P. 75.

( _ ^ /7m + 3 _ 3 /21a; 4y \\

-^ y)\u-\-\ 2lH4y\Hl «+l/J'

Hence, by the question,

o / s f 4 3 /21.r 4m \l ,

\ yj Im+1 21# + 4y\ii; + l M+l/J v y/

7m+3 3 /21s 4y \i _ / ; / 4x

m+1 21# + 4?/V^ + l + w+l/J " |/ U m+1

!

4?/a; \ /_ 21 x 21 vx\}

4 5. T . ( 3.5. r + S. T ) > s

M+1/ V t?+l W+l/J

/ 12 9_ f2lx Ay \ 7_m + 3_

or,(a?-y; \u+i+2lx+4y\v + l~]rir+l) + m+1

3 /21a? 4 7/ \i //I2x+4ux 63a; + 21paA

2T^+T^V^+T+27+T;i " j/ V m+1 ' zT+T J'

, flu+15 6 /21a; 4y \\

or' C*-sO i^+r + 2T^T4m WT + ^+t;j =

»- l/Sri-SS) (2)-

Lastly, B would have pumped out the same quantity of wine

as he did before of spirits. But after A had pumped,

. „ 21 vx 4 mm

wine in Q = — — r 4 f=,

v+\ M+1

, , „ 3x—3y (2\vx 4uy\

.'. wine pumped by Q = —. ~- • I T -\ ^ )»

^ l : 2lx+4y \v + l m+1'

Also, spirits pumped by Q = ^ ,

2\vx 4uy _ 21# + 4m

r+1 M+1 M+1

or,

2\vx 21x + 4y— 4uy .^

v+l M+1

Although there are but three equations for the four unknowns x, y, m, v, these may still be determined. For v (1), (2), (3) are homogeneous with respect to x, y ; making

y — iv x, we obtain

P. 75.] (l-w) {

IN ARITHMETIC AND ALGEBRA.

159

7u+\5 w+1

4-

6

21

4-

*»U1

21 + 4?y\w + l M-fl/J

y V m + 1 y + U

S

21 a

21 + 4io — 4 wz^

(5).

(6).

#4-1 w-V- 1

from these u, v, w being found, the quantities required may then be obtained ; it being observed that (6) gives

21 uv—l

w =

4 ' (l-w) (1 + v) '

ST. JOHN'S COLLEGE, 1826.

1. First multiply by x + 1, then reduce, then multiply by 3 x -f 2, and we have x = y .

2. Ans. # = 3, y = 2.

3. Since aibx—2a4b^/x 4- a46 + x —2^x + 1 = 2.r + 2, /. x (1 - a46) + 2 (1 + a46) v' x =-(1 - a46), 1 + a4fi

rr + 2.

I

a

4 6

v' x = — 1,

.". completing the square, &c.

>/ x +

1 4- «

1 -a4

4 & ^ /0 + ? 4 6)2— (1- - a 4 *)2

1/'

2a26

(1-a46)5

*/ x = —

1— 2a26 + aib

1

a

46

and

1 4- 2«2i + aib

1-a4

(l-"«26)2aTld (l+«26)2 r ana_ 1-a4* '

1 4- a2b

1

a

46

1-a

26

# =

"l4-«2& 1 -«26\2 1 4-«26

and —

1

a

2 6'

,1 4- a and -=-

2 6\2

n + a2b\ \\- a*b)

id y3 + 3a?2y =252-/' ^'

160 HINTS AND ANSWERS [P. 76.

4. First divide the second by the first, and we get

9 y fx + yy_ x 2 x — y^

2x+</(x2-y2)~u\.~~2r -*.vi«-y;+-2- J'

(Private Tutor,) p. 98.

3a-a+y* ~252l '* ^(* y ^i

2a? = a/ (x2 — y2)\ and ?/ 3 4- 3x2y each of which must be combined with either of the given equations.

The former gives y2 = — 3 x 2, which, combined with the second of the given equations, leads to an absurdity. But squaring the second of the given equations, we get

.-. (a? + y? + (# - ?/)3 + 2 (;r2 - y2)^= 648,

/. ar 3 + 3.r*/2 + {%- - ?/T = 324 (2).

Adding (1) to, and subtracting it from (2), we get

(x + y)3 = 576 - (x2 -^ y2Y

{x—yY- T2-{x2-y2f\

(X2 _ ^a _ 41472 _ 648 {x2- iff + {x2-y2)3, , . 2Ni 41472 4608 " 512 _. ' ■ ^2-^2 = -648" ="72- =-^-==64' .'. (#+2/)3=576-64=512l («-y)3= 8 / '

' . " " « V .". a?= 5 and ?/ == 3.

x — y — 2J J

Those who are skilled in the general Theory of Equations may find two pairs of imaginary simultaneous values of x and y from the equations (3).

5. Let x£. be the daily demand on A's Bank at first ; then y£ B's

P. 76.] IN ARITHMETIC AND ALGEBRA. 101

3 x -f 3 x x 2 == A's cash, and 9x + 3?/ = A's cash + B's, .". by the question,

3 y = 7y - 4000

and 9/ + 3y = 7 (* + y) /. ?/ == 1000, and x — 2000/.

.}■

6. Let x — sum paid for each small burner for 6 nights in the week, then the whole sum paid, as at first agreed, would have been (3 -f 5 + £) x. But the whole sum actually paid was

(3 + 5 + 4 + 2) £ x, .". by the question,

45 37 11 31[

24X 4 x ~~ 20 ~~20'

315 x - 222 x = 31 X^=?L^J,

Hence, x = f/.

45 x 7 2 63 21 _-, _ iU

.'. — kt~- X -= = T^i = -7 =5/. 5s. the answer. 24 5 12 4

7. Let x be the miles he travelled an hour at starting,

2y the distance from Cambridge to London. Then, by the question,

x + V • f ~ 4 ~ 4

x

= 5 CI).

Again, if z be the gain of the watch per hour, then during the actual 5 hours of riding the first half, as shown by (1), the watch will have gained 5 z, and he thinks he has been on the road 5 + 5 z hours. Hence he supposes he has only 4 — 5 z hours to ride y miles in, and, consequently, that his rate must

Y

162 HINTS AND ANSWERS [P. 76.

-. — ^-r— miles an hour : so that 4—02

= * x ... . (2).

4- 5z From the equations (1) and (2), we easily get

% = ^ hours.

14

Lastly, the true time to the fourteenth mile-stone is — hours:

x

and, consequently, the apparent time would have been, on the

second hypothesis,

14 J. 14 1J33_

x 20 X x °r 10 x'

133 Hence, he would suppose he had 9 — ^k hours to ride

y — 14 miles in. He, therefore, would make his pace = y — 14 miles an hour ; so that, by the question — 133

9 " 10*'

14 , ,„ v - 14 n 7

or

10"? ) - 133 7

$

x 6

and x =i 6 miles, Hence, 2 ?/ — 6 x 10 — 60 miles.

ST. JOHN'S COLLEGE, 1826.

1. The product = 0000 208 ; the quotent = 15000.

2- ©* = (I? = i- ' M = 1-837

and V-012 = 109, &c.

3. Ans. 4/. 16*.

4. Ans. 800/.

P. 77.] IN ARITHMETIC AND ALGEBRA. 163

5. Answers

2^2+3 /3 + 2-v/6 ,3^2 + 4-^/6-2^/3

6. The sum is

19 1

and

a*-x4'

7. See (Wood, art. 111).

S. Dividing the first by the second,

fx\*~ 3 c ,yj " d'

fc\ «-3

d\*-&

■■•5-0

i

Hence, x mav be found in a similar form.

9. If|(ar+2) + ^arbe<|(a;— 4)+3, and >± 0+l) + i,

then is 7a; + 6< 6x+12, and > 6#-f 10, then x < 6, and > 4, .*. ,r = 5 is the integer value required.

10. Let x be the radix of the system ; then

5 + 4 a; -fa;2 = 290, .-. x = 15.

11. (1). Here the first term should be ■§-.

The series is Geometric, its common ratio being — v"4, a (r»- 1) _ . (-x/jf-l _ . (4)4-l

164

HINTS AND ANSWERS [P. 77.

= - i x ¥• (V4 - 1) = - W (1/10-2), which is easily exhibited in decimals.

(2). The common ratio is r,

a—b

1 V ,

1 . fa— 6)n_1- a— 6 — 1

(a_6)«-i. a -6-1 ' a — 6

a + 6 (a— b)n — I — (a— 6)n-2' a— 6—1 '

(3). Here the common difference is — 2 a.r, .'. S= |2a+(»-l) b\%

— \2 (a + xf— (n— 1) 2ax\^ = \{a-{ x)*—(n— 1) a^ £n, — \a2+(3— n) ax + x2l n.

12. The Arithmetic mean = — ^— ; the Geometric = y («6), and, by the question,

— 2~ = 2\/(ab),

• - -4 '•6

r/

- = 7±4^3 = (2±-v/3)2=(2±^/3)(2±^/3;,

2± y/3 2 + -v/3*

13. Ans. The roots are 2+^/3, and */(a2— x2)-\-x.

P. 77.] IN ARITHMETIC AND ALGEBRA. 165

14. Ans. The G. C. M. is x3— 2 a •- + .'', and the reduced fraction is

> *2 + 3

3* x2 + x — r

ST JOHN'S COLLEGE, LS27.

1. First multiply by 39 ; reduce ; then by 4, &c. x =. 11.

2. Ans. x — 16; y — 25.

3. First iT^— o (3 t-4 */3a0 = 16-ar— 8 a/3^ + 3,

= (4^-)/3)', . ^^3(4^^3) = (4 /,_v^j

Hence, 4 s/ x = /3 + (81 */3)* = v/3 + 3 /3 = 4/3,

. . x =^ o.

4. From the first,

3^2 _ 11 + 2a- = (7 + 2?y - if - xf, = x*- 2x (7 + 2y - if) + (7 + 2y - f)\ '. x*-2x(S + 2y-y*) =3^- 11 -(7 + 2*/-y2)2, •. <r = 8 + 2y - y* ± V &V2 - H + (8 + 2y - y*)*

- {7 + 2y-ym,

= 8 + 2y - .y2 ± (y + 2),

= 10 + 3t/ - y2 and 6 + y - t/2 (1).

Substituting these values in the second, we have

^(r-3)-10 + 2y_?/'

This rationalized becomes 3,6 _ 4^5 _ 20 y4 f 60 y3 + 152 y- - 200 y + 200 -0.

166 HINTS AND ANSWERS [P. 78.

But this cannot be resolved by the First Part of Algebra. Again, substituting the second value of x, we have, in the second equation,

6 + 2 y- y*

V (y2 + 2y + 1)

6-3,3

. y- - 2y + 6 or y + 1 = 3. r-^-f

.-. y3+y»_6y + 6 = y2- 2y + 6, t/3 - 4y = 0, .'. y = 0, and = 2, and = — 2, .*. a: = 6, and = 4, and = 0.

The simultaneous values "J- __ > satisfy the given

equations when the sign of the symbol ^ is considered + and

x = 0i

" v- belong to those equations, when the radical is taken

negative.

5. Let # be the number of balls in the large bag, y that in the smaller,

z the number in a handful ; then, by the question, x — z—iy — zf .

x - z = z2 I . (sr - *)■ = * i

f " * (*/ _ ~\2 = z I i yj ■

andy-5r + (y-2)*=y+fyj vy ~' ^*y Let y = mx/ then

(w - l)3s = 1 1

(m - l)2s = 1 + fui '

M — 1 = - ,

1 + f «,'

.'. |-Ma + U — 1 — § M = 1,

2w2

.".— 3- + 4 « = 2,

,-. a2 + £ u — 3,

19 - 1 + y 19 16 ~ 4

whence ,r, t/, z, are easily found.

M = — I ± V

P. 78.] IN ARITHMETIC AND ALGEBRA. 167

6. Suppose the stream to run from A to B, and make x miles the rate per hour at which one waterman would row in still water; also lety be the distance between the two towns; then the rate at which he proceeds from A to B, is x -{- 4, B to A is x — 4,

.'. the time of the first voyage = — — — -\ ^~r>

X -\- ~k X — t:

Also, had the water been still, the time would have been — - ,

x

.'.by the question

?/ +i = *?+2 a).

# + 4 x— 4 " r 60 Again, the rate at which they go down the second day is

-I- -^ 4- 4, and the rate of returning is x + - — 4 : and

3x had the water been still the rate would have been -^ ; .*. by

■ ■ • (2).

b question
y		i	V	3r +	8
3x	{ Sx	- 4	60
2~ +	4	2	2
Equations	(1)	and (2;	) become
		,r3 9 .r3	- 16. v — 64 x 32 y	= 30j	i
			:. 32// :	1 O 4 — Tb	x\
		9x'2	-64	= 13,
		X2	- 16
		which	gives x -.	= 6.

7. Let t = actual time = time occupied by B, since, by supposition, he is the slowest workman, t — x = time occupied by B.

Let y = work done by A in one month,

*= B

.". (f — x) y = A's actual work, t z = B's . . . .

168

HINTS AND ANSWERS

[P. 78. • • (1)

s

J .

.". (t — x) y -f t a = whole length of foundation

— i t (y ' + ' *) by hyp =

Also, | £ z = 3 y and i ty = 12 z = (t — x} y + 36 •'• i t [y + 2) = 12a + 3 y by addition, tz — V */, (t-x)y — 12 s — 36, .'. substituting these values in the first equation, 12z-36+ V5 V = 12 sf + 3Z/, .*. |y = 36, ••• # = 48, .*. substituting for y, in the remaining equations, we have f X 48* = 12 « 12 z =48(« — a) + 36j» .-. £ *2 ,= 144

.-. 16? = 5^

z = 4 {t — x) -f 3 tz — 180 from the 1st of these,

9 — - >_6 / * 5 ^

.'. substituting in the 3rd,

(2) (3)

i 6 +2 - 5 ' "	= 180
.'. £2 :	2 2 5 — 4
.*. t .'. z	»_5 5 l	— 7-^1 = 24i
. substituting these values in	the 2nd equation,
24 =	30 -	4# + 3
.'. x —	1 = 2	i 4,
.'. t —x =	5*
. length of foundatior	i = («.-	- x)y + tz
	2 1 — 4"	x 48 + V5 x 24
	= 252	+ 180
	= 432	yards.

P. 79.] IN ARITHMETIC AND ALGEBRA. 169

ST JOHN'S COLLEGE, 1827.

1. Let x be the number of yards required ; then, v

. , diagonal

Slde= y2~'

i • i a (diagonal)2 .'. square yards in the floor = ^ ,

At

(5 + f+ii)2 /214V , /lOTY* .". by the question,

x X * = W87 X *'

/107V 11449 and.r=lx^J = -^

— 23 yards, 1 foot, 8— inches.

2. \/°§^ = a/ 420 = 20.4939015,

also #(97+28>/12) = V 1^(97+28-/ 12) =2+^/3.

3. Ans. Rate per cent, is 5 f .

4. Ans. 15'. 36'Vt- past 10 o'clock.

_ ml . #3-2.r2 + 3.r-4

O. lhe sum is . , „ ,,

x4— x3 + .r2— x + l

ai * + 2_A'BC«.

Also, assume — — — + T + ? ;

xA— x x x-\-l x — 1

clear it of fractions ; collect coefficients of the powers of x, and

put each of them = 0 ; the equations thus got will give

A = - 2, B == h C = h

and.,f+2__2 + l 1 +3 1

x3— x x 2<r-fl 2 x— 1

6. See Wood, art. (76).

170 HINTS AND ANSWERS [P. 79.

1 1—1 —1

7. The quotient is x4, -f|ailcHi«^*+ &c

The coefficient of the ninth term is — b (fi + y).

8. Their G. CM. is x2+l, and .'. the L. C. M. is x4—\.

9. The common ratio of the first series being; if— be a

x x

x4 proper fraction, the sum ad infra is -j .

The common difference of the second series is — a-\-b,

n and sum = \{n-\-\) a + {n — 3)b\ -.

The Arithmetic means are &, %, &, $, TV, —\, —&.

10. The roots of the first equation are x = — and

a2 4a2

of the second x = . 0 2-

y = 2.9.

of the third z = 9 and 4, the third being solved by first clearing it of fractions and adding x + 1 to both sides, thus producing

x2 + 2x + 1 .= 49 + 14 sf x + x,

jj j

11. (1). Ans. an 4- nan^bs/~ 1— n. — -_a"-262 —

1. 2 3 -'^PV-l,

andthe Pterin, if ! ,isL3'5- • -J2n~3) x2„-2

' -/(l-^2) '2.4.6 . . . (2ii— 2)

(2). The n* term of _|A + &c. is (2w + 1)(2» + 3), v y l2. 3* w2 (w + 2j2

and putting w = 0, 1, 2 &c. the series will be

-35, -V. -VS&c.

12. From the given logarithms it is found that

log 2 = 3010275 (which is not accurate), and thence log 50000 = log 1000 00 - log 2,

= 5- .30 10275, == 4. 6989725 nearly.

P. 80.] IN ARITHMETIC AND ALGEBRA. 171

13. Ans. 7.

14. Assume l~3x + 2f = i + ftx + qxi + &c.

1— x + x2

Multiply by denominator, collect coefficients of x°,x, x2, &c. and put them each = 0. From the equations thus made, we easily get

B = - 2, C == - 1, &c. Observe, the same series will be produced by actual division.

ST. JOHN'S COLLEGE, 1828.

1. Multiply by 24, reduce ; then multiply by 13 — 2x and get x = f .

2. First, be x — x2 y = xy2 — acy,

c (bx + ay) = xy (x + y) (J).

also, from the second, axy2-\-bx 2y—x2 y3 = abc (x + y) — abc2,

.'. xy (bx + ay) = x2 y2 -f abc {x + y) — abc2 . . (2).

.'. xy —.{x + y) — x2 y2 -f abc (x + y) — abc2,

x2 y2 (x + y) — ex2 y2 + abc2 (x + y) — abc3,

x2 y2 (x + y — c) = abc2 (x + y) — abc3,

■=■ abc2 (x + y — c)

.-. x2y2 = abc2\ (3).

and x + y — c / (4).

Combining (3) and (1), we get

/, ac >J ab\ , , ., / c \f ab\

c[bx± vx j=± c V (ab) i^x ± — v— J

.". &r2 ± ac \/ ab = ± ^ ab x2 -{■ abc,

.'. x2 (b + -y/ a6) = «6c h= ac V a£,

abc + ac >J ab \f b+ \/ a

x2 = — j —, — = ac-1—, -— = ac,

b + */ ab \f b -+- \/ a

.'. x = ± -V (ac)>

Hence, ?/2 = — or,

J ac

172 HINTS AND ANSWERS [P. 80.

y = ± V {be).

which are two pairs of simultaneous values.

Again, combining (4) and (1), we have

c \bx + a [c — x)\ = x (c — x) c,

.". bx + ac — ax = ex — x2,

.'. x2— (a — b + c) x = — ac,

a— b + c / (a — b + c)2 — 4 ac ••• *= 2±~± V 4^

a — h — c ± ^ {a2 + b2 + c2 — 2ab — 2ac — 2bc)f _ 2

and .'. y = c — ic — a + 6 + 3c ± \/ (a2 + 62 + c2 — 2a& — 2ac — 26c).

2

3. First, (4^ + 8)*+ 7??J^8) = «**+*, then, squaring (4.r2+ 8)3 + 6. (4aa+ 8) +^^-^=256^ + 768^ + 576,

or,64T6+384.r4 + 768^2+512 + 24,r2+ 48 +4-^3,

.-. 64^+128^+24^+^-5—5 =46

256 a;8 + 512a;6+96^4-64^2

512 *6+ 1924^+192 r2- 128+9=0, .'. 256 ^+1024 x6 + 1120 x* + 128 ;r2- 119=0,

or, (2^)8+16(2^)6 + 70(2^)4 + 32. (2.r)2-119=0,

/.119 |(2^)a — If +70(2^)4-87(2^)2+(2;r)8+16(2^)6=0,

.-. 119 \(2xf-ll+87{[2xf- 1^-17(2^)4 + 16(2^)6

+ (2 x)*= 0, .'. 119 ^(2^)2-l^+87(2^)2 ^(2^)2-l^ + 1782(2^)4

|{2^)2— 1| +-(2^)6 l(2x)2-ll=0,

... l(2x)2-ll K2^)6+17(2a-)4+87(2^)2+119^=0,

.-. (2xf— 1 = 0,

The principles of the Second Part of Algebra would give six other roots of the equation.

P. 80.] IN ARITHMETIC AND ALGEBRA. 173

4. From the second we have

a;

.-.(** +y*) {a** /+-|i-^/(^+ yh}=ay-(^+y2)

= a)3- (** + vh ( ^ - *V + y*\ >

2 2T4. 2i 4 91 /•• - 4x1 /«\!

.*.(** + yT).{ & + 2x3if +i/+—\/(x3 +yi)\=(i)3

.-. (** + if) {{a* + /)2 + -^rgVCf* + /)} - (I)3,

-91 ± v/^8281 + 108000^

- 432 — 91 ±341

"~ 432 " '

250 , 432 -^32 an " 432'

125 . . = -2T^and-1'

- (|)3 and - 1,

.*. (a? + y^f = i and — 1 (1).

Substituting the first of these values in the first of the given equations, we get

25 + 48 (xyf = 241.

(xyy . putting u = xy,

. 48 u - 241 J + 25 = 0,

. 2 [21 u— 1)— 6 m— 241 w? + 27 = 0,

, 2(27m-1)-6w-27x9?/^ + 2w* + 27 = 0, ; 2(27w-l)-27(9^- l)-2?<?(3w*-l)=0.

174 HINTS AND ANSWERS [P. 80.

.-. (3#- 1) (18 u? + 6itJ + 2-81 J -21 -2 J) = 0, :. (3 m*— l)(16«*-75 u*-25)=0, :. 3 iF - 1 = 0 (2),

and 16 u^-15 ^-25=0 (3).

1 125

Equation (2) gives u= -^r- , and (3) gives m=125, or qFo".

1 2. ± 25

Combining xy = ~= with .»3 + 2/3 z= o« ^ w^ easily be

8 1

found that x = ^= and y = — . Other simultaneous values

may be obtained by combining the other values of u with

a 2 r>£

x + y =3g.

Also, other simultaneous values may be found by substituting (#3 _|_ y3^2 __ — J in the first of the given equations.

5. Let # o'clock be the hour they are striking ; then the a?th stroke of the faster clock is at 3 x— 2 whilst the #th stroke of the slower is at 4 x— I.

Let the yth stroke of the faster coincide with the 2th stroke of the slower; then

3y— 2=4s— 1. Of this the solutions in integers are

z= 2, 5, 8, 11, &c. y = 3, 7, 11, 15, &c. .". by the question, the coincidences are 3,

and2.r- 3 = 19, /. x = 11.

6. First, 5£ points = \ of 32 points = £ x 360° = 60°,

and 2f = TV of 32 = TV * 360° = 30°.

Also, before the first tack, the distance of the vessels is

3 a ±Hp P or 3a 4- Q-l)ff

P. 80.] IN ARITHMETIC AND ALGEBRA. 175

Again, after the first tack, the distance of the cutter from the first direction is

s/ 3

p sin 60° or p

9

that of the smuggler is np sin 30°, or ~.

Hence, (by 47th prop, of Euclid, b. i.) the distance of the vessels, at the end of the first tack, is

The vessels now changing their tacks, but still preserving the same angular distance from windward, at the end of the second tack, they lie in the very same line as at first, viz. to windward ; and this will evidently be the case also at the end of the 4th, 6th, &c. tacks. Their distance is now

71—1

Similarly,

n~\ di=3q -f — — p— 2p(l — «V 3) the distance after the 4th tack

de=39 + ^V-3/>(1-kV3) 6th . .

&c.

Also,

+

wpy'3 p\2

2

in

A = -l/K"2^3)^ &+ I "' ws- s> - «,

(1-«V3)}2} dl = |/[(^+|^3)2 + {3?+ f (■ + „V 8 - 2) - 3 p (I-b/3)}*],

&c.

or r

176 HINTS AND ANSWERS [P. 81.

Now, if the cutter come within shot of the smuggler in an odd tack, let that tack be the (2x+ l)th ; but if at the end of an even tack, let it be the 2 yth ; then

r = d2y = 3q+—j-p—yp (1— n^3,

• = ^+1= (*+§&)*+ {3? + | (» + nVZ- 2) [

-^p(l-w^/3)p, -J

whence x or y may easily be found, and we get

6<7 — 2r + (ji — X)p y = 2p(l—ny3) '

6? +?> O + n</3— 2) T V I4 r2— ^ (w + ^3^ °r' * ~ 2p (l-w/3) *

and y or x must, in specific cases, be found in nearest integers.

7. Let x minutes be the time of filling in the second irruption, y feet the length of the tunnel and shaft at that time, z the breadth and depth of each tunnel ; then the area of a horizontal section of the two tunnels and the bottom of the shaft is

2 y z square feet, and the area of a section of the shaft is 4 z2.

Also, at the second irruption, the volume of the tunnels and shaft was

2 y z x z + 4 z2 x 4 s, or 2 y z1 + 16 z3 cubic inches, and the time of filling, the first time, is x— 10 minutes.

Again, let v be the velocity of the water, or number of cubic inches per minute entering at the first irruption, and v' that at the second ; then, since volume filled oc vel. x time,

volume, at first, : volume, at the second irruption, : : v x

(x— 6) : v'xx,

.'. volume, at first, = —^-7 -. (2?/s2 + I623),

v x ^ J

P. 81.] IN ARITHMETIC AND ALGEBRA. 177

and .*, length of the tunnels and shaft, at first, _ v (x— 10)

V X

(y + Hz) - 8 z.

Now, time of filling 2 y z*+ 16 z3 with v = 2Vz2+16**

v

and the time of filling v (x~10) (2yz* + 16 g3) with

v x

r— 10

»= -r£ (2 V *2+ 16 z3),

by the question,

2yz2+l6z3 _ 3 x— 10 v 2 v' x

(2?/22+16a3),

and # = ,5 — 7 (1 .

A , ., c ,. ,, velocity of influx

Again, velocity ot ascending surface a , — : — ^—^ -. — ,

° ■* horizontal section

v v

: : velocity of the levels : vel. in the shaft,

2yz ' 4 z

:: 1 : 8

V

(2).

y 16 v'

Again, had the tunnels been 1 10 feet longer, the horizontal section of the levels would have been

2 (?/-l- 110) z square feet, and the ascending velocities, at the first irruption, the second, and for this supposed section, would have been respectively

v x v' vl

2£[X— 10)(2^2+16z3)' 2y~z 2(y+110)z '

and the velocities of ascent in the shaft, in the first and second irruptions, are as

v , v'

and

4.2 —" 4z2' .". by the question,

v' x v 1 v— v'

2z~{z-\0) {2yz2+\6z3) 2yz '9 4z2

A A

178

whence Also, 7

x

HINTS AND ANSWERS

2

g (#-10) (y+8 z) v V

[P. 81.

1 v — v' 9" H?'

yz 2{y+\\0)z

z z \ (v \

y w + 110 18" W ll

1980 ^ _v_

y (y'+ U0) " |/

(4).

To solve these four equations, make w = — ; then they become

30

x —

3—2u

V

= 16 m

.r

s (a?— 10) (y+8z) y

1980s

* = I (l- - 1

9 \u 1

5>

y(y+HD) +1- u

■ • (5).

• • (6).

• • (7).

• • (8).

From the (6) we get n = ^—, and substituting this in (5), (7), (8), we get

x =

16;

240,

9xy

24s-y

1

= 34z + y

z(x—l0)(y + 8z)

260 z- 16 ys +y (?/ + 1 10) = = 0

hrom (9), a; = —

v " 24g— y

= 342 + y

• (9).

K . .(io).

J. ..(11).

, which being put in (10), gives

also, z =

y+82

y(y + U0)

16 y -260"

|. ..(12). /. ..(13).

whence x, y, %, and u may be found.

P. 82.] IN ARITHMETIC AND ALGEBRA. 179

ST. JOHN'S COLLEGE, 1828.

1. The quotient is 5020.

and/(6+-v/8-Vl2-\/24) = ^3-^2-1. This third is done, by assuming

y/(6+A/8-Vl2-A/24) = ^v+Vy+Vz, then, 6 + -V/8--/12- ^24 = s+y+s+2 V(>#)+2^(*»0

.*. ^(^) = >/2, ^/<» = -1/3, s/(yz) =-V$,

.'. xyz = 36, &c. Again, 2 + v/3 = 2 + 1.7320508 = 3.7320508,

and .'. lj(2 + l/3) = 1.55, &c.

2. The five quarters of wheat are worth

|| X 60 x 5 = **? x 20 shillings,

and the barley

32 EJ 0 32x54 .... _ x 54 x 3 = n shillings.

Ans. 22/. 8s.

3. Let x be the pounds he had in the 3 per cents; then the

. ■ 100 800

pounds stock he gets m the 4 per cents are j^ x or ^ x"

100 o

The income on the cash, when in the 3 per cents, was gp * X o,

.*. by the question,

3200 1200 _ .n

•829" ^ ~ "337" ^ ~ ' .-. x (337 x 32-829 x 12) = 829 x 337, 829 x 337

'"• ''' ~ 836 ' which is easily reduced.

1^0 HINTS AND ANSWERS [P. 82.

4. First, add the two last fractions.

a 3*2

Ans. — - — T.

X"— 1

13 5

The quotient is x~* — \ x~^ — | x~* — &c.

Also _L=ii__iJ 1 J_

' x4-l " *"a?-l 4'^+l 2,a:2-i-l'

This is done by the method of indeterminate coefficients ; first assuming

1 _ A;g + B C D

*4-I *2+l + x^l + x~+l'

5. */£ is < (|F, 4/2+4/7 is > */3-+V&

and a -f \/ a is < 1+a2.

6. Ans. G. C. M. = at*— 1.

7. Let S = 2.34545

/. 1000 S = 2345.4545

and 10 S = 23.4545

2322 1161 129

Again, 1 — \ +

990 495 55 '

a 1

5

, n— 1 , rc — 2 rc— 1

and + 4 . . . n terms = — =— ■

n n 2

Finally, let x, y, be the two quantities ; then

\/(xy) . .

— ^-^- is given = 2a, suppose ;

* V

2 x

r - — -.

y a

a V y

whence -'

y

P. 82.] IN ARITHMETIC AND ALGEBRA. 181

8. (1). x-3 + X 2 = 2,

.'. x~* = - i ± VI = ~2± = 1 and -2,

.'. a- = 1 and(£)3.

(2). 4^+6^(1+4 2^ = 27 (l + x), /. 2ar+31/(l+a?) = ±^I36(l+;r)| = ±6V(l+»), .-. 2a; = 3^/(1+4?) and - 9 J(l+x), .'. 4*2 = 9 + 9* and 81+81*,

.'. *2_ 9 x = 9 and ^2_ y ^ = »T. ,

9+^/(81 + 144), 81 ±^(81a+ 16x81)

a? = — q and x — q »

9 ± 15 , 81±9-v/97 .'. ar = — q — ana x — — — g ,

&c.

.-. A._4 = l + ar+l-2-v/ (!+*)»

and * — 8.

(4). From </* (1— xy)—y— x, we get

2;+*/ a; , l + V* Vx

y = =V*- i =

1+aV* 1 + x* \ — Jx + x

Hence, 1— y = -,-- ^ — — ,

' ^ 1 — \/* + .r

, , 1+*

and I + y = ^ — -- — >

, . (l-A/a?)(l + a?)

and y— v= J x - — , — ■

* 1 — \/ x + x

Now, dividing the first equation by the second, we get

a x^{\— y*) (l-Vx) V(l+x)

s/ x ~ y—x \/x{ I -x/x) (1 + x)f

:. <n/(l+x) = x,

.-. z*—a*x = a2,

182 HINTS AND ANSWERS [P. 82.

. j,__a2±^(a4+4fl2) a2±aA/(a2 + 4)

. . JU — c

2 2

Tj X + J^X

Hence, v = ^ - — .

J \-\-X\/X

It also appears from

a (\—^x) ^(l+x) = x(l — ^/x), that s/x— 1 = 0, or, a? = 1

■, x = 1\ ., ?/ = 1/

Hence, y

9. Divide repeatedly 2304, in its own scale of 5 by 11.

11) 2304

11)104- 10 2- 7 .'. in the scale, whose radix is 11, the number is (t meaning the digit 10) 27 t.

To make the number consist of 1, 3, 32, &c. we must trans- form it to the system whose radix is 3.

3) 2304

3)414 - 2

3)121 - 1

3)22- 0

4- "0

.'. in the ternary system, the number is

40012, and it .'. consists of 2 x 1 , 3, 0 x 32, 0 x 33, 4 x 34.

10. Let x be the number of guineas, y of crowns ; then

2\x+5y "= 20000, and this solved in integers gives

x = 5, 10, 15, 20

y = 3979, 3958, 3937, 3916

Also the number of solutions {Barlow's Theory of Num- bers, p. 325), is the greatest integer in 20000 4000

5iT2P °r ,n "ST' or 19a

1 1. For the rule, see Brtrlmrs Theory of Numbers, p. 33.

P. 83.] IN ARITHMETIC AND ALGEBRA. 183

Since 2160 = 24. 33. 5. Number of divisors = (4+ 1) (3+ 1) (1 + 1),

= 40.

12.

n. (re— 1) . . . (n — r+ 1) : n (re— 1). . ,(re— r + 2) : : 10 : 1 -,

and

re(w— 1) . . . (re— r-f 1) p w(»— 1). . .(w— r + 2) „ q

= ;=r * z 7Z i 11 O ' t J

r ' 1. 2 .... r-1

1 . rc = 15 w— r+1 5 I- •

and = o I r = 6.

r 3 J

13. Amount = PR" (Wood, art. 397.) /21\B_1050 21 ' ' W _ 100 20 X '

.-. n — 1 =

log 21 — log 20-

But log 21 — log | x f X 10,

— 1 + log f + log f = 1 + . 17609,

+ . 14613, = 1. 32222, and log 20= l+log2=l + ±logl6=l+£log^x 10, = f + i log 1.6 = f + .05103, = 1. 30103, .log21 — log20= .0 2119.

1 _ 100000

" n ~ 1X02119" 2119 '

i to 407

and n = 48 ■ years.

*z\ 19 '

14. When x is small,

•* 1+i.r+l— f.r . 2— 4-./:

it = -,— - — —. j— -~— nearly = s — nearly

= 1 — * ■§■ x nearly by actual division.

When x is very large - is very small.

184 HINTS AND ANSWERS [P. 83.

Let .'. u = -: then x

(1 + uf (I— m)J 1+m + y/(l + ?/)~ ~l + w + /«*/(! +««)

l+w+ v/m. (1+i u—^u2) — r~ — I j »

l-f-ttT-f tt + £WT— £tt2

JL J. A J. !_'

= m3 + m2 - m - | a3 + £ w2 -f f & 6 nearly, the terms containing w2, and the higher powers being omitted. Hence the value of the function, true up to small quantities of

nearly

the order

1 115,1

T+T71T— - !" + *• T+"

/a? a; o i # * " V'

ST. JOHN'S COLLEGE. 1830.

1. Multiply by 210, &c. Ans. x — 4. 2 \ns * =01* = 4l -r = tV1

3-First^-r+7^2=8(^/^2>

Q

..x—2^x+l = -(*+-/ ar-2),

a;

.-. <y.r-l)2 = -(y-r-1) (^/.r + 2),

<V*- 1) |a/*-1 ^ — -j = o,

.'. (y/X—\) (Xa/X — X— 8 a/X— 16) = 0,

P. 83.] IN ARITHMETIC AND ALGEBRA. 185

.*. ^/#-l=0 i (1)

and x\/x— x— 8«/x— 16j_ ^ (2)

From (1) #=1 ; from (2) a?=16i &>~4{i) 4

•c

C«^ vzc-

^3

4. From the first,

4 + 4yi/x + xy2 = x2y+xy2 + -j-,

.-. 2+y^a> = ±tfy U + |) (1)

This equation splits into two others. Taking + y\/y—2y^x+2x\/y—4: = 0 (2)i

yVy+2yVx+2xVy+* = o (3)1

The second of the given equations gives

x\fy—y>^x = x^/x—3 (4)

Adding (4) to (2) we get

3. X X 3-

y2 — 3yx2 + 3xy2 — l = x2

.-. y* — 1 = Xs (5)

Substituting y2 =\ + x2 in (4), we get

x + x\/x— (1 +x'2)'2.^/x = x\/x — 3,

.*. x2 + af + x2 = 3,

and — 4,zT + 2 a?^— 1 = — 4 x + 2 a?*— 1, .*. adding these equals,

(aF— 1)3=2 (1— £*+«*)=— 2(V*-1)(1+^V«)

/. (^#-1) (ar— 2^/^+1 + 2 + 4 -/a-) = 0, .-. (Var— 1) (aM-2 v^+3) = 0,

.-. a? = 1 1 (6)

and x + 2x/x + 3= 0 J (7)

From (6) and (5)

y = 4 (8)

Also from (7) </# = — t±*J — 2,

.-. ^=-1+2^-2 (9)

And from (9) and (5),

y'T/ = l + \/^ = ±V~ %

B B

186 HINTS AND ANSWERS [P. 83.

.', y = -2 (10)

.'. simultaneous roots are

x = 1\ x = — 1— 2*/-2l a- = — 1+2*/ — 2\

y = 4 J ; «/ = - 2 J ; y = -2 J '

By combining (3) with (4) other sets of simultaneous roots may be found.

5. Let x = A's rate per minute,

y = B's . t

.'. 6.2? + 2 = length of course,

= % + y + 20,

ai a , n 6x + 2 Also, 4 (.*•-#) = 4^-,

.'. 6x — 6y — 18t S80y - 874a + 2 = 0 J '

.*. 880 (a?— 3) - 874 a^ + 2 = 0J '

."., 6a? + 2 = 2640 yards = length of course.

6. Let x == original revenue,

y = interest of debt, z = expense of collecting x, .*. far = increased income, 4 s = expense,

•*• fa-— y — | z : a; — y — s : : |f : I and -?T x — y — %z: x — y — z :: -i\: 1 and T9T x — y — £* = 4000000

.'. 117a- - 52y -78z = 204 (a;— y —2) 207a? - 368?/- 276s- = 48 (* - y 9%—16y—12z= 64000000

.-. 87a- - I52y - 126s = 0 ....

159a-- 320^-228* = 0 (2)

9% _ i6y_ 12* = 64000000 . . (3>

Multiplying (1) by 2, and (3) by 21, and subtracting, we have

15a: -32y = 1344000000.

y

P. 83.] IN ARITHMETIC AND ALGEBRA. 187

Multiplying (3) by 19, and subtracting (2), we have

12# + % = 1216000000,

/. 3r + 4y = 304000000,

and \5x + 20# = 1 520000000 ^

But \5x - 32y = 1 344000000 J '

.-. 52y = 176000000,

Also \5x - 32y = 1344000000 1

24# -f 32?/ = 2432000000 i '

.'. 39a: = 3776000000,

.-. x = 96820512 ff/.

44000000 QQQ,,K 5 2/ = --j^ = 3384615 ^l.

z may hence be found.

7. Let .r yards per minute be A's velocity at starting,

y ...... B's

z C's

to yards the length of A's stroke, v that of B's. Then, since velocity a number of strokes x length of them, dato tempore,

: . x : z : : w : w — 1 . . . . (1). Again, whilst A takes 42 strokes, B takes \ x 42 or 49, and since A has then got 42^ + 20 yards from B's starting place, by the question, we have

» + 0 + tV) + 0 + tt) +...'..» + #. = 42m; + 20-1 + 16, or (2v + 4)V9 = 42w + 35, '

or 70 = 6w - 9 . . (2). Again, A having now gone 42w yards; C, 42m?— 42 yards; and B, 42iv — 1 yards ; it follows that C is 61 yards behind B. Also now C's strokes are exactly as quick as B's. Hence, by the question,

v + ff + O + ff) + O + H) + • • • • 28 terms = w~~ l + (w— 1+4] + (20—l+f.J + . . . . to 28 terms + 11—61,

or (2v + 8 — $ x 27) 14 = (2ic - 2 + i x 27) 14 - 50,

which gives

7v = 7w-16 (3).

188 HINTS AND ANSWERS [P. 84.

From (2) and (3)

w = 7, v = 4 f yards per stroke. Finally, from (1)

x : % :: 7 : 6 (4).

x : y : ■: \ w : v x & :: 42 : 33 :: 14 : 11. So that the velocities of A, B, C are respectively as

14, 11, 12.

ST JOHN'S COLLEGE, 1830.

1. Ans. 1 I,

2. Ans. 13 \,

3. Ans. In 104 days.

4. The Danish stock is worth 2350 / ^L or 47 * 30/.

I UU 8

This will buy Russian stock =

100 47 x 301 100 x 47 x 301 x

JlOi 8 887 1414700

or

887 ' Now, interest of Danish stock

3 x 2350 = 34±7 = !«*

100 ~ ~ 2 _ 2

and that of the Russian

J_ 1414700 70735 6Q2

20 X 887 ' ~~ 887 ~~ "887 .'. difference required

662 1

887 " " 2'

- 9^

1774

5. The G. C. M. is cr* + 2a?"1 +1., or (— )!

P. 85.] IN ARITHMETIC AND ALGEBRA. 189

6. Each set of three consonants and one vowel may make

4. 3. 2. I words ; hence, with three given consonants, and the

five vowels, there may be made 5. 4. 3. 2. 1 words.

19 X 18 x 17 Again, there are — ] ~ — different sets of 3 consonants.

Consequently, the number of words required is

19x18x17x4x3x2x1 0™,.

. — — — or23256.

1x2x3

7. See Private Tutor, vol. i, pp. 22, 262.

8. (27^-25^) (27^ + 253) = 27-253~ = 27-6253, But*;3— 1 == {x-l) (x2+x+l),

■■■& »}{(i)^w^}=e-i>

.-. if , = (27* + 25*) {(|f)* + (|f)* + 1} ,

1 i

= (27^ + 25^) I25 x 253 + 5x53 + lV

v r ; V 272 27 ^ /

., n, , .• iv- »* 625 19058

the expression will oe rational, being = zl — -=j = .

9. (1). Gives ans.

J*

/on a o yi—xi-\-2x2y2 (2). Ans. 2. « 4 4 •

± i (3). Extracting the root, it is ± (1— f a?1 + #2).

5a4

10. The first quotient is -^5 b2-\-3b3.

The second is a^(«-1) + .n'(«-8) + . . . . a?0, .". the last term is I.

11. *y(-4-10V-2) = 2-V-2.

190 HINTS AND ANSWERS [P. 86,

Also, ^/(|-V2) = | (^ 6-|/3).

12. Let a; be the common difference of the reciprocals of the terms ; then

x+ ^_ + _^ =11 2 + 2 + # 2+2# 12"

„ 2 + 2a? + 2 +# _ _5

ence' 2(a?* + 3ar + 2) 12'

which gives a? = 2 and — £ and the series continued both ways

4,5 25 ^ 5 25 4» 6) 8

nnfl 5 5 _5_ 15. 5 5_

cUIU. 3U 245 175 25 35 45 11

13. Let r be the common ratio of the terms, when n Geo- metric means are inserted between x and u ; then

x rM+2_1 = m.

•■•^(jh"

and the nth mean is a; rn = a; f - W1? Hence, M, = x. (-)"+h M2 = 2^+i x (-J5*1* M3 = 4*H a; (^ Wi +

&c.

Mi_ 1 M2_/lN^ M3

•-T2- ^ HC=W*H' ^^y^ to, terms,

, . /1\— /V+i\_l_

.'. their sum = n. \o)n+1—\~Qjr)n+l'

14. Let C and C be the combinations taken r together and (r+1) together respectively; then

~ m (m— 1). . . ,(m— r+2) p, w (?rc— l)....(m— r+1)

C _ ' 1. 2 (r-1) "' U ~ 1.2 r ' "'

C m — r+1 m-j-1

C r r

P. 86.] IN ARITHMETIC AND ALGEBRA. 191

But, if for any one value of r, C be < C ; then, for all higher values of r, C is still a fortiori less than C,

.'. when the number of combinations is greatest,

1 is = or < 1,

r

m-\-\

and .'. r is = or >

2 '

.. ,-,. , «, ,„ „ (a2-{-b2—c2~d2)ab

15. First, a2-{-b2— a2 = v— - — z 1— — ,

ab—cd

/a2+62— aA2_ (a2 + b2-c2—d2)2

' V 2ab ) ~ ' 4 (ab-cdf '

_ 4 (ab-cdf— (a2-\-b2—c2-d2)2 4 (ab-cdf

\2(ab-2cd) + ti2-\-b2-c2-d2l (2ab-2cd-a2-b2 + c2 + d2)

4 (ab-cdf '

{(a+bf-(c + dfl \(c-df-(a-bf\ 4 [ab—cd)2

(a+b + c + d) (a + b—c—d) (a—b + c—d) (b + c—a—d) : 4(a6— cdf

QUEEN'S COLLEGE, 1820.

1. This is the best method in practice;

Let S = .1293131

then 100000 S = 12931.31

and 1000 S = 129. 31

.-. 99000 S = 12802

12802 = 1067 •'• ~ 99000 " 8250"

2. The G. CM. is x—2.

3. The root is .00135.

192 HINTS AND ANSWERS [P. 86.

4. At simple interest, at five per cent. ( Wood, art. 392.)

M

P =

1 + nr'

, . , 90 735 450

. . present value required == , — + ^ — |-

1 l JLT1 4__2_T 1 _J_ i ' '

1 T To L "T TTo L +2?o

800 . 240x735 45x2400, -\ —- — i.

~ 9 r 247 ^ 251 which is easily reduced.

5. Number = a+6. 10 + c. 102 +

= a+b. (3+7) + c(3+7)2 +

= a+b. 7 + c. 72+d, 73 + . . . 4- quantity x 3. .'. &c.

6. ^n is > 3/ 0 + l)ifrc3is >0 + l)2.

Let w = l-f?rc, in which m may be any quantity, however small or great ; then

-v/rcis > 3/(ra+l) if (l+m)3be >(2 + w)2, whence it is evident the proposition is not true ; for, even when m = \, (l + l)3is <(2+l)2.

7. For

(a+ky-ib+ky^at-fr+S (a2-62) £ + 3 (a-b) k*.

8. {\-x) * = \+±x+\at*+&x* + &fa*+&,c.

9. (1). Sum = :

14 x 3"

(2). Sum = »(*»+l\

(3). Here is an error of the press. The series should be

l + ^a? (1— ^(l + v'o?) (1-«)2(1-1-a/^)

orr^(1 + T=^ + (T^+&C-)

P. 87.] IN ARITHMETIC AND ALGEBRA. 19S

Sum of n terms =

i (ns) " 1 i-(i-^)»

l + V^ 1 "^(l + v'ar) (1 — .r)""1

I— a?

The sum to oo = oo .

(4).

Ill .

s=-+^-+7-x + •] 1 11

12. \- r x_\_~r b'

«2 ^»3 .*»

+ I5-+ZT + • ■ whence S =

\r-Xf

10.

n[n-\) n(n-l)(n-2) , w(w-l)...3. 2. 1

JNo— n + I. 2 + 1. 2. 3 +'" "1. 2...(m-1)w

But (1 + 1)" = 1 + n + -j— g- + 1,

.-. No. = 2»-l.

11. (1). x = \.

(2). r; = f and — 3. (3). x = 2.

(4)- x = m±/(2»-m2)' 2

(5). x = 9 and 4.

(6). ,r3-f l=p. (.*•'+ 1),

.-. o+i) o2-.*+i-p) = o,

l±v/(4p-3j .*. a? f= — 1j and —9

(7). #3+ 1+^z 0+1) = 0,

.-. fa+l) (#2-#+l+jFD = 0,

l-p±^(y*-2jp-3) .*. .z = — 1, and — L 2 '

c c

194 HINTS AND ANSWERS [P. 87.

12. Make?/ = -; then ( Wood, art. 381.),

and v3 + - V2 + - V + - — 0,

whose roots are in Arithmetic Progression. (See p. 84, of this work.)

13. Let x and y be the digits ; then

x2 + if = 25, xy = 12, ,\ x + y = ± 7, x — y = ± 1, ■'. x — 4, y = 3, .". the number is 34 or 43.

14. Let x be the number of moidores, y that of guineas ; then

27 .r- 21 y = 12/20,

or9r-7y = 80 1 . ' = ]g 4

and .r -j- y = 16) ' ' •'

15.

^(-4 + 2^-5), = 1 + A/ — 5, and ^(2^ — 1) =1 + A/-l.

16. Let « be the first term, / the last term, and s the sum. Also x the number of terms, and y the common ratio. Then, since

yx— i )

o -	— t*.	y-	-1	!>	i
and / -	= «ya	-1	J
mf =	= %>
.'. 9 =	Jy- y-	-a r
•'• y =	a— ~ 1-		and	log / — log a X — - ^-, 2_ logy	+	1,
				log /—log " log (a— 5)— log	a	=*)	+ 1.

P. 87.] IN ARITHMETIC AND ALGEBRA. ] 9.5

QUEEN'S COLLEGE, 1826.

1. Ans. 0.8332 moidores.

2. sf (379.864) = 19. 4901

and if (.0000574640) = 0. 03S

3. Every odd square number is of the form 4 » + l. Let m = 4 n -f 1 ; then

(m+3) (m+7)_ (4rc+4) (4rc+8)_ Q+l) Q + 2) 32 ~ 32 2

one of whose factors must be divisible by 2, and .'. &c.

4. n may = 10«_1, but cannot = 10*, m . . . = lO^1, = 10P,

71

.*.—... = 109~p, or may have q—p+ 1 digits,

&c. &c. This is also proved reversely from the like rule in multipli- cation.

5. x—a is the G. C. M. required, and the reduced fraction is

3x-a—2b

(x—a) (x—b)'

6. Ans. x = 12 for the first equation,

x = 8 for the second, x =■ 5 and — 4 f for the third, x — 4: and — VV f°r tne fourth, r = 16 for the fifth.

x = 1 i

for the sixth,

.^

For the seventh, make = war, and we easily get a? + b~ ± J |4a2 62 + (a2 + b2f\

u =

2^"- Whence x and y are easily obtained.

196

HINTS AND ANSWERS [P. 88.

7. Let x and y be the numbers ; then

x2 f y1 = 3xy

and 2 *

^2 - 2/2 = - J

By subtraction, 2?/2 = a; /% - -J, .-. 2;/3 = 0 (3y2 - 1),

and subtracting in the first

(32/2-l)2 + 2T- 3*/2-l' •'• V + %4 - 6«/2 + 1 = 6?/2 (3y2- 1) = 18«/4 - 6y2

.'■ %4 = I, •"• V — (t)5 whence * = £• (£)* (3 + </ 5).

8. The Harmonic Mean between x and % is >

T3 « + 6 , , ,

But j? = — 7r— and s = \/{ab),

__ Q {a + fyx/ab a + b

~'(Va+Vb)2' ' J / a / b x2

a +b

= 2.

/?+v si-

9. (1). Common difference = ^,

Sum = (2 + V ) 6 = 45,

(2). Common ratio = f,

(f)10-l 3,0-210 58025 Sum = —^ = __ = __

(3). Common ratio == — f,

« 1 5

hum = -^ = — •

1+1 7

(4). The general term = (2?? + 1) (2n + 3). (Sop Woorf, art. 427.)

P. 88.] IN ARITHMETIC AND ALGEBRA. 1(.)7

q™ - (a»-l)(2n + l)(2n + 3j

Oil 111 77 "+■ g.

10. These are both recurring series, whose scales of relation

{Wood, art. 414.) are

f+g = 2- 1, and/+g =-2-1,

. _ l+2#— 2a? 1 . _,, 1— 2a: + 2*- 1

• ■ o = -i s »= 71 st, and o =

1 _2ar+ a?2 ( 1 -.r2)' 1 + 2x +.e2 _ (1 + xf

l 7 14 14 7

11. (1-^f = 1 - g*8 + -^ - -./•« _ ^+ &c.

12. It = a (l- n + n ^ -&c.)

_ b (n -2n.n-^.+ X n-(nill^~2)- &c But (I - 1)" = 1 - n + rc.^=^ - &c. = 0,

ml

.*. It == — bin — 2n.—^- 4- 3. — * . '\ ' — &c.)

13.

= _ bn (I - 1) "-1 = 0.

(a+6)*

(l + -)"*= 1 - - * nearly.

14. ^ (45 + 28 x/ -1) = 7 + 2 yV-1.

and 4/(16 + 8^5) == 2^/(2 + ^ 5) = 2. ^tl=^5 + l.

15. (1). Let o, — a, /3, be the roots ; then

a — a + /3 = — 4, or/3 = 4, Hence, the roots are 3, — 3, and 4. (2). Let a — b, a, n-\-b, be the roots ; then a— b 4- a 4- a 4- 6 = —9, .'. a = —3, and (a—b) (a + b) a = —15. .-. 9-fi2 = 5, b = ±2,

198 HINTS AND ANSWERS [P. 89.

.*. the roots are — 5, —3,-1.

(3). heta, b, c, be the roots ; then

a + b + c = 7 "i and a+b = 8 J '

.'. c = — 1,

and abc = — 15,

.*. ab — 15,

a2+2a6 + 62 = 64,

4 «6 = 60,

.-. a— & = ± 2 \

«+6 =8 /'

.*. a = 5, 6 = 3, and the roots are 5, 3, — 1.

16. Let the origin of co-ordinates be at the extremity of the base, the axis of x being along the base ; and let base = a, and given ratio = b ; then, if s be one side,

y1 — s2—x2 = b'2 s-—(a—%)2,

s2 =

(a — x)2 — x2 b2-\ '

. 2 (a-x)2~x2 2_ a2- 2ax-{b2~\)x2 ' • y ~ b2-\ X ~ : b2-T

.'. the locus is a conic section, which it is easy to describe. See Wood, Alg. pt. iv.

QUEEN'S COLLEGE, 1827.

1. s/ .008281 = .091, and fy .000405224 = .074.

2. Let x be the number of days required ; then, since for the same work,

number of days x ^ ~ ,

number or men

1 1

x : m : : T : -,

a + b a

P. 90.] IN ARITHMETIC AND ALGEBRA. 199

ma

X =

a + b '

There appears to be a superfluous condition in the enuncia- tion.

82-7-

3. The value required = ~ x 1743/. sterling.

= 1444/. 10.5. 2^d.

4. See p. 49.

5. At simple interest, ( Wood, art. 391.),

M = P(i + nr).

At compound interest, ( Wood, 397.),

M'=PR»;

.'. by the question,

P(l+2r) = |-P. (1+r)2,

,\ 8 + 16r = 7+l4r + 7r2,

,. . . 1+2^/2 3.82 . which gives r = = = -=— nearly,

54*

~ 100'

6. Every odd square number is of the form 4 n + 1 , ( Barlow 's Theory of Numbers) ; but any number, whose two last digits are 9, is the form 9 +9 x 10+ 100 x N ;

if this be a square number, and .". of the form 4;? + 1, 8+9/ 10+ 100 xN, it is divisible by 4, and .". 9 x 10 is divisible by 4 ; which it is not. .". " No number whose," &c.

7. See Barlow's Theory of Numbers.

8. The G. C. M. = .r - 7,

x1 + a1 and the reduced fraction — '

:r + b

9. Firsts =i (r + I) (y + i),

y x 1 x y xi)

I l«y+?+=+ ™

iV-.^-iy,

200 HINTS AND ANSWERS [P. 90.

Also, 1-^=1 -i (,:+ i)2= -i(. and 1-* = 1 - 1 (y + I)' = - ■ i (y - I)', /. ^(1-*). /(I-*) = * (x - i) (y - i),

3"

* \Xy~x~~ y + xyr

10. The index being n, thepth term being P, and (p + 2)th — Q, we have

n. {7i- 1) (m-2) .... O— yj + 2)

1. 2. 3 (J»-1)

a?-1 = P,

, n (n— 1) O— 2) .... (»— « + 4) n , _

and -1T-2 3 .... (^-3) ^ = Q' . (n-p + 3)(n-p+2) _ P

•• (p-2)(p-\) ■* -Q'

whence « is easily found from the quadratic.

11. See Private Tutor, vol. i. p. 263.

12. Let # be the first term, and y the required ratio; then

a being the given sum to oo , and b the sum of the cubes, we

have

x a

and b = x3 + x3y3 -f x3y6 -\- x3y9 + &c. to oo ,

= X3 (\+y3 + jf + lf + .... QO ), I

= X3.

\-y3

l-y3 • 1+y + y*'

.". by2-{-by-{-b = a3—2a3y + a3y2,

whence y, and then x, are easily found.

P. 91.] IN ARITHMETIC AND ALGEBRA. 201

13. Let x be the common difference of the reciprocals of the series, which are in Arithmetic Progression ; then these reci- procals are

2, 2 + x, 2 + 2a, and, by the question,

1 1 1 1_ 13

2 + 2+x + 2 + 2^ 12 - 12' .'. (4+3ar) 12 = 7 (4 + 6x + 2x2),

which gives x = 1, or — "T°,

.'. the series is either

— 1 oo 1 i i A &<*

or -7- A 2- 2.

14. (1). The Common Difference =—-f, and Sum = 0.

f&i « 105469

^- Sum= 354294'

(3). Common Ratio

and Sum

2s/-\

i 4 + 2/-1.

1 - 15

2-/-1

15. (1). Ans. a; = 16.

(2). ar = 6, and- *&>, (3). a- = 7, and - 5T8-

(4). First a;2 + 9+V(^ + 9) + i = ^4-a;2 + |, .*. -/(a;2 + 9) + £ = x>-~\, .-. a;2 + 9 = (.r2-l)2,

= x4—2x2+l, Hence, ar = ± £ -/ (6 ± 2-y/4l). (5). Cubing the second,

x~y+3^/(xy). (i/x-^/y) = b,

b — a

(6 -a)3

D D

202 HINTS AND ANSWERS. [P. 91.

Hence x + y = ± i/ |«2 + ( ~^> } ,

but a- — y = a. Whence .r and y are easily found.

(6). These being homogeneous, make y = m<- then w2+ V w = V w^ giye u'> whence x and y.

16. • (21 + •-400) = 5 + 2^-1,

V (76 + 44^3) = 1 + ^/3.

17. ?/i (3* + 52) = m. K34)^ + (53)^L

and /. m = (34)^ - (34)* . (53f + (34)* (53)^ - (34)^ (53)*

+ (34)^ (53)* - (53)^,

= 3'T° - 3*. 5* + 32. 5-3^. 5! + 3y 52 _ 5-^

= 27/^3-9^9 + 45-34/3x5^/5 + 254/9-25^5, = 45+27^/ 3+164/9-25^5-15. ^5. </3.

18. See the Preface to any System of Logarithms.

The same series, although it may be the best for a certain set of numbers, is not the best generally.

19. Since ^19 = 4+^/19-4 = 4 +

-y/19+4'

.-. ^/19 -.; 4 +J_

2+1

1 + 3

3+1, &c.

20. (1). By the Method of Divisors, its roots are found to be 3, 4, and — 1.

(2). The equations x4— 5x3— x4 + 5x3-

have a common measure of the form x2—a2, and .*. their difference is divisible by a quantity of the form

!— 5^2 + 45^— 36 = 0| 5-5#2-45.r-36 = 0* '

P. 92.] IN ARITHMETIC AND ALGEBRA. 203

x2— a2. Consequently x2— 9 is that factor, and two of the roots are ± 3.

(3). This is a misprint for ax x b~x = c.

logc

Ans. x =

log a— log b'

21. (I) The (w + l)th term of 3. 5 + 5. 7 + 7. 9+ is

(2rc + 3) (2»+5). /. Wood's Alg. art. 427.

3. 5 + 5. 7 + 7. 9 + . . . n terms =

2rc + l)(2n + 3)(2rc + 5) 6

5. 2 J

. , 4^+18rc2 + 23rc-30 .". sum required = — ^ .

(2). The (rc + 1)* term is ^—-1^^-^

1 n i l'

.'. sum = — ^ Q n + C or = — ^j-

2(2» + ir 2 2(2« + l)'

4»_ i

(3). This is Geometric, and Sum= — „ — .

22. See p. 1 13.

QUEEN'S COLLEGE, 1828.

1. Ans. Quotient = 21. 4.

2. Ans. .0198, and .54.

3. Ans. The accumulated dividend on the 106/. stock amounts to more than that of the 100/. stock by 1/. 4s. %d. f .

4. Since (2m+\f— l=4m2 + 4m=4m(m+l),and either m, or m+ 1, must be even,

.*. (2w+l)2-l is divisible by 8.

5. Ans. 18. ~4 years. .

6. Ans. 1828 in the septenary scale is 5221.

7. SinceV-l=/(0+/-l), the Ans. is/K1 + /-!)■

204 HINTS AND ANSWERS [P. 92.

8. If P be the pth coefficient, then P. n~P+l [s the (p+ l)th ;

.". their difference is 3 = — I p — — pr— ) .*, &c.

2p \ 2 /

See also Private Tutor, vol. i. p. 260.

iG^r-G^r-o-sr-

2x n+l 22 x2

1 + n- r+i+n- ~ (TT2? + &c'

, ~ «. i i n—p n—p—\

10. Since a:b :: 1 : ^ • r 0 , we get

jt?+ 1 jo + 2

And, in this particular case, n = 12.

11. Ans. £rc — (p— 1)^ (w— p)+l-

12. If a + ar1', and arm + arn_m be the respective sums, and c their difference ; then

3 = a (1 — r») (1 — *-"-"■), which is always positive, when n is > m.

13. Let 2 w be the number of terms, a the first term, r the ratio, s the sum of the odd terms, and s' that of the even terms; then

.<? = a + ar2 + ari+ .... n terms,

r

n

1

= a. t-3

r — 1

s' — ar + ar3+ .... n terms,

ar.

r

n

I

r-V

.-. s : 4 : : 1 : r

14. (I). Ans. Sum = 0.

(2). Ans. Sum of 12 terms = £$&, and to oo = |.

P. 93.] IN ARITHMETIC AND ALGEBRA. 205

(3). Ans. by Method of Increments; the

(2n+l) (2n + 3) (2n + 5) . rTXT . .onN

sum = v — ±-AS — _H_ LI 1 _ o m ( ^ood, art. 420.)

„N . 3n*+Wn* + 9ni + 2n ,, T (4). Ans. - — |o— (by Increments).

15. Increment of the sum S of the series, is

a S = A n. A„+1.

... S = An-y ^ A>i+1 + correction,

An-!- A„. Aw+1 — A0. Ar A2

~ " 36

6 being the common difference of the progression.

16. (1). Ans. x = 8. (2). Ans. x = 7.

(3). Ans. # = -|, and T3o. (4). Ans. x = 5, and — 2, (5). Ans. a? = 7\

it = 21'

(6). Ans. x = 49 1 , x — 25 \

W y = 25}' and y = 49/'

17. Since ^(12-4/3) = 2/3^/(1/3-1)

= 1/6. ww^+v^-vw*-*/^,

the extraction, as required, is not feasible. The question is most probably misprinted in the paper.

The required cube root is \/3+\/2.

18. The roots are 1, — 1, 3 and 4, and are found by as-

12 suming them of the form a, — , b, c, and making use of the

coefficients —7, —12 and 11.

19. Supposing the roots a, —a, b, c, we have

b + c=p, bc—a2= q, and a2bc = s.

Wl1e„cc&C=g+/(|2-4;),

206 HINTS AND ANSWERS [P. 94.

and .". 2b = jo + /|/-2g-2/(?2~4s)|, 2 c = p—Slp2-2q-2\/{q*-4s)\,

mda=1//(g»-4l-gf

20. Ans. A = ._ i B = _ Aj C = fcwdD = - «.

21. Log 512 = log 29 = 9 log 2 = 2. 70927.

22. Ans. ,= ^J^_V^±?)j

log a

oq o- k 7x2+1 ^o. Since 5 = ^-^- ,

7x22-1

9 =

3

19 = 7-2i|±J, &0.

.'. the nth term of the series is 7x2"-'+(-l)" 3 '

.*. the series consists of two Geometric Series, whose general terms are

i X 2*~\ and [-£± ,

whose common ratios are

2, and (-1), and first terms

-?-, and — j.. Also, the sum of n terms is

7x2"+1-15-(-l)»-i 6

Observe. The series is recurring ; its scale of relation being 1+2.

24. Ans. x = 7, and 7 = 5.

P. 94.] IN ARITHMETIC AND ALGEBRA. 207

QUEEN'S COLLEGE, 1830. 1. Ans. 21. 10.?. 9d.

O A no ' 7

3. Ans. Is. 6d. a day.

4. Ans. 251.

5. At simple interest

P

present value = -^ -■ ( Wood, art. 392).

At compound interest

P

present value = ^ .(^Fooflf,art. 398).

Hence, in the former case,

p. valUe = 8^f 9Q^ - 734/. 3*. Od. J?' 1 -1- 5| x 3J_ 53

100

and, in the latter,

875. 9. 6 35019 x 1010

F. value _ - 4 207,, I-

which may easily be completed by reference to Logarithmic Tables.

For log P. value = 10 + log 35019-2- V- log 207, = 8 + log 35019 - V -log 207.

6. Ans. See Wood, for the proof.

The quotient = 12500.

7. The quotients are x — sfx, and (a?— 2) (>— 4).

8. Ans. The product

a4 — x4 d2x

9. See Wood, art. 92. The quantities have no common measure.

208 HINTS AND ANSWERS [P. 94.

10. (1.) Ans. x = 7.

(2). ar = 2, and — f

(3). x — ac + b2)

ab — c

(4). To the first add 2 x xy; thence x + y = 5, and — 6. Square this, and subtract 4txy, &c, then x — y — ± 1, and ± \/ 12, .-. 2x = 6, and 4 also — 6 ± -/ 12, and 2y = 4, and 6 also — 6 + a/ 12, .". the simultaneous values are # = 3, x = 2. j? = — 6 + ^/ 12. x = — 6 — </ 12i y = 2/y = 3J y = -6- ^ 12/ a: = -6 + -/ 12-* ' (5). From the second ^2 + 2xy + if — b2 — 26 </ (a?y) + #y,

and XV = ~^~

Add this to the first of the given equations, which will give

(b2—a)2 x + y; and subtract 3xy = 3. y from that equation, and

it will give x — y. Whence * and y.

(6). Squaring the second, we get, by means of the third, xy -f xz + yz = 56, .'. xy +j/2 + yz = 56, 56 56

^ x + y + 2 14 Hence x2 + z2 = 84 - 16 = 68,

and 2xz = 32, .'. a: + z = ± 10, and x — 2 = ±6.

11. Let # = number he purchased; then they cost him

— a head, and he sold the x — 3 oxen at + 8 a head,

x x

P. 9.5. ] IN A KITH M E T I C AND A L G E BRA. 209

.". by the question

/240 \

(x - 3) f— + 8 J = 240 + 59 = 299,

from which, by the solution of a quadratic, it is found, that

x = 16.

12. The sum of the first series = — 275,

3 that of the second = j^.,

13. Let x be the first term, y the common difference ; M the mth term, N the nth term; then

x -f (in — 1) y = M,

x + [n — 1) y = N,

M-N

.*• y — ,

m — n

, ' , lx M — N (m— 1)N— (n— 1)M

and .\ x = M— (m — 1). ■ = v }- * }—.

m — n m — n

Hence, the sum of^t? terms of the* series is

r2> (m-l)N-(«-l)M M-N |p

\ ' m—n m — n J 2'

or, \(2m — p — 1) N - (2n + p + 1) M*

P

2{m—n)'

14. See Private Tutor. 199, vol. 1.

The second form is proved by expanding the Binomial according to the Theorem, and then collecting the terms equally distant from either extreme.

15. y/\\ -L- Va - m*)\ = y — §— + V -g—

16. The base of any number N is « in the equation

N = «J-, in which x is defined to be the logarithm of N in the system whose base is a.

For the determination of the base of the hyperbolic system, see Private Tutor, vol. i. pp. 24, 25.

/y.2 //*3 /yi4

%JU w f€/

JYap. log (1 + a?) = x —g + -j - -j +

E E

210 HINTS AND ANSWERS [P. 95.

X% X3 X*

.'. Nap. log(] -an) =«— g _ 3" _ T ~

•*• Nap. log J±£! =2 { x + y + y + &c } .

1+tf" 1 1 Let 1-±— = 1 + -, i— x c

thence .r = ^~r

and log(l-fc) = logc + 2 { JL + 1. ^Lj. + &c. } .

1 7. Let x- -\- y* = 2% and make 2; = nx -\- y ; then s2 = »2a;2 + 2 nxy + y2, I— »2

?/ = ~2^ *'

and

*2 + 2,2 = *2 {l +(-1^2}&c.

= a;

,2(l + rc2)2

4w2 '

which is a square, whatever may he the assumed values of n and x.

18. See p. 46, No. 7.— p. ,59, No. 6,— p. 61, No. 19.

19. This is a recurring series, whose scale of relation is

/ — 13, g = — 44 and the

c 1—8 ar

hum to qo = ar. , — ==-?: . . „-■„•

1 — 13ar + 44a2r2

To ascertain the nth term of the series, let Cv C2 . . . Cn be

the coefficients of the successive terms; then, making a = 13,

and 6 = 44, we have

C3 = «C2 - 6C, = aC2 - 6,

C4 = «C3 - 6C2 = (a2 - b) C2 - a6,

C5 = (a3 - 2a&) C2 - a*b + 62,

C6 = (a4 - 3a2 6 + ¥) C2 - a3 b + 2a62,

C7 = (a5 - 4a3 6 + 3a62)~C2 - a*k + 3a2 62 - 6»

C8 = (a6 - 5a4 0 + 6a* 6* — 63) C2 - a56 + 4a362 - 4ab\

C9 = (a7-6a564-10a362-4a63)C2-a66+-5a462-6a263-f 64,

P. 96.] IN ARITHMETIC AND ALGEBRA. 211

C,0= (a8 — 7a66+15a4&2~10a*63+64)C2-a76+6a5#l-

10 a363 f 4a&4, &c. and, generally,

C„ = {a»-2-(« -3) a"-* b + (---^-Z^)a«-« p

_(,-7)(,-6)(,-5)aB.86; + &c}Ca

+ (»-8)(»-T)t»-6)a...ft,_ftc-

whence may be found any term of the series, a being 13 and b = 44.

But the sum of the series to infinity, commencing from the (rc + lth) term, is {Wood, art. 414),

Cw+1aw+i r«+i + Cw+2 a"+2 6*+2 - 13ar. Cw+1 a»+x b"+\ 1 _ 13ar + 44a2 r2

and, consequently, this sum from that of the whole series to oo ,

we shall have the sum of n terms of the series.

20. Let a — b, a, a -+ b, be the roots ; then

3a = 3 and .". a = 1. But this does not verify the equation; consequently the roots are not in Arithmetic progression.

21. See p. 94, and Wood, art. 341.

22. Wood, art. 352.

CORPUS CHRISTI COLLEGE, 1827.

1. The duty is ^Vo X (2377 + j- + ^) + 5763 per bushel,

^109157 °r> 10768400 PenCe-

2. Amount a number of men x number of days x wages per day,

amount

.'. number of men cc

number of days x wages per day'

212 HINTS AND ANSWERS [P. 97.

and if x be the number of men required ; then

X * 24 x j * "* ' Iff which gives .r = 45.

3. i — ix is the sum paid yearly as premium for the assurance ;

i f

- x 100 is the sum his executors will claim at his

a decease ; the annual interest of this is

? — % T

l x 100 x y^r = ix by the question.

whence ?', =

1 " " a + r

4. Wood, art. 20 and 377.

5. The G. C. M. is & + 7,

and the L. C. M. is 36.

6. This is evident from the continued multiplication of (a ±b)(a ±b)(a±b) . . . , to n factors.

Those who are not satisfied with this sufficient proof, may consult Barloiv's Theory of Numbers on the subject.

7. See Private Tutor, vol. i. p. 262.

8. Let x, y, z be the gallons per minute discharged by the first, second and third spouts respectively ; then

O+20 32 = 384^ (x+z) 24 = 384 and (x +#-f- z) 16 = 384.

.'. x = 4, y = 8,z = 12,

9. (1). x = 13, y = 3. (2). ~- = | and £ .\ x = 2 and - 3.

10. x = 8 and y = 4.

11. (1). The four values ol'.r are ± y7 (1 ± l/i)-

P. 97.] IN ARITHMETIC AND ALGEBRA. '2 13

(2.) The first equation (the mistake is in the original paper) should have been

3x /x+y _ n

V x+y ' V

S.r

From this °^1 = 3 ± 2 V 2,

y= (8 ±6/2)*, (8 ±6/2) «2-(9±6 v/2) x = 54, 9 + 6/2 54

'. x

X ~ 8±6/2 X ' ' 8±6/2 ' a 3±2y/2±/209±156y/2)

*' 4 ± 3/2

which comprises four different values, each of which may easily be computed in decimals, after first rationalising the denomina- tor.

12. (1). The common difference = — j

3 5

and the sum = I 1 ^ )10 =

3; 3

(2). The common ratio is f ,

1

and sum = — = %-. 3

(3). The series is of the form

x (l+3.£ + 5^2+ &c.) which is recurring its scale of relation being/'+g,=2— 1,

.'. the sum of n terms _ . l+«-(2.+ l)-.+ (9.-l)^H) (froorf>art.4140

		0	— ,r)2
	A 2	2/* + l + Ore '	2/2-1
1	2n+l
a*		(i)2
3	—	2 ft + 1 2 ; 2*-i +	«- 1 Oft
3.	2"	— 2rc-3

13. Let S == <7prrr .... then

i0?+»+« S = qpr. rrr .

and 10/f'" S — qp. rrr .

214 HINTS AND ANSWERS [P. 97.

• • O

qp. (r— 1) qp (r— 1)

10'+'" (10" — 1) 999... n nines 0000... (l + m) zeros'

14. Let r, -r-r , — -.. &c. be the Harmonic

a + 6 a + 2 b a + 3 b

series ; this form being assumed, because the reciprocals of

quantities in Harmonic, are in Arithmetic progression.

Now any two pairs of consecutive terms are of the form

1 1

and

l and

a + nb ' a+(?2 + l)6'

1 1_

a+mb ' a + (?«+l)6' 1 1

a + ?ib a+(n + l)b (a + nb) \a + (n + 1)6£' 1 1 b

a-\-mb a+(ra+l)6 (a + mb) \a + (m+\)by whence the proposition is manifest.

15. Let x and y be the two numbers ; then

£+2 = -/(ay) +13

>•

and y/(xy) = — '$■ + 12

v ai %+y

Hence, immediately

x + y — 338l . a/x + y/y = ± y/650\ and 2 </xy = 312J * ' */x — </y = ± -/ 26J '

whence x and y.

16. /(2V- 1) =/(0+2^ — 1) = by the rule 1 + ^/ — 1.

4/(11 + 5 w) = M.

17. See Private Tutor, vol. i. p. 262.

18. Transform the number 1319 to the ternary scale ; in that scale it is 1210212,

.'. 2xl + lx3 + 2x32+lx34+2x3Hlx36 = 1319.

19. Adding the two equations together, we get

3x — y =. 12,

P. 98.] IN ARITHMETIC AND ALGEBRA.

215

.*. y = 12 — 3x, .'. making x = 0, ± 1, ± 2, ± 3, &c. the values of y = 12, 12 + 3, 12 + 6, 12 + 9, &c.

= 12,

and x = 9, 6, 3, 0, —3, —6, &c.

y = 15, 18, 21, 24, 27, 30, &c.

Hence z = 5 + 2 y—x = 29— 6 x,

= 29, 29 + 6, 29+12, 29+18, &c.

= 29,23,17,11, 5, -1, -7, &c.

35, 41, 47, 53, 59, 65, &c.

Whence the simultaneous values of x, y, z are

# = 0, 1, 2, 3, 4, 5, 6, 7 y : : 12, 9, 6, 3, 0, -3, -6, -9

2 = 29, 23, 17, 11, 5, -1:

13

and x

y

z

-1, -2, -3, -4, -5, -6 15, 18, 21, 24, 27, 30 35, 41, 47, 53, 59, 65

)

20. See Private Tutor, vol. i. p. 24.

CORPUS CHRISTI COLLEGE, 1828.

1 3

1. -I of 2. 6 = 1 shilling. .\ the Ans. is -y^- or^.

<¥H 1 3

2. The cost is 650 x -j^=^ * 181 = 588^- 5s-

„ „x „ , length x breadth x depth

3. No. of days ce — — r —

men x hours

. if x be the days required, we have,

420 x 5 x 3 _ 230 x 3 x 2

24x9 '• * 248x11 '

25x42 248x11

x

/,

24x9 ~ 23x2 '

13

= 288 days 2 ^ hours.

*216 HINTS AND ANSWERS [P. 98.

4. Let P be the present value ; then in n years the amount of P is M=P + rcrP, {Wood, art. 391.) and the amount of the annuity in that time is

a+2a + 3a+ .... n terms

+ r £a + 3a + 6«-i-10a+ [n— 1) terms|

.,„ n (n—\)n.(n\-\)

= I2a + a. (n-l)\ ^ + «**. — i 2 3 — '

6(1 + nrj l

5. 5^2_i is the G. C. M. required.

6. See Wood, for the proofs.

The quotient is 12450.

7. (1). Multiply by s/{x+*/x)', then

x+y/x— ^{x2— x) = 4 v^-

Whence # = -f-f .

(2). Multiply the first of these by 60, and the second

by 24, reduce, &c, and there results

32 x — 15 y = 84) 12 x - 13 ij — - 46/ '

whence x and # are easily found.

8. (1). V(a + *) + V{a-*)=-jQ>

/p2 \2 ^4

= (j2 ~ aJ ="T44 6«^24-«2,

a?4

(6a-l)#2= 0,

144

,*. a? = 0 ; but this does not satisfy the equation. Again, x2= 144. (6«-l), /. x,= ± 12 -v/(6a-l). (2). From the first,

P. 99.] IN ARITHMETIC AND ALGEBRA. 217

XV

.'.-4-- = ± ^5,

y x

.'. x2+y2 = ± a/5, xy. But x2 + y2 = 4 + 2 xy,

:. 2xy = 8 (2 ± -v/ 5), and x2+y2 =4 + 8 (2±</5). •"• 0+*/)2 = 4+16(2+^/5),

= 36+16^/5 = 4 (9±4v/5), /. # + ?/ = ± 2 (\/5±2),

= 4 ± 2^/5. But #— -y = 2,

.'. a; = 3±2 ^5, y = 1 ±2 </5.

9. Let a:, y, z be the numbers ; then, by the question,

x (y+z) =s 26^

2 (x+y) = 56J

Adding them together, we get

xy + xz+yz = 66,

.'. yz = 40^ 372 = 16 V. xy = 10J

Multiply these together, and there results,

xHfz2 = 400x16,

.'. a?y2 = ± 80,

.-. ar = ± 2, y — ±? 5, z = ± 8.

io. s,=$a+(»-i)$j£ s2=i4+(«-i). 3^ i

S3 = \6 + (n-l)5\ | + &c. .'. S, + S2+S3+ ... Sp= 2 x P + -g- (1 +3 + 5+... jo terms),

= ¥ + ^ |2+2)c(p-l)||,

11. See p. 66.

F F

218 HINTS AND ANSWERS [P. 99.

i« T , o a a + \b a + 2b

12. Let S = --I V-+ z h . . . n terms,

a a a+ b = -+ -o-+ — r- + . ■ . n terms,

+ -r+ — + . . . (n— 1) terms,

,y»2 <*• 3

Vry 1 ra a

a b

r r2 1 .

+ b

1 ra a+ b ... 1

«--- {-+-j3r+...(»-l)tenns}

r

b l_r"-i 1 j a+[n—l)b\

6 r«-i_l g + (w-l)6 . . S. (r-l)= « + -^-r- -— j p. 1

/•" 1 & frn-\_ 1 w. — J>

ft /rn-\_ 1 W._l\

*=f' V r— 1 r~/'

= a ^— + - . 1V (rB— »r + »— 1),

rn_l rn—nr + n — \

•'■ S = a- r»(r_i) + 6l rMr-1)2 * To apply this form to the numerical example, we have a — \,b —2, and r = 3, 3"— 1 3"— 2 rc— 1 •'■ b~ 273~"+_ 2x3" ' 3"_n_l — 3S >

13. See Private Tutor, vol. i. p. 262.

1 2. 3. 4 .... w*

14. The number is -1* — —^ — 5 — -.

la 4/. O a a a P

Thus, the factors of a3. 6. c. may have 1. 2. 3. 4. 5.

1. 2. 3

different arrangements.

= 20

15. Let x be the common ratio ; then

1 13?

xn = ,rn+1 x

P. 99.] IN ARITHMETIC AND ALGEBRA. 219

' -^--1 *• 1-3? ~ '

and x = ^,

.*. the series is 1 + \ + \ + • • • co .

16. See Wood, art. 259.

4/(2/7 + 3^3) =£(V7 + /3).

i i ax c

17. Let ckc be the number, and make -r = y + t,

then ax — by = c, in which integer values of x and y will give the Answer. In the particular instance,

\0x-\7y = 9,

7m+9 •'• * = y + "To- '

Make 7 m + 9 = 10 w,

3iv — 2 .'. y — w — 1 -I =: — .

Again, make 3 m>— 2 = 7 v,

o L v+2 . . w = 2 v 4- — k— .

Make t> + 2 = 3 m,

.*. v = 3 m — 2,

in which m may =1,2, &c.

Hence, 3 w = 21 u— 12,

.-. w — 7 u — 4,

.-. y = 10w — 7,

- 70m - 40 and x = 10m — 7 -f ^j '

= 17m - 11, .-. ax = 170m - 110, which is exactly divisible by 10, and whatever is the value of m when ax is divided by 17, it will always leave the re- mainder 9 ; and these numbers are

60, 230, 400, &c.

220 HINTS AND ANSWERS [P. 99.

18. If possible, let

a a

have the same remainder r ; then if q, q are the quotients,

(m <f>)9' = qa + R,

[n<f)2 = q'a + R,

, d>2 (m? — n2) •'• q — q = x-3 — -an integer.

But (f, is prime to a, and .'. <f is.

Hence, m2 — n2 is divisible by a; that is, (m — n)(m + n) is divisible by «. But a is prime, and .". either m — n or m + 7i is divisible by a. But each of them is less than a

.: &*»* and (M2 a a

cannot have the same remainder.

19. See Wood, art. 383, or Woodhouse's Trigonometry, p. 211.

CORPUS CHRISTI COLLEGE, 1830.

1. His tax amounts to

19

(1384 + f) + (^ + i|),

or - ,^ or 193/. 5s. 9$d.

5539 x 67 1920 the net income is 1191/. 10s. 2%d.

2. Their Least Common Denominator is 90, and the sum of

the fractions is

36 + 60+50+63 209

' ' or

90 90

3. The quotient

510000 42500

~ 342537456 " 28544788 '

- 10625 - 0014 &c ~ 7136197 '

V 25405534 in the senary = 1503.51 nearly.

P. 100.] IN ARITHMETIC AND ALGEBRA. 221

It does not terminate,

z cwt. = 3^ quarters = 3q. 31b. £ = 3q. 31b. 1| oz. .42857 month = I week, 4 days, 23 hours, 59 minutes, 56 seconds.

2 3 10981

4. The area of the field ~ 52 ^ x 34 j or — - — square ft-

10981 25. 54905 OOQ_ .. _, .-. the price = — - ^— x ^l. = -^— = 2287/. 14s. 2d.

5. Present Worth = . SUm ( Wood, art. 392.)

1 -f nr

7 543 + 20 10867

130 37 , 481

365 800 ' 29200

10867 x 1460 16166

29681 "29681 which can be reduced to shillings, pence, &c.

6. See Wright's Self-Instructions in Pure Arithmetic on the subject of the Cube Root.

7. See the last quoted work for these subjects in Arithmetic, and Wood for them in Algebra.

8. (1). The G. C. M. in the first, being x2 + xy + y2, the

fraction becomes

x*— xy+y2

x2 + xy + y2'

m This - (^-i)(*3+i)

(ex-l)(ex+l)(x2—x+l){x-bl) (e*+l)20r-l)0z+l)

ex — 1 x2— x-\-l z ex+V x—\~ "

9. Wood, art. 377. I 1

10.

(a— b) (a—c) (x + a) (a—b) (b—c) (x + b)

222 HINTS AND ANSWERS [P. 101.

1 (a— b)x + a2— b2— c(a— b)y a—b' (x + a) (x + b)

x+a+b—c (a—c){b—c)(x-{-a)(x + b)' Ai 1 f 1 x+a+b— c -]

■■ *e a^eSate = (a-c){b-c)* lFi^-(»+a)(»+a)}

1 ab—ac—bc + c2

(a—c) (b—c) (x + c) {x + b)

1_

" (x + a) (x+by

11. (1). Ans. x = 13, (2). Ans. x = 20, (3). Cubing both sides,

2 + 3a ^/(l - x) = a3,

/a3-2\3

(a3-2)3 * 27a3 •

a

(4). Dividing by (a71 — #*)»*we get,

an + a?"\i-

r =— 1,

xnJ

fan + xn\±. /an \an — xnJ ~ \qF

/an + xn\± 1 \/3 • ' \an-xnJ ~ 2 ~ 2

/l a/3\w

a" + <p* = (an - xn)(- ± ^-J ,

-a*

.-. arH = a

. /I v^"1'

_ „ 0±v^T=2m " a * (l±A/3)"t+2w'

;(l±^3)»-2-U •• * - a \(i±A/3)»+2'"J •

(5). Taking the logarithms of both sides,

m log a x x -{- n log 5x«/ = log &,

»*' log a' x r -f ri log 6' x y — log #,

P. 101.] IN ARITHMETIC AND ALGEBRA. 223

whence, by the usual method,

ri log k — u log k'

x

mn' log a. log b' — m'n log a! log 6' rri log # — w log k' y ~ m'n log a! log b — mn' log a log 6' '

(6). Let the roots be- a, ar ; then a3 = 64 ,\ a = 4,

and the other roots will be determined by

4

- + 4 + 4r = 14, which gives r = 2 and ±,

whence the roots are 2, 4, 8.

(7). Let a be the first root ; then the others are 2a, 5a, and 8a = 16 ; a = 2, and the other roots are 4 and 10.

12. ( Wood, art. 224).

32415415 - ^ 4- 415 - 32383-

13. Wood, art. 228, for the Permutations.

For the Problem, see Private Tutor, vol. i. p. 260. ex. 41.

— — f (b c\t —

14. (a — b x — cx2)n = a» <l — ( — 1 — x jx I « .

Of this, the 6th term is,

m . m ~ ??i n m .

1 2-— 3 4 ,

— m n n n n (b c\5 .

— a». — — — -. — . — -. - + -) xb,

n 2 3 4 5 \a a

the 5th is,

m in m

_ 1 _ o Q

(b c V - + -x ) x\ \a a )

m m m

--1--2--3 ™ m n n n

a"" n ■ 2 3 4

the 4th is,

m , m

1 - —2

" 7ft rc w foe

— an-

n ' 2

(b c \3

-75— • I - -1- - X J X3,

3 \a a / '

224 HINTS AND ANSWERS [P. 101.

which are all the terms in which ar5 will appear. Hence the coefficient required is,

r m m m m

a I a5' n' 2 3 ' 4 ' ~~5 '

_ — i _ o _ o

. V*c m n n n~

a4 n 2 3

a 2 - — 1 - —2

+ a5' rc' ~2 3~

)

--1 --2

m n n ^~b ,

or — — • • • #« o

n 2 3

r-_3 __4 w n

b* ( - _ 3) «62c + 3 a2c2 j ■

15. /(9 + 2 / 3 + 2 / 5 + 2/15) = 1 + /3 + / 5. To effect this, we try

/ 2/3x2/5, /2/3x2/15_ V 2x2/15 " V 2x2/5 ~^' , . /2/5x2/15

Similarly, if / (a + / & -5- /c 4- /(/) can be thus extracted, the root is

/^bc_ /ybd /vcd

V 2/d + V 2/c + V 2/a" For the other question, see p. 202. No. 17.

16. Wood, art. 190.

17. Let P be the number, whose number of digits is p,

Q q,

then P is < 10p, and Q is < 10«,

.-. PxQis< IOp+9;

that is, it cannot consist of p-\-q-\-\ digits.

Also, P is not < IOp-1, and Q is not < 10?"1,

P. 102.] IN ARITHMETIC AND ALGEBRA. 225

.*. P x Q is not < 10+9-2^ or it cannot have less than />-f q— I dioits.

N5-N_ N(N4-j.)_N(N-l)(N-fl)(N2+]) ~~W ~~ 12" 12

B„, ffl+l)N(S-*l) H an integer,

and v N is odd, N2 is odd, and .'. N2-f 1 is even, and divi- sible by 2,

.*. N5— N is divisible bv 12.

19. (1). Ans. 6$. (2). Ans. §. (3). T tc a aJrb a + 2b a + (n—\)b

L,et O — xm XmJrV #"M-2p + • ■ • ^.w-H/n-l) f *

a-\-b a+2b a+(m—Y)b

.'. S. xm=a + ~^-+ x2p • + • • • " ^.-ijp— *

S,r'»_ a a + b a + (m—2)b a + (m—l)b_

s%m(xp—}) , 6 fi , l , * , __L_\

a + (w— 1)6

no* *

4\(«-i)p

(J -1 a + (n-l)6

1

b x(H-*)r — 1 q+Q — l)j = a + a;(»-i)P rP— 1 #np

_«(^P-1) 6 ^-W^ + B-l

~ ~. «■* znp ' XP—l

I , #»p — 1 a-"?— nrP+n— 1\

(4). This series is

1 J_ J_ T7~5+3. 7+5. 9 +

G G

226

HINTS AND ANSWERS, &c.	[P. 102
LetS=\| + i + 4+ 4 + i + ■••	CO,
S — f = i 4- T + T + TT + TT + • •	. CO ,
•••^=4 (l. 5+3. 7 + --00)'
. series = f.

20. See p. 42. No. 21. Also, p. 61.

21. If v be any integer whatever, positive, negative, or zero,

~ a\ will give all the simultaneous values.

Also, x 4- y = 4 (56— «). .". when # = 56, x + y = 0 is the least.

22. See Wood/house's Trigonometry. Art. on Log.

ALGEBRA.

PART II.

CORPUS CHRISTI COLLEGE, 1880.

[P. 103.]

1. See Private Tutor, Alg. pt. ii. p. 15.

2. Wood, art. 311.

3. Private Tutor, Alg. part ii. p. 111., &c.

4. Wood, art. 307.

5. Wood, art. 318.

6. (1). Wood, art. 325.

(2). Since ^6-l = 03+l) (x*-\)

= O2— %+ 1) (#+ 1) Cra + # + 1) (*-l),

.-. ;r+l =0, x-l =0 1

a?— # +1=0, and x2 + x+l = 0J '

whence the roots are

i i l±j/-3. -l±:l/-3.

-1, 1, —2 2

(3). The roots are -2, 3± ^3- See also No. 6. p. 97., and No. 12. p. 101.

7. See Private Tutor, Alg. pt. ii. p. 119-

8. All the roots are imaginary. See p. 81:

228 HINTS AND ANSWERS [P. 103.

9. The expression is ( Wood, art. 327.)

x =

which is essentially of the form,

x = 4/{a-lb) +4/ («— b)T

or = \f [a + b) — \/{b—a) when b is greater than a. In the first case,

b\ } , ,/ a

a

Make- = tan2 0 ; whence 0 is found by logarithms, and a

x

= «c*fc {l + /;^}=SeclML4(cos2<^.

#« "• V ' V l+tan*fl

Again, make (cos 2 0)3 = tan 2<p, which gives ^>; and

x 2.

— = sec3 0. sec2®,

3 / "'

v a

and log # = '■! log sec 0+21og sec $ +3 log a. Similarly may the other case he ti'eated, vising sin 20, instead of tan 20, as is obvious.

10. Assume

24-^2 = I + B x + C x2 + D arV-B *'4 + &**

1— a?+a««

then

1 + 2 # = 1+Ba?+ Caf+ Dz*+ Ea*+ F #5 + . .

— r— B^2- Ca? + D.r4 — Ea;5 - . . + 3i'2+ 3B*3+ 3Ca;4+ 3D#5 + ... ..*. B = 3, C-B + 3 = 0; .*. C = 0, .-. D + 3B ^ 0, E-D = 0, F-E + 3D = 0, &c. .". D = - 9r E = - 9, F = -9 + 27 = 18, &c.

P. 104,]

IN ALGEBRA.

229

1 + 3 x-9 a?— 9 .r4 + 18 *5+ &c-.

1 — .r+3.r2 The scale of relation is/-f g = 1—2,

11. See p. 102.

12. (1). See Wood, p. 352.

(2). S^xS, = (aP + bP + cP+...)(«i + l)U-ci + — aP-ri + aPb'i + aPc<i+ ... + bP+i + aibP + bPcZ + . . . + cP+a+afe* -*-&%? + ...

\!)Pa(i-\-bpc(i+ . .

2«p&f'

&c.

&p &q — tSp^-q.

P 1 r — ^ P-f4

Again, multiplying by Sr = ar + br -+- cr + . . . we get

ap+rbi + aib?+r -f aP+rci + . . . j ai+rbP + aPbi+r + a<i+rcP + . . . V apbicr + a^Pcr + cflbrcr + . . J The first line = SP+r S9 — S^+g+r (as is easily seen), the second = Sq + rSp — Sp+qjrr,

and the third is 2 (a?¥cr), :. 2 (av¥cr) —

Sp Sq Sr — Sp S^ _|_ r — Sq Sp 4- r — Sr bja 4- g +- /-> Sp _(- 9 4. r.

13. Let 05 = a+u, u being a small quantity; then

x"

a4 + 4a3 m +

px3 = pa3 + 3 pa*u 4- . . . 9a-2 = qa? + 2qau f . . .

r# = ra-|-r«, .*. g = at+ (4 a3 + 3 pa2 + 2 qa + r) u nearly,

a- a,

?< =

4a3+3j?a2+ 2^ + ?' But a4+^a3+^a2 + ra = cv

nearly.

.". a3+pa? + qa + 7-

230 HINTS AND ANSWERS [P. 104.

$ — \

. . U —

<5 ' aQ3a2+2pa + q) + -1

.*. &c. 14. See Wood, art. 363.

EMMANUEL COLLEGE.

1. The quotient is xm(n-1) + xm(H-*) + . . . 1.

2. The quantities have no Common Measure.

The L. C. M. =

{a— x)(a2 + ax-{-x2){a-\-x) = a4-f a?x— ax3 — x4.

a2

3. The root is x — a — ^ &c.

2a;

4. (I). Ans. x = 4.

(2.) Raising the first to the 4th power, &c.y it will easily be found, that

2 2, , 2 a'~bi

xy + j-a?xy = H ,

a2 i/(18a4-1464) whence #?/ = =--+- ^-^ =-. • .

Squaring the second of the given equations, and subtracting 4 x xy, from the result, x — y will be obtained, and thence x and y.

(3). First

5 — x 4 — x x + 2'

. 27±<v/57 Ans. a; = —■ .

(4). x2y2 + 4xy = 96,

/. xy = — 2±10 = 8, and —12.

Hence a — y = ± 2, and ± 2^101 and x + y = 6 j

.*. a- = 4, 2, 3 ± 2-v/lO) y = 2, 4, 3 q= 2 V 10/ *

P. 104.] IN ALGEBRA. 231

5. See Wood, art. 127 ;

or, in the equation x'2+px = q, assume

*+pa,- + Q = [x + |) = x2 + px + ^, .-. Q = £

■2

X

6. Wood, art. 162.

7. Wood, art. 182.

8. Jfood, art. 190.

9. The sum of 1 + 3 + 5+ . . . 2 rc+ 1 = 2 rc. ^ = n\

10. Let r be the common ratio of a, b, c, d, Sec. ; then

b = ar, c = ar2, rf = ar3, &c.

. 1 _ 1 J 1 _ 1

• * a2— 62 ~ a2 (1 — r2) ' b2— c2 ~ a2/-2 — o¥~ «V2 (1 — r2)'

Hence the series -5 — t^-, -^ -, &c. is Geometric, its

a2 — b2 b2 — c2

1 a2

common ratio being — ~, or 7^.

Hence the required sum

('a2 \ »

x

(—)'' - 1

V627 a2n — b2n

&-b2 A a2 "S2!"-1) (a2-62)2'

11. The reciprocals of the terms being in Arithmetic Pro- gression, let x be their common difference ; then

6

1 1 a-b

b a~~ ab

Hence these reciprocals are easily found, and thence the terms of the progression.

12. See Wood, art. 230.

13. See Private Tutor, vol. i. p. 23.

14. See Private Tutor, Alg. pt, i. p. 200.

232 HINTS AND ANSWERS [P. 105.

15. This is evident; because no part of a rational quantity can be destroyed by the opposition of sign of an irrational quantity.

16. Wood, art. 258.

17. Ana. f, 36, &c.

18. Private Tutor, pt. i. p. 98.

EMMANUEL COLLEGE, 1826.

1. Wood, arts. 38, and 40.

2. The sum = 1/. 10s. 2d., and the decimal required is 7541, &c.

3. The present value

210 42000 _ ld 2 142

See Wood, art. 392.

142

.*. Gain = 30/. 19-9. Id. 2~oKq7-

4. 35 yards : 32 metres : : 1760 : -^- x 32 = 1609 f.

5. Let # feet be the breadth required ; then it contains

36J x x square feet,

•"•2§ xix^x'T= 12 + To + 40 + geo^geo'

491 whence as = 11 .nqo feet.

■

37 1

6. First, .037037 &c. = ^ = ^,

.'. V-O3^037, &C- = T = '3333- ' •

Also, .1111.. . = * =d (1)2= (-333. ..)».

P. 106.] IN ALGEBRA. '23:*

7. Let the number of the reinforcement = x ; and y the quantity of each man's ration ; then, ».*

i n food

number or men x ^ li—. ,

number or days

m 30000^-1000^ : I00(): SOPOOj,

5 30

or 1000 + .r = |x 10000 x V° = 4000,

.*. # = 3000 men.

8. The quotient is a4— a3 b + a?b~—ab3 + b4.

9. Interest a principal x rate per cent, for given time,

.'. Interest on 350 : Interest on 450 : : 350 x 35 : 450 x 27,

: : 245 : 243.

10. Amount = PR" (Wood, art. 397), .". by the question,

V20/ ~~ VToooV '

• „ - lQg5 = log 5

" log 1050 - log 1000 .0211893'

and log 5 = log 10 - log 2 = 1 -.3010300 = .69897,

6989700 „_ 209124

•'• n = -2TT893 = 32 2TT893 ^

1 1. The number of terms is n + 2 ; then, if b be the common difference,

1 + 0 + 1)6 = 31; /. b =

ra + 1

.-. the 8th term = 1+76 = 1 + IIP- = ? + 2**,

w + 1 « + 1

and *fh term = l+(»-l) 6=1 + 30^-3Q = 31"~29

w + 1 //, + 1

w + 21l 31rc— 29

n+ 1 n+i

: 9 n +1899 = 155 w— 145,

1022 whence n = ~ =14.

: : 5 : 9,

H H

234

HINTS AND ANSWERS

[P. 106.

12. (1). Ans. x = 5. (2). Ans. x = 5. (3). Ans. x = 12 and -2. (4). Dividing by xy, xz, and yz, the equations become

1 + 1 = 1

x y a j: % o

1

!-

i + i

y z

i

Adding these, and dividing

1

x If

Hence, - — ~ I- + -r x 2 \a b

i a

z 2 \a b

S-

i = i (i _

y 2 \a

s 2 V «

i +

c„ '

J>

1 1

6 C

.'. T

2 abc

y

z =

T1 ac + oc — ao

2 a6c

a6 — ac + 6c j

2 a&c |

ab Jf- ac — be J

(5). Clearing the equation of fractions,

r

1 -+-

2 —

|/(l

,r3 — x2 = 56, + 56' 1±15

8 and — 7,

x = 4, and 4^49. (6). Ans. ^ = 3 and — |. (7). See p. 125. Ans. x = 5, y = 3. (8). a?*-2a* = 8,

.*. a* = 1 ± 3 = 4 and —2, log 2 — oo

\ ar= r-^L- and log a

log a

log a'

P. 106.] IN ALGEBRA. 235

13. See Private Tutor, vol. i. p. 22.

14. The reason consists in this form (±\Za)2 = a> and •'• the square root of any quantity has two values.

15. The odd combinations

n— 1 n— 2 nln—l)ln—2)[n—3)[n—4) , = n + n. _£-._- + — !, 2. 3. 4. 5 +'"

and the even combinations

n (» — 1) m (n— l)(ro— 2)(w— 3)

~ 1. 2 1. 2. 3. 4 + • • •

,, m (m— 1) . n [n—\) (»— 2) _ .'. odd — even = n ; — ^ — + ^ — ^-q °cc'

„ , n(n— 1) m (m— 1) (m— 2) 0 ButO^l-l)" = l-w+ ^ 2 13 3 i+&c-

.". odd — even = 1.

16. Seep. 218. No. 15.

17. Let x be the number ; then

% 3

—3 — = w an integer, 4

, .e— 8 and jr — = v,

4?^ + 3 = 9v + 8, 4m? — 9z? = 5,

.*. m? = 2v + 1 +—5- •

Make v -f- 1 = 4m where m is an integer, .*. w — 9m — 1, .-. ^ = 3 + 36m — 4 = 36m — 1,

in which u = 1, 2, 3, &c. and .'. * = 35, 71, 107, &c.

18. /(14 + 8/3) = y/1-^"+ ^/¥^=V*W*,

236 HINTS AND ANSWERS [P. 107.

and v/(v/8 + / 6) = ^/^±^- + \Z^^>

= Vh + 1/72 = vx i + 1/ <?

=* (^72 + y 8)-

Assume y/ \\mn + 2 (w2 — n2)\/— 1| = A + B^ — 1, .*. \mn + 2 {rtf -w2) </ - 1 = A2 - B2 + 2AB y/- 1, ,'. A2 — B2 = 4 m», and AB = m2 — w2,

.-. A4 — 4mrc A2 = (m2 — rc2)2, .*. A2 — 2mn = ± </(2m2n2 + m4 -f w4) = ± Oa + rc2),

.'. A = m + w-i .". B = m — n) .'. the root required is m + ra + (m — rc) v' — 1- This may also be resolved by the common method of extracting the root of a binomial surd.

19. Let x be the number of guineas and y that of the moidores, then 21a: + 27?/ = 7020,

or Ix + % = 2340, for the resolution of which see p. 42. No. 21.

20. 15 = 25 x 3 = 100 x |,

.*. log 75 = 2 + log 3 - 2 log 2, But log 3 = i x .95424250 = .4871212, and log 2 = | x .9030900 = .3010300, &c. &c.

21. It =

3 + 141592

3 + *

1000000'

I

\~, nearly,

1 16

P. 108.]

IN ALGEBRA

113

237

3 +

15. 113

355

ne

arly.

22. See Private Tutor, Alg. part 1. p. 98 ; and make ym = u, thenun— 1 is divisible by u—l, &c. &c.

23. (2» + I)2— 1 = 4rc2 4- 4rc = 4fe (ra + 1) and either w or (ra + l) must be even .'. &c.

24. Dividing continually by 12, the remainders will be the digits, and the numbers will be

1024, and 249. Their product is 52710.

25. The sum of 52 cards

= 4(1+2 + 3 + 4+5 + 6+7+8 + 9+ 10) + 30, = 4x11x5 + 30 = 250, .". if x be the value of the last card 10ft + a + x = 250, .-. x = (25 — n) 10 — a, Now x cannot exceed 10 and a is supposed less than 10, .'. 25 — n cannot exceed 1 nor be less than 1, .'. x = 10 — a and n = 24.

EMMANUEL COLLEGE, 1827.

1. Wright' '$ Self-Instructions in Pure Arithmetic, p. 136.

2. -f- of 15/. = y?l. ; 3 -a of 1/. = y/.

£ of f of -3- of 1/. = f*. and f of f of 1 shilling = -fel. . sum required is V° 4- V + r + yV- = °? r^- = 7^- 17*. 5^rf.

Again, 1.0

5=k 1 = J

6rf= J

3<*

ii — i

x2 — 2

1603

801 .	10
400 .	15
80 .	3
40 .	1	6
20 .	0 .	9
10 .	0 .	4i ^2

1352/. 10s. 7M Ans.

238 HINTS AND ANSWERS [P. 108.

3. The square root is 121. 96, &c, and the cube root is 103 exactly.

4. Interest = 120± x 2± x^ = 14/. 6s. 2\d.

.'. Amount = 134/. 16s. 2\d. Also,

Interest = 15* x R- 3i x f 1 x -5*Y- -fl 4- -1-Y,

2 x V X 100/1 - 2 V + 200/

31 A 63 \

= t(1+200 + 36 X-J_ nearly, V 40000/ '

= — (\ — 63 ^

2 V + 200 + 10000/ '

26426x31 „

40000

5. Product = a;4 — § x4 + ± x2 -f 1^,

The quotient = px2 + qx — r.

6. The G. C. M. is 9a3 b2 (a— I) and the reduced fraction is

3a+l

4o2 + 2a-l"

7. The root is x2 — x + \.

8. The product is ± a* — f aT° + 1 a* +± a^r _ ^ a^

— -1 --2 ' 3

The quotient is am -\- am b + am b2 -j- &c.

x -\- 1

9. (1). First multiply by • and reduce;

At

then multiply by a? + 3. Ans. x = 3.

(2). Consider - and - the unknown quantities ; and find x y

them by any of the common methods, &c.

A mc — nb nb — mc

Ans. x — 7 ; y ~ j- *•

ac — no ' J bin — »/)

P. 109.] IN ALGEBRA. 239

EMMANUEL COLLEGE, 1828.

1. See Wright's Pure Arithmetic. The product is .1487992. The quotient is .005.

2. The square root is 15367.

The cube root is 41. 1.

3. 11 = 23-^=23:'

999 " 333

4. (1). Interest == ^234 + 1 + — "\ x — * _ = &c.

(2). Discount = future value — present value,

200

1 + 2T

200 - J55- = 200 (l - J??),

= 11. 13*. lOtf. T%.

13

5. The G. C. M. is r+6, and the reduced fraction

x+5 9¥2— #— 3"

6. The remainder may equal, but cannot exceed 2 x root. Similarly, in extracting the cube root, the remainder may equal, but cannot exceed 3 x root ; and so on for higher ex- tractions.

The most direct and simple proof of this is, perhaps, this : It is obvious, that 99, 9999, 9999*99, &c. leave greater re- mainders than any other intermediate numbers; for if 1 be added to each of them, they become perfect squares ; but the roots of these are

9, 99, 999, &c. and the remainders are

18, 198, 1998, &c. /. &c.

240 HINTS AND ANSWERS [P. 109.

7. (I). Squaring, &c.

d2^ax y Katx2^ xV'

Squaring, &c. again,

JL _i_ .JL

a* d3x "lci>

which cannot be solved except as a bequadratic. But if the given equation were

x+a = V (tf + V VaW + x-*)}' 1 2 // 4 9\

and ( - + - \x a

1 2V 4 9

aa'+#2'

, 1 4 9

and— -1 =—5,

xz ax xl

.'. x2 = 2 a#, and # = 0, and = 2 a.

(2). From the second -——•—.- = 1.

y* 15 y

x 8 ± 17

• _ — — i. and 2-

" y~ 15 - 3' anQ *' whence x = 3, and 5 y — 5, and 3. (3). #,= Y> and 5. (4). 65 = a + 5 a-# O3 -f y3) 4- 10 xhp {x+y),

= a-\-5xiy \b3—3xy. [x + y)\ +10 b x2y2, = a-f 5 b3xy—5 bxhf,

..xhf- b\vy=-^j-,

•'• ** = -2 ± v U + -sr-J'

_62 /4a-65

"T11 V 206 ' &c. &c.

8. Let a, b, be the extremes ; then — i— is > i/a& isa + 6— 2 ^abif >0, or if (\/a — */b)2.

.'. &c.

P. 110.] IN AI.GKBIM. 241

_ % + y , . . '2xy

• a = ~2 ' = ^'r2/' C ~ aHhy'

, b2 xv 2xu a : b : : b : — = — — = — =-.

a r+y v + y ~~2~

10. See p. 233.

Sum = (14-138) 10 = - 240.

11. (1). Private Tutor, vol. i. p. 21., and p. 262.

(2).

_ aVl - I J 1 ** _ ^ ^ _ 10 ^ \

V 3 " a 9 a2 8] a3 243 a4 ' /

12. 2" ±= (1+1)" = 1 + w + y-p + &c.

13. Seep. 235. No. 15.

.. _ . w (ra-l) (n-2) . . .3. 2. 1

14. The number -

1.2. 3.. .px 1.2. 3... qxl. 2.3.. r &c. For, if ^ be the number required ; then, it is clear, if the p things, q things, r things, &c. become all different, instead of the same in groups, that the whole number of permuta- tions is

Pxl. 2. 3 ... pxl. 2 3 ... 7x1. 2. 3 ... r, &c. But this number is also 1. 2. 3 . . . (77,— 2) (77 — 1) n,

1. 2. 3 ... (77-2) (77-1)77

•'" F _ 1. 2. 3 . . .pxl. 2. 3 yxl.2.3. . . rx&c'

In the word Constantinople, there being three t?.'s, two o's, and two J's ; the number required is (in this case)

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13

" 1. 2. 3x1. 2x1. 2

= 5. 6. 7. 8. 9. 10. 11. 12. 13,

= 15120x17160 = &c.

15. Private Tutor, Alg. pt. i. p. 98.

16. The root is 2 - /3. </-l.

1 1

242 HINTS AND ANSWERS [P. 111.

17. The numbers being in Arithmetic Progression, let them be x—y, x, x+y; then

x — y +x -|- x-\-y = \5\

and (x-yy + xt+ix+yY = 495J '

.*. x = 5, and y = %, .'. the numbers are

si £> «9 T y °> T •

18. For, in that system, if d, d! , d\ &c. be the digits of any number N, we have hence

N = d + d'. a + d". a? f- &c.

= d+ct (a-\ + l)=<f (a-1 + 1)2 + &c. = d + d + cF + Scc. f (a— l)x(flf + &c) &c.

CHRIST'S COLLEGE, 1824.

1. Here is a misprint; for 12 p-{-p read 12jp-j-^,

» - ~l40~ = 20 + 240 = Psh'U'"Ss + ? Pe"OT-

*» = -& = •«■

3. .1111 = \ = (£)« = (.3333. . . .)2.

4. Reduce them to the same denominator, &c.

a i+ w P nRt — m9 — nP

n q nq

6. Instead of (6a > 6) it ought to be b (a > b), and the (h) should be (p).

The form is true ; for if u, v, be the remainders,

a ■= bp + u,

b = uq + i\

u = rr.,

P. 111.] IN ALGEBRA. 243

u _ yr+1

b — uq \ = 3 . u.

7 r r

ftr

.*. M = r.j

gT+l

1*1 P9»"+l

.*. a— bp -\ r = 6. " .

^ gr-t-l r

7. The G. C. M. is ^ — 1 ; and the reduced fraction is

5x'i—3x—2 5x-l '

9. ffW, art. 380.

,_ rr„ a+b , 2a6

10. lhe means are —^ — , yao,

2 ^""'a+i Let a = 6 -f 2, c being a small quantity, a+6 26 + S

then 2 - 2

v/a6 - V 6 (6 + 2)* = ^ ( 1 + I)* = l> (l + 55) nearly>

26 + 3 ~ 2 '

and 2^ = ^(W = 6

a + o 2 o + c)

= (ft + !)(!. + ^"H-

= (6 + 3) (l - Q nearly,

(26-3) (6 + ^ _ 26*+ 262-68! ~ 26 ~ 26

26 + 3

o &c.

nearly,

244	HINTS AND ANSWERS	[P. 111.
11.	The sum =f_&£./l - - + TL _	-&C.\
	_/ /~jL 1 _ /	/-g *
	£ g* ' 1 + *- g	g '£+*'
	_/+*

Observe. It ought to have been stated that the series is convergent.

12. Wood, art. 209.

13. Wood, art. 102.

14. Let S = y/(a + by/—\) + ^/{a-b \/ — l);

then (b is omitted in the first term). S2= 2a + 2 V'(«2 + ^2), /. S = /|2a + 2A/(«2 + 62)^.

15. (.). *=3± }/(£+*).

(/3). This should be ^/(x + 9) — \ + ^/x. Ans. x = 16.

/ \ a abc (ab + ac—bc) ahc (ac—ab—ac)

(y). x ns. .* : - tt262 + tt2c2_62c2' 2/~ tftf + aW—bW

(2). The answer is .r = § and — -§. a b x

(e). First

y^a-f *) a? «+#"

#4 ,„«.'. *2

Squaring, &c. — — - _ (a2 + 26). — — = — b2, (a + x)2 v r J a + x

. x2 a2 + 2b ± a tf(a* + 4b) OA

- 5?» "' •; — 2 = 2 A supp°se^

.-. j;2-2A^ = 2 a A, whence the /p«r values of x.

(r,). First*3—* == 2 (.t+1),

.-. ,r O+l) (;r— 1) = 2 0+1), .*. 0+1) 02-#-2) = 0,

P. 112.] IN ALGEBRA. 24fl

	IN	A l-GEBUA.
X —	:-l,	and	x2-	-a; =	= 2,
X -	= *±	I =	2,	and	# =	—	1,
the	roots	are -	-1,	-1,	and	2.

16. For, by means of the (n+p) equations, only p of the unknown quantities can be eliminated so as to leave n inde- pendent equations.

17. If the roots be possible, they are both negative when all the three terms have the same sign, and both positive, when the three terms are not all alike.

The roots are both impossible, when the last term of x2-\-p x + q = 0 is positive, and p2 < 4 q.

18. Let x be the distance from A at which they meet, and y the number of days in doing it ; then

ay = x \ by = D-.rJ

D . aD

.'. y = T, and x = r.

y a + b' a + b

a and 6 would have different signs if A or B were to run after the other ; that is, if they both went in the same direction.

The negative sign is merely introduced by the operation, the velocities a and b having no reference to direction.

19. Let x be the first term, y the number of terms ; then

*+(r-i)x2= 19,

and \2x + (?/-!) x 2^ | = 91,

/. (l9+x)y = 1821

x+2y == 21 J '

whence x = — 5 and 7 \ ■> ■ u i • u i

which explain themselves.

// = 13 and 7 ' They are easily verified.

246

HINTS AND ANSWERS

[P. 112.

20. Let x, y, 2x be the digits ; then

2x+\0y+ 100 x

22

r

:)

y+3x 102 x + 10 y

xy ~ ~

5lx + 5y = lly + 33a- 102a; + lOy = Hay /. a = 4 ]

which give the answer, although 12 is too large for a digit.

21. See p. 219. No. 17.

22. See Barlow's Theory of Numbers.

23. Wood, art. 405.

24. The amount = P R", ( Wood, art. 397.)

,.3P = PR.,„rR.= (l+4)"=@"=3,

.*. (1.035)" = 3,

log 3 _ 4771213 3 317

149403 149403 years'

rc

log 1.035

25. This question is erroneously stated ; for the first is the only term of the expansion that can involve an, and no part of the expansion (at least according to the Polynomial Theorem) of (a-f b-\-c)n, can be negative.

26. The number is see p. 241.

1. 2.3..

1. 2. 3. .

Qrc-f n—\) (m + n) m x 1 . 2. 3 n

27. Were the judgments not supposed already given, the probability that they would agree in awarding a just decision, would be

a

/

or

ac

a + b r + a" (a + b) (c + d)

P. 112.] IN ALGEBRA. 247

The probability of their agreeing in an unjust decision

would be

bd

(a + 6) (c-fd)"

But after it is known that their decisions accord, the proba-

, bility that both are just, is

ac

ac-\-bd'

and the probability that both are erroneous, is

bd

ac + bd'

Also, the probability that the first delivers a just, and the

other an erroneous judgment, is

ad

ad+bc

and the probability that the second is just, and the first the

contrary, is

be

ad~\-bc'

CHRIST'S COLLEGE, 1828.

1. i of a pound = 'r shillings I the difference of which I of a guinea — \* shillings J

is exactly V? shillings = 7s. Id. 3^ farthings.

2. The sum is 3.125.

3. The second quantity in the original paper is

xA — x2y — x~ y2 -f y:i. The G. CM. isa-2-y-.

4. The square root is - -f - — —py

t1-*' =1-3-t- ST" &c-

5. Wood. art 377.

248 HINTS AND ANSWERS [P. 113.

6. See p. 33. No. 2.

7. Let x, y, be the numbers ; then

3±y5

x + y — *y i * = — 2 —

x + y — x2 — ?/2 i ?/ ^1±^/5 '

8. 8-5v'2 (8-5y/2) (3 + 2^2)

3-2^2 ** 9-8 - *+VA

>/ — r. x/—y = —s/ (xy), ^/—xx^/-y=V\ — V ixy)\

l/r=xx^/-y= VC\/ - z) x V ( V -30 - W—yf#)

9. m>o^, art. 162.

10. Wood, art. 252. This is imperfectly expressed in WW/. It aught to have been ; the square root of a rational quantity, &c.

1 1 . WW, art. 222.

The fraction — | = f ,

12. WW, art. 182.

13. WW, art. 220.

14. Wood, art. 189.

15. If possible let a -f b = wz/n , . , .

. , f w being their common

ana a — o = mq>

measure,

then 2a = ?» (;? 4- q) and 26 =± ffl (p — q),

.'. 2a and 2b have a common measure m, but they can have

no common measure but 2, for a and 6 are prime to each

other; .'. a + 6 and a — b can have no common measure

but 2.

P. 114.] IN ALGEBRA. 249

16. For all numbers, prime or composite, must be of one of the forms

4m, 4m + 1, 4w + 2, 4m + 3; but prime numbers, greater than two, cannot belong to 4m and 4m + 2,

.'. all primes must be included in the forms 4m -j- 1 and 4m + 3, or, 4m -f 1 and 4 (m -f 1) -1; that is, 4m -f 1 and 4m — 1.

17. The series will never terminate excepting when n is a positive integer; and then the (n -f l)th term is the last. (See Private Tutor, Alg. pt. i. p. 199.)

IS. Wood, art. 380.

19. Wood, art. 285.

Also, -/ — 49 = 7 </ — 1. This would seem to have been a mistake of the original paper, and we suspect instead of — 49 it should have been ^ — 49. This being presumed,

,/(0+ ^-49)= l/^ + l/-^0. V-l,

= 1/1 + VI '-1-

20. The first series is recurring ; its scale of relation is f+g-{-h-\k^= 4—6-f4 — l; and the sum is

(l-#)4 • The second series may be summed by Subtraction, Multi- plication, or by the Method of Differences.

n #

Its

•5 sum is

n+ 1

21. See Wood, art. 228, and p. 241.

22. See p. 196.

:: R

250 HINTS AND ANSWERS [P. 115.

23. (1). * = _^L. x -v/a-t-2

(2). x + y + 2V (*y) = 36> a; + ?/ = 20, .-. 2s/ (xy) = 16. Hence x = 16 and 4, y = 4 and 16. (3). #2- 16 + V 02-16) = 12;

whence x1 — 16 = 9 and 16. .-. x — ± 5 and ±4^2.

(4). Let y, a, ab, be the roots; then

		a3 = 27, .'	. a =	3,
		,.\|+3 +	3* =	13,
		62-	3 "	1,
b =	5 + 3 ""	/t =	5±4 3	=
the roots	are

3 and ^,

1, 3, 9.

ST. PETER'S COLLEGE, 1827.

1. The square root is 203975.

First, Ty = .058823529,

the root of which is .2425, &c.

2. In four years the Amount would be

760i(l4-^V)4(M=PR"), 1521 264

or,

254 '

, T 1521 /264 . \

.-.the Interest = —^- Iggj — 1 J,

1521 264-254 - 2 / 254 '

1521 66351 = ~2~ * 390625' which is easily reduced to pounds, shillings, &c.

P. 115.] IN ALGEBRA. 251

3. j- cwt. = v6 qrs- = 2f qrs- = 2qrs- su>s.

Also £lb. = V oz. = 13 | oz. Hence the sum stands

qrs. lb. oz.

2.8.0

8 . U

it

• y

i

2 . 17 . 1^.

4. Wright's Pure Arithmetic.

5. Wood, for Rule,

42237 19

Also,

75582 34'

8 aW - 10 ab3+ 2 b4 2b 4a+6

* 9 a46 - 9 a362 -f 3 a26a — 3 afr1 3a 3 a2 + &'

6. (1). First multiply by 12, &c.

8785 Ans. x = qy.

(2). Ans. a? = 5\

(3). By reduction, we get

2191 4065

	226 " "	226'
whence z	by the solution of the quadratic.
(4).	From the first,
	xy =z — 3±	:5 = 2, and -8.
	From the second,
	y =	= x-\-l,
	.". X2+ X -	= 2, and —8,
	' r — — i ». • * — 2	± •§- — 1 and — 2 ;
	also = — \	V-31 ± 2 '
	• >/=!.-	1.1 .-'781.

252 HINTS AND ANSWERS [P. 115.

(5). The first gives, by reduction, x4—§Vy. x1 = 9 y, ■'■ x2 = Vy ± 5</y = Vy, and ~^/y . . . (1). The second gives, by reduction,

3- 10

y3 + «y =■ oi x\

81

^2 n

•"• y2 2 * "¥ 'r2 r : 5^2j and "~6 r2 • • • (2)-

Hence, */f = 5 / 9 \fy, and — 6 x 9 i/y, .'. y = 45, and —54.

Also, y2 = 5 x — a/?/, and —6 x — */y, •'• 2/ — — 5 and 6 ;

whence the corresponding values of x* are 9 v"45, 9 j -54 ; and — >/— 5, —\/6; or, 27 a/5, 27 V -6; and —a/— 5, —^6. .'. the simultaneous values are

^=±V(27A/5),±A/(27A/-6),±v/f-v/_5),±^/(_^/6), 2/=45, _54, -5, 6,

of which the imaginary and surd values admit further reduc- tion.

7. For a proof as general, and far easier of comprehension, see Private Tutor, vol. i. p. 22. Those who wish to see Eider's Demonstration, may consult torn. xix. Novi Com- ment. Acad. Petrop. p. 103; or Lacroix's Complement des Elemens d' Alegebra, p. 159.

8. See p. 46. No. 9.

9. (1). Sum = (2 + 99x8) 50 = 39700. (2). Sum = (30-15xi) 8 = 200,

(3)- S = y^ = *

1 +4

P. 116.] IN ALGEBRA. 253

(4).

a \n

1

( — a\ y 0V#)n_1 ' a+x^x

\x\/x)

-3«-8 {xsfxf— (—a)" a-\- x\/x

10. Private Tutor, vol. i. p. 200.

11. V (2+ a/3) = Vi+Vi= I a/6+1 V2.

^(3^6 + 2^12) ^v/^^ + Z^-6^6

= V(V6) + -/(^6), = n/24 + V'6. s/ \xy — 2 x s/ [xy — x^)\ — V^- a/ &/— 2 V{xy— x'2)\,

= */#■ iV(^— ^)— V^I-

12. Let a? be the number of the 21s. 6(/. coins, and i/ that of the other coins ; then

43 x + 34 y = 4000,

22-9.r

Make 22— 9x = 34 tc,

4—7 a? .'. $ = 2—3 K> -t — .

Make 4 — 7?/; = 9 r,

4-2^ ,\ iv = — v + — = — .

/

Make 4— 2r = 7«,

.-. r = 2-3 m- ?4

Make V) = t,

.-. r = 2—7*, .-. iv =9t-2,

254 HINTS AND ANSWERS [P. 116,

.*. x = 10— 34 n .*. y = 15-t-43^J '

Hence, it appears, that 10 coins of the first kind, and 15 of

the second, give the only solution.

13. (5a*fz^Ax*r = 625*y«< (l - ^-2)\

8H/ _16s 96 a2 256a3 256a4 \

~" ^ 2 V 5 a?/2 + 25 x2*/4 1 25 xY + 625 a?42/«y"

: 625 xY*A -2000 3?Y25+2400 xhfz^- 1280 afyV

+ 256 *4a8.

3/ . 4- 1 x x~ ox \Q x

• (1-*) = (U-*)' = l -5_T--ir- m-

(1+a;2)-3 = 1— 3 a?2+6d?4~10 *6+ &c.

14. The present worth of P pounds to commence after p years, and continue thence to q years = its P value to continue from the first for (p + q) years — its P value from the first for p years

i-_ L i_l i_I

A— ^ r— A = =r-

R 1 ' R-l Rp" R-l •

When ^ is infinite this expression becomes

A

Rp(R-I)'

and the required present value = ^ R — ^-. 15. The cost = 31. Is. 5d.

ST. PETER'S COLLEGE, 1828.

1 . The approximate fraction required is ^ .

7i ( n 4- 1)

2. The general form of Triangular Numbers is — y — ^,

1. 2

whilst that of Pentagonal Numbers is — j ^ — -.

P. 116.] IN ALGEBRA. 255

Hence, to resolve the question, let the xth triangular number = yth pentagonal number ; then

x[x+l) = y(3y-l), or, x2 + x = 3y2— y, ;. x*-y*+x+y = %f, , 2</2

x+y

2y2 .'. —— must be-an integer; which can only be when x-\-y J

x = y. But, then,

x (.r+1) x (3x— 1) 1. 2 1. 2 '

• • *£ ~\~ X 0<£* — X 7

x2 = x, or x = 1, .*. 1 is the only triangular number which is a pentagonal number.

3. First,

/ffll+/a3+ . . . .fa2„ _, = i|M«+M3fl+ M(2»-i)«|

— HM~a + w"3o+ u-C2H-J)al,

(«2a)n— 1 . (?/-2a)"_l

•'• A(A+/«3+ • • • -An-i) = I (tt2ffl» + W-2*»-2)

4. Let x be the first term, y the common ratio ; then, if * be the sum of the odd terms, and s' that of the even terms, we have

yn— 1 , n ?/"— 1

//

-#*•

*

256 HINTS AND ANSWERS [P. 117.

A A p-1 o v

Jn _ sn C — : 2-i-

.'. the pth term=z xyP~l=" — -, s' "• s n — P suppose,

and the 9th term == xy^ = !_._ / M. 5 w =Q suppose.

Again, let z be the Common Difference of the terms of the Arithmetic Series of which P and Q are the extremes ; then

P + (r + l)^ = Q,

r-f 1 and the r* mean = Q

r+l r+l

1

1 s'n — sn s2

r+\' s'-s ,±' 1 V*

s

1 / i\ p

+

©'

5. See Private Tutor, Alg. pt. i. p. 260. Ex. 41.

6. In this y5 should be y3 and # ice versa

J_ 1_ 1_

a/3 + I + a/3 + 2 + t/3

— — L _L _JL

a/3 ~ l + i/3 + 2+V3 + 3-1-V3 1 1 1

Make S = — - 4- 1 _l . „> ,

+ 2 + t/3 + °°

1 1

a/3 v3(1 + a/3) + (1 + ^/3) (2+ a/ 3) + ' " •* >

1 ' ' 'Vs ~~ 73* Similarly,

_L l 1

•'• S35 + (^'+-;(n-1)tem = 3 + 5 + "ai=I'

which cannot be generally summed.

7. This question properly belongs to the Theory of Sym- metrical Functions. See Private Tutor, vol. i. p. 271.

P. 117.] IN ALGEBRA. 257

It may easily be proved, by showing it to be true for any simple case, such as the squares ; then assuming it for a general index p ; and proving it true for the index p+I, provided that assumption be true, and so on, as will be evident.

8. Since, Pz~Pz = n (n~ !) (n—2)~n 0—1) = n (n— 1) x (»— 3), ¥\—V* — =n[n— 1)0— 2) x (w — 4),

&c. = &c.

jpn-i— Pn~2 — n (»— 1) («— 2) 4. 3x 1,

•'• (P3— P^ (P—PJ ■• ■ • (Pn-i-Pn-z) =n{n-l)xn (n—l) 4-(ra-2)x...«(w-l)...4.3x(n-3)(»-4)(»-5)...3.2. 1.

= „ « „ X £» = #*#> Pn-rPn^ Pn

*' ' ' P3 P3 Prv-x

P

P3 P«-!

•'• P = ft?»-l (^3-^2) (P4-P3) (Ps-Pi) ■■■■ (Vn-l-P*-*)'

9. First, ( Wood, art. 224)

a a2 a3

*i - J^p s2 - Y327' *3 = r^3r> &c-

111 1 — r l-2r l-3r

5X 52 ss a a2 a?

Let S = + — j- + &c.

a a1

1 — r r r

a a2 a3

l_r , l-2r l-3r 0 a2 «3 a4

1-r _ 2^ J 1 s

a a2 , I a'

• c CT~ 1 « (1 — ^) — 1 a a [a — I)

. q- a(l-r)-l

* • s _ " (a- 1)2 '

L L

258 HINTS AND ANSWERS [P. 117.

10. The general values of x and y are x = 269- 19 w, y = 2 + 7w, w being any whole number. .*. x+y = 271— I2w, in which, if w = 22, x+y = 7, which is the least positive value. This value will not give the series of the question ; but, if the least value of x + y were 103, which would correspond to w =14, then we should have

3 3

103

1-12° 1- !

103 1.03

and 103" = 3" (l - ^j\

= 3"{i+"to3+ 4r5J-(nH)+*°r

It is manifest 103 is not the least sum of x and y which will satisfy the equation Ix+I9y = 1921.

It must be obvious to every one, in fact, that this paper is not only erroneous, but exceedingly injudicious. It is much beyond the experience of those for whom it was designed.

ST. PETER'S COLLEGE, 1830.

1. (1). Ans. 195/. lis. Ifrd.

(2). | of T8T of a moidore = If X 27 shillings, and 2£ of 8.9. 5d. = \5 x W shillings,

24x27 , 25x101 , 100 . •"• SUm = -te- + -108- + "9" shlUingS'

= 21. 4s. 5%-ftd.

2. Wood, art. 93.

The G. C. M. of 36.595 and 5.7980 is 65.

P. 117.] IN ALGEBRA. 250

The other G. C. M. is a (2ax—'3y2), and the reduced traction x1 (3 a2— 5xy)

is

if (5a2 + 4.T7/)'

2 7 1 87

3. .02777 = ,1 +

an<H-£ + f....

9000 9000'

1 _2_

l + l 15'

T-

the reduced fraction = ^7^^ x -=-

9000 2 1200'

4. (I). The root is x2 + x— \.

(2). The root of the other quantity is

s/{ — a™. bn. c~r) + \/{ — a». c». b~r).

5. 2 tt 9) 29 t 965 80 (e 7 t 7

27e<3

1 fe25

18 4 3 3

26 e 28

2 5 0 e 6

1*320

18433

1 £ £ 9 Again, let r be the radix of the required system ; then 9 + 9r2+9r4 _ 22059, • r4+r2= 2451-1 = 2450,

r2

i±1/^=^?=49j

.'. r = 7. This question is imperfectly stated, the digits in 90909 exceeding the radix of the system.

6. (1). First,

2 ax + x2 = (b—a—xf = ft8 + (a -+-./■)* — 2 b (a+x),

260 HINTS AND ANSWERS [P. 118.

b—2a

(2). First

(a + xf — 5 (a2- a?)* = — 4 (a-xy .

_2

Divide by {a — x)3 ; then

/a + arUp/g + a-y = _4 \a—xJ \a—x)

' \a—xJ

± £ -f 4 = 4 and 1 .

.*. a-f.r = 64 (a— #) and a—x,

63 a . . * = — — - and # z= 0. bo

(3). First,

/ . ,io, sfte— 18) , a/#

••■ ^(18-» + v/(J-18)=-7(^,8y

.*. /£18.r (#— 18)^+#— 18 = — x, .'. IS a: (#—18) = 4r2-72#+182,

x

n /729 . 27

= 9 ±y'-T = 9±-r

(4). Cubing the first, and substituting

216 = 126+18#V^' .-. 4x*y* = 20,

i 1

.\ x* — yb = ± 4,

i

.-. ,r4 = 5 and 1,

j/* = 1 and 5,

.'. x = 625 and 1 \ y = 1 and 3125^ * (5). From the first,

(*+„) (.r'-fy2) = i3f

P. 119.] IN ALGEBRA. 261

and the second is misprinted (in the original paper) for

x2if {x^ + y2) = 468. Make x + y = u, xy = v ; then

x2+y- == u2 — 2v, and substituting, we get u (u*—2v) = 13 1 v2(n*-2v) = 468 J '

v*

= 36,

u

,2

and u == 3^' • ^

■1

■■•^ (lie -*») = 468-- ••«•

which, being in the form of a quadratic, will give v3, and then

v; then, by (1), we shall have u; thence x + y and xy, which

will give x and y. Two pairs of simultaneous values of x and y

are

x = —2, y/ = 3 \

y -, -2, **= 3 J "

7. See Private Tutor, vol. i. p. 23.

(a2_^2)3 _ aT M _ M3)

_ a V "" 3' a2 9' a4 "81* a6 "243" cP~ )' Also, ^y-V(3^)I* = (*y)* {l - (J^)*}*

= W I ~ 4 ^2^ 32- xiy 128' *2?/ V .^y

35 & B 1

2048 *V J"

8. •(•«-•!») .y/v^+^-^/^t^4

4/(26-15 /3) == 2-A/3. (#W, art. 259.)

262 HINTS AND ANSWERS [P. HO,

9. The series is Geometric, the common ratio being — 0 :

x2

tl _ i

0 , s x2n x2n—y2n

. . S = (x—y). = — ?- ,

y x2n-x

x x—v

S =

X

.'. S : 2 :: x2,t—y2" : x2n.

] 0. The enunciation of this question is ambiguous. c ' Dif- ferent throws" may be understood both as not producing the same aggregate of the numbers shown by the dice, and also as giving faces of the dice not all the same in any two throws. We shall adopt the latter signification.

Now, each of the faces marked 1, 2, 3, 4, 5, 6, 7, 8, of the octahedron, may appear with each of those of the dodecahe- dron, thus producing 8x12 combinations.

Also, each face of the hexahedron, or common die, marked 1, 2, 3, 4, 5, 6, may appear with each of the preceding 96 combinations ; so that the three dice produce 96x6 combina- tions.

Lastly, each face of the tetrahedron may appear with each of these combinations, producing in all 4 x 6 x 96, or 2304 different throws.

11. See pp. 42, 46, and 190.

12. (a + b x -f ex2 + &c. . . .)" = an + n. {bx + cx2-\-dx*

+ ex4 + )

+ ILfa^il (bx + cx2 + dx3+8tc. ..)2

1. 4*

+ "J-!)(2':-i24(H"3)(,-+ ),+&-

P- 119.] 1N ALGEBRA. 263

.*. the coefficient of x* is n (n — 1)

1- ~' 1. &

n (n-l) (n-2) (n-3) "*" I. 2. 3. 4 ~ ' ° ■

CATHARINE HALL, 1828.

1. f of 9* =f x V = V,

61 of^ of* = V5 X^xi = T'o = .05.

2. Ans. 283/. 11*. lid} $.

3. Interest
= iwP = 2J-x-^x284i =	5121 160
4. It = 3.9. 4d
_ 0 a2—ab4-b2 5. Sum = — =_ . a2— b2
.rp ab merence = T, a — b

= 32/. 0.9. Ud,

Product = a?—b3,

Quotient == a»-*+o»-*6.f «*-3 b2+ b"~\

6. (1). Ans. x = 7. (2). Ans. x = a and b. (3). First,

x*— 2x+5+6\f (x2— 2x+5) = 16, ••• \f{x2+2x+5) = -3 ± 5 = 2 and-8, .'. x2-2x = —1 and 59,

.". x = 1 and ± y^ 60.

(4). First, *-l = ^r (l/*+l),

.'. (V«+l) (x-^x-2) = 0, •'• a/# = —1, and a?— </# = 2,

264 HINTS AND ANSWERS [P. 120.

.'. y/x = I ± $ = 2 and — 1, .'. a; = 4 and 1. (5). See p. 240.

7. See JFood, art. 90.

8. The lowest fractions are

c + d x—\

/T2iand^T2-

~ a/« + 2' a-/ a + 8* a*y/a +

10. Seep. 218.

11. Seep. 200.

12. Wood, art. 402.

13. See Private Tutor, Alg. part i. p. 291.

■/ 15625 = 125, ^/ (1+07)= l+i^-i^2 + .1-V^3-&c.

i4. •(•-3-i)= y7^- + y^- </-!>

= •* + \/|/— 1, 1/ (10+/108) = l + ^3.

15. Jfbod, art. 452.

16. Let # be the number of crowns ; y that of the guineas ; then

5x + 21 y = 20 x 20 = = 400, y = 0,5, W, 15, 20 ... . !r = 80, 59, 38, 17,-2 . .

> Ans. 4 ways.

P. 121.] IN ALGEBRA. 265

1 5 = A

2. 6 = *

JESUS COLLEGE, 1828. 530

132 . 10 66 . 5

198/. 15s. Od.

2. Euclid, Book i. Prop. 35.

3. The product is

pa^x^-a'x12 - «3 ar**—{p—2)- 3.

4. First to the straight line apply a parallelogram = 2 x the given figure ; half of this parallelogram will be a A on the given base, which is the a required, excepting the vertical angle.

Upon the base describe a segment of a circle containing an angle = the given angle of the vertex. This segment will be met by the upper side of the parallelogram in two points P, P'. Join P or P' with the extremities of the base, and there will result the A required.

^ 5 - l 2 3 ■ 375

5- 1374 -4^3 = 8- '61b-

6. (1). x = f,

(2). x = i, y = |,

64 10 / 8\

(3). First * + -?? 3" {' + ;)>

169 9

8 . 5 ± 13 _ , 8 x + -= — ?j — = 6, and — ^

M M

2G6 HINTS AND ANSWERS [P. 121.

.'. x'2 — 6x = — 8, and x2 + %x = — 8,

4 y/-56

.'. x — 3 ± l,and.r =

or,.r = 4, 2 and 7. Euc.

3- 3

— 4±A/-56

8. Let x—y, x, x + y be the numbers ; then

3x = 15, (,r2-y2).r = 120, .*. x — 5, y = ± 1.

9. Similar figures a squares of the homologous sides; .*. if A, B, C be the areas of the figures on the sides a, b, c, re- spectively, c being opposite the right angle, then we have

A : B : : a2 : b2,

.-. A: A+B :: a2 : «2 + 62, and C : A : : c2 : a2,

.-. C : A + B :: c2 : «2+/;2 :: 1 : 1. (Euc. 47. b. i.).

10. Wood, art. 189.

11. Woodhouse's Trig.

12. Woodhouse's Trig.

13. The root is

.r2 a2

-3 'J

14. If a be the area of the a , A, B, C the angles, and a,

b, c the opposite sides ; then

, 0 sin A

a2 — 2«.

62 = 2 a.

sin B.	sin C
sin	B
sin A.	sin C
sin	C

C2 = 2 n.

sin A. smJ>

15. Log .225 = log 225-3,

= log 25 + 2 log 3-3, = 2-2 log 2 + 2 log 3-3,

P. 122.] IN ALGEBRA. 267

= 2(log3-log2)-l,

= 2x .176091-1= 1.352182.

lc 9#3-72 „ ./;3- 8 , , n

17. Euc. B. xi.

18. Since cos P + cos Q = 2 cos — -^— - cos — 17—^3

.'. cos (604-A) + cos (60— A) = 2 cos 60. cos A.

= 2 x i cos A = cos A.

19. Ans. f| — f*. = 4 _ |. = i shilling, or 2 pence.

20. Let x be the common ratio ; then

i«4= 128 ; /. x = 4, and the series is

\, 2, 8, 32, 128.

MAGDALENE COLLEGE, 1828.

1 . f of 4tf. = 'T6 of a penny = ff of five-pence.

2. The value = .878125/. = 17s. 6fd.

,, xt i_ p j length x width x depth

3. Number of davs oc — s ; c_

men x hours

.". if a; be the number of days required, we have

64.27.18 lfl 72.18.12

*• : -^r^-T^ — : : 16

32.12 " 36.8

64.27.18 8.36 8.9.18

" ''' ~ 24 X 72.18.12 ~ 2.9.3

4. For the proof, see Wood.

131 (1). The G. C. M. = 127, and the fraction = -jgp

{2). The G. C. M. = x-2 = — «•

2«/2+3v— 5

(3). The G. C. M. = y-\ - -^f- < ■ ,

268 HINTS AND ANSWERS [P. 123.

5. The quotient is

am + 3am-1 bn—6am-z b2m.

6. That a m = — is a definition : not a theorem.

am

See Wood, art. 53.

7. The square root = 3y2-\-qif—™„ Tlie cube root = 258.

8. Wood, art. 189.

204

9. (1). Ans. x =

251

(2). Ans. x = -. c

(3). From the first,

\x + y = 17;

from the second,

2a: - 3y = —9, x = 3, y = 5. (4). Ans. a? = 5 and V- (5). From the first equation, x2 2x x

? + 7y + y + y + ^ = 20'

| + ^2/) + (| + 1/ y) = 20,

or A - + Vy — - \ ± # = 4 and - 5 . . . . (I),

y

. 4y-8

4- |/?/ = 4 and — 5,

2/

y2 = 8 and y = 4,

P. 123.] IN ALGEBRA. 2(Ji»

.\ x = 4y — 8 = 8. If the — 5 be taken, we get

Sf* + 9y = 8,

y-

.-. i/2 + 1 =-9. (y-1), *. dividing by V*/ + I,

- Vy + 1 = -9(^-- l),

y + 8/3/ = 8, and s/ y = — 4 ± -v/ 24, whence 7/ and .'. ar.

10. Let -, a, ax be the numbers: then x

/I + 1 + x)= 13]

1 + i t

l+x - ' J , the numbers are

1, 3, 9.

1 1. Let x be number to whom he gave 9 pence ; then

9^ + 15 (n—x) = 12/?,

\5n~\2p 5n—4p •'• x ~ ~~6~ = 2 '

■

12. ( — -— ) = (l +-) =1 +»» — nearly, if x be small.

13. Private Tutor, vol i. p. 22 gives the proof.

<ya + y^)4 = (^/a)4 + 4(A/a);i </b + Q{^af{^bf

= a2 + 4a-/ (a*) + 6ab f 46-/ (a6) + b2.

i^ /i , »(rc— 1) »(«— l)(n— 2) , 0

14. (1 + 1)" = 1+^+ \ 2 ' + } 23 -+&c.

15. ^(7 + 6^-2) = 3+ ^-2, See Wood, art. 258.

270 HINTS AND ANSWERS [P. 123.

M

16. The Present Value = y~= — (Wood, art, 392),

1 + nr

25| -= 713/. 6s. 4d. '

3^~ 407'

100 Whence discount = future value — present, is found.

+ 2 100

17. 1'he present value of an Annuity A, for p years, is ( Wood, art. 402.)

P= il^-.A, R- 1

and for p + q years it is

I

1 -

RM A,

A - R-l

.*. the present value of A, to commence after p years and continue q years, is

i »

A Ri p' __ p _ _ .

r ~" Rp" R-l

Hence, if V be the value of the Estate for n years, and V that of the reversion,

v-1"^ P l- P l

r_1 • x' V'~ R»" R-l'

.-. V: V :: 1-i- : i-, R" R"

:: R«- 1 : 1,

in which R = 1 -f y^-y

18. Log 720 = log 10 + log 9 + log 8,

= 1 + 2 log 3 + 3 log 2 = &c. log 7. 2 = log 72 - log 10 = 2 log 3 + 3 log 2 - I = &c.

P. 124.] IN ALGEBRA. 271

19. First (a*-m*-* = ^a^Z^

_ (aa— b*)*x {a—b)°~*

•'• (a2_52)2 — (Vf^'

.-. (a +6)2a: = (a+6)2, .*. a: = 1.

20. Let N be the number, d0, dv &e. the digits, then

N = d0 + d,r + d#* +d3r3 + . . . .

= rfo + ^(r+1-1) +rf2(r+ l-l)2 +&c.

= ^0 - rfl + rf2 - <*3 + ■ • • •

+ (r+ 1) x Q, whence the proposition is manifest.

21. See Barlow's Theory of Numbers on this subject.

19 . 6 14 . 7

273 . 0 11 . 4 . 6

284 . 4 ' . ~6 Ans. 284 square feet, and 54 square inches.

TRINITY HALL, 1828.

1. | of T6T of half-a-crown = f XT6T X|- shillings,

= ~ shillings, .'. fraction required — ~fr.

2. 3. 11 i = 3+ft+& = 3 ft = 'TV shillings,

.". decimal required = £f£ = .1997

3. Ans. 169/. 12*. Id. ft.

4. Interest = »rP = 9x^ x 999,

= 404/. 11.5. 10^/. f.f.

272 HINTS AND ANSWERS [P. 124.

_ TT pounds

5. Horses oc — i ,

days

.'. if x be the number required,

y .'. x = 40.

.*. • p .• • • *7 • _L_4. y . . I • 2 O *

6. ffW, art. 76.

7. (1). Product = a:2 -4.

(2) = at— ll#4+38#3— 45«2+27a;— 10.

77 71 a? 1

(3) = ** - - 60^3+ "120 r2~ 15 +120 *

(4) = a3-b\

8. (1). Quotient = x + a.

(2) = x*+5x + $ (for 24 write -24).

(<Jj = .17 -J" 3" ^" l" 9* ® TTf •^'T"!!-^

(4) =(2y/.+ |)(4^+^).

(5) = a1- b*.

9. JFood, art. 93.

2

10. Sum = T.

a—b

11. For proof, see Wood.

„ . #4- a #— a x2—a2

Quotients — ^Xjp= £!=&.-

12. Wood, art. 110, &c.

13. -v/5. 76 = 2. 4; v'O. 576 = .758, &c.

\/ (a* 3a* $ a , qja\ a |

V \— J- 6rt6-8 62J = rj - 2 6 .

14. (1). Ans. x = 12.

(2). Ans. x = 25.

(3). Ans. x = 3 and 2 i y = 2 and 3 J '

P. 125.J IN ALGEBRA. 273

(4). Ans. .=***£=«

(5). Dividing by V (1— x)2 we get

— 2 — O '

• 1±£ \ — x

l±5A/-3+10(-3)±10.(>v/-3)3+5(V-3)4±(>v/-3)5

~ 32

_16 + 16-y/-3_l + /~3

32 ~ 2 '

.'. 2 + 2# = l=p-y/— 3— a?±>\/— 3. #,

.-. # (3=j=\/— 3) = -1 + \/— 3,

_l=PV-3 (-l+^-3)(3±V-3)

.r —

3q=V-3 ~ 12

^-1

_ ^/3 '

15. Wood, art. 162.

16. JFood, art. 189.

17. Sum = (4-9) 5 = - 25, 1

Sum = 4.

1 o

1+f

18. The Common Difference is evidently 2, .". the series is -5, -3, -I, 1,3,5, 7,9, 11, &c.

rn — 1

19. s = a.

-1'

.-. r»= 1 + (r-1) j,

.". rc log r = log -j 1 + (r— 1) - 1 , &c. 20. See Private Tutor, vol. i. p. 21, &c.

NN

274 HINTS AND ANSWERS [P. 126.

21. Cubing the first

a + b\/a = m3 + 3m2n^a + 3mn2a -\-[n3a^a, and since the surd terms cannot be destroyed by the rational terms,

a = ?n3+3mn2a, b \/ a — 3m2?l ^/a + fl3a \/ a, .'. a — b^/ a = m3—3m2nA/a + 3m?l2a — n3a\/a,

= (m — n \/a)3, .-. &c.

22. V(S + 2V5)= a/^P+ l/^-4- V5 4- 1,

^(2+^/5) = i (1 + ^/5)

23. See Private Tutor, vol. i. p. 26.

24. See Barlow's Theory of Numbers.

CAIUS COLLEGE, 1830.

1. Ans. 317/. O.s. 2M

19 1

2- ")• 190000 = 10000 = •°001-

(2). ~° = 2040000.

(3). The quotient = a2— -f a+f.

(4). «*-* _(oi)6- wfl_ (a^y-^y (a+fty

aT—bT a? — 6T aJ—b1

i

iT

= (aT + aT 6T + ^) (a+6¥).

3. -/17 =4.1231056.

4. JFbod, art. 78, 109, e* sey.

5. Wood, art. 92.

P. 127.]

IN ALGEBRA.

275

a?+b*±(a-bf _ 2(a*+62-a6) 2ab

<#-& W^¥ ' or tf=F

7. (I). #+12=^ (2). x = 6 and 1.

]91 12

(3). From the second, it is found that xy x-y = ±7, and x — y = 9, /. x = 8 and 1, ?/ = 1 and 8.

8 ; whence

(4). First, - + - x y	= l]
1 + 1 x %	1 2	-.
1 1 - + - y z ling them together,	_ 1 3 J
1 1 1 - + - + - = x y z	*0'+i +
	n = 12'
1 1 1 ■•*- 12' y	5 1 7 12' # 12
12	12

5>

X =

7'

y

5'

12.

8. WW, art. 182.

2 '

9. JfW, art. 222.

(1). Sum = (4-10x|) V =

(3)"— 1 (2). Sum to n terms = 4. ^_, ■ -

which is easily computed.

Sum to oo = 4. ^- 16.

410 310

"IS-'

1-f

(3). Seep. 218.

276 HINTS AND ANSWERS [P. 128.

10. See p. 240. No. 8. The required means are *■£ and 3.

11. Discount = future— present value = 153 — ,

1 + nr'

( Wood, art. 392),

153

= 153 — = 3, by the question,

. 153 r 3r

" ~2~ -d+ 2' 150r = 6,

1 4

" r = 25 = 100' °r per '

12. See p. 254. No. 14.

13. The whole number = n^ ^ — — - + . .

But.(l + 1)»= l+«+*^H +&c. &c.

14. See Private Tutor, vol. i. p. 22.

The fourth term of (a+b)n is

n(n-l)(n-2) . 3

1. 2. 3 '

and .*. that required is

f-(t-l)(f-2) /^*\n /»/„\*.

24/2 «/

2187 #f

15. /(10/7 + 22) = 1 + ^/7.

16. See Private Tutor, vol. i. p. 327.

17. The possible parts cannot equal impossible parts, being essentially distinct by definition.

P- 128.] IN ALGEBRA. 277

18. N = .afiyaPy

1000 N = a/3y.a/3y = „/3y + N,

• N — ^X — y+10/3+102« ~999 ~ 999 •

19. 6 5 4 1) 1 4 3 3 2 2 1 6 (1 4 5 6

654 1

446 12 3 6 124

5455 1 45 6 65

55536 55536

20. See Barlow's Theory of Numbers.

1

21. \/5 = 2 +

4+1

4+&c.

22. There is no difficulty in this.

23. See p. 209. No. 16.

FINIS.

ERRATA.

, 4s. 5d. 53 4*. 9d. 19

P. 43 last line for — - — = — read — - — = r~.

J 5 9 5 20

,53 19 . „ 53 . 19

— 44 first line for — read — , and for — read — .

— 44 line 5 for 16874 read 16875.

— 47 line 17 /or 10x2 read 10 X 3, and for 132 read 187.

,24 „ 3 ,68 „ 5

— 53 //«e 21 for 48 raid 44, and 22 for — = 3 - ra»d — = 3 --.

— 54 fine 19 /or 38 read 38 and - 30.

,R ,. n , 115 ,115

— 56 fcne 11 /or — rmd — .

— 57 line 6 for x*+3P-iy + &c. read aiM-i-fa**-* 2/ + &c-

19/ora:2»— 2 readj-2»-».

— 58 /i«e 17 for 12 y read 84 ?/.

= li6=15_l read =2. 11 11

ftne 18 /or 19 - + — = 19-„ read 5. J 7 11 17

— 61 lines 30 and 31 for 1. 471213 read 1. 4771213.

— 63 line 11 for If read It.

21

— 64 line 10 for 12 read 6, and for - 21 read - — .

7 47

— 65 line 5 for 487 read 507, and 7 /or 1 — read —.

— 68. Question 24 is conducted on wrong principles.

tt' I" 3. 4. 1 _ 2

The probability required is —-^=^ ^ ^ L g - -.

See Lacroix's Probab. p. 249.

— 103 /me 13 refer to p. 79.

— 240 line 5 for bequadratic read biquadratic.

Errata in the Questions of Euclid, Arithmetic and Algebra.

— 37 Question 10 a for —ab in denom. read +ab.

__ 49 8 (2) for x— 3 read -jp.

42 - - - - 8 (4) for x2 -\-2 x - - - - read x * — 2 x.

42 - - - - 10 (2) for d—b - - - - read a—b.

45 ... . 6 /on2-!/"- - - - read x2n—y2".

50 - - - - 2 for least - - - - read last.

86 - - - - 9 (3)- The series should be

1 1 1 &c qv7r+ (l-a-)(l+vV) (l-z)Ml+v/.t) + 93 17. This is erroneous in the original paper.

-W.-.-UWJIr-^^tt^i-^.

This a misprint in the original paper.

_ HO 1 for 12 p+p read 12p+a.

_ H2 3 "for xi— x*y— x*y+y3 read x*—x2y-~x2y2-\-y3.

_ 118 6 (5)/or ,r'-!/ readx*y*.

This error is in the original paper.

BINDINU 5t.ui. julo ••" ■

QA	Wright, J.M.F.
43	Hints and answers
W74
1831
Physical &
Applied Sci.	*

l) of T bignmnd Samuel Library mm*****™*™*™"*"™******"

l*b Dec 95

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