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CHAP, xii]                     CKITEKIA FOR DIVISIBILITY.                             341
Noel (ibid., 378-9) gave tests for divisors 11, 13, 17,..., 43.
Bougon (ibid., 508) gave several tests for the divisor 7. For example, a number is divisible by 7 if the quadruple of the number of its tens diminished by the units digit is divisible by 7, as 1883 since 188-4—3 = 749 is divisible by 7. J. Heilmann (ibid., 187) gave a test for the divisor 7.
P. Breton and Schobbens (ibid., 444-5) gave tests for the divisor 13.
S. Dickstein31 gave a rule to reduce the question of the divisibility of a number to any base by another to that for a smaller number.
A. Loir32 gave a rule to test the divisibility of N, having the units digit a, by a prime P. From (N—a)/W, subtract the product of a by the number, say (mP —1)/10, of tens in such a multiple mP of P that the units digit is 1. To the difference obtained apply the same operation, etc., until we exhaust N. If the final difference be P or 0, N is divisible by P.
R. Tucker33 started with a number N, say 5443, cut off the last digit 3 and defined u2 = 544—2-3 = 538, /w3 = 53 —2-8, etc. If any one of the u's is divisible by 7, N is divisible by 7. R. W. D. Christie (p. 247) extended the test to the divisors 11,13,17,37, the respective multipliers being 1, 9, 5, 11, provided always the number tested ends with 1, 3, 7 or 9.
R. Perrin34 would find the minimum residue of N modulo p as follows. Decompose N, written to base x, into any series of digits, each with any number of digits, say A, Biy C/,..., where B{ has i digits. Let p be any integer prime to x and find qi so that q&=l (mod p). Let a be any one of the integers prime to p and numerically <p/2. Let ft be the ifh integer following a in that one of the series containing a which are defined thus: as the first series take the residues modulo p of 1, q, q2,. ..; as the second series take the products of the preceding residues by any new integer prime to p] etc. Let y be the jth integer following ft in the same series, etc. Then N' = Aa+Bift+Cjy+... is or is not divisible by p according as N is or not. By repetitions of the process, we get the minimum residue of N modulo p. The special case A+BiQit with p a prime, is due to Loir.32
Dietrichkeit35 would test Z = lQk+a for the divisor n by testing k—xa, where lOrr+1 is some multiple of n. To test Z (pp. 316-7) for the divisor 7, test the sum of the products of the units digit, tens digit,.. . by 1, 3, 2, 6, 4, 5, taken in cyclic order beginning with any term (the remainders on converting 1/7 into adecimal fraction). Similarly for 1/n, when n isprime to 10.
J. Fontes36 would test N for a divisor M by using a number<N and = N (mod M), found as follows. For the base B, let q be the absolutely least residue of Bm modulo M. Commencing at the right, .decompose N into sets of m digits, as Xm,..., aw, and set f(x) = a,mxn+t$mxn-l + - - - +X«, whence N=f(Bm). By expanding N=f(q+M^), we see that f(q) is the desired number <N and ^N (mod M).
S. Levanen37 gave a table showing the exponent to which 10 belongs for
"Lemberg Museum (Polish), 1886.     "Comptes Rendus Paris, 106,1888, 1070-1; errata, 1194. 5SNature, 40, 1889, 115-6.                 "Assoc. franc, avanc. sc., 18, 1889, II, 24-38.
"Zeitschr.Math.Phys.,36,1891,64.  »«Comptes Rendus Paris, 115, 1892, 1259-61. 370fversigt af finska vetcnskaps-soc. forhandlingar, 34, 1892,109-162.   Cf. Jahrbuch Fortschr. Math., 24, 1892, 164-5.