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CHAP, vi]                    PERIODIC DECIMAL FRACTIONS.                           163
If M is divisible by p, we may take n = l and conclude that A*/p2 differs from l/p2 by an integer. If M is not divisible by p, S must be, so that n is divisible by p and the length of the period is pt. In general, for the denominator px, we have n = I if M is divisible by px~l, but in the contrary case n is a multiple of px~1. If the period for a prime p has an even number of digits, the sum of corresponding quotients in the two half periods is p.
An anonymous writer25 noted that, if we add the digits of the period of a circulating decimal, then add the digits of the new sum, etc., we finally get 9. From a number subtract that obtained by reversing its digits; add the digits of the difference; repeat for the sum, etc.; we get 9.
Bredow26 gave the periods for a/p, where p is a prime or power of a prime between 100 and 200. He gave certain factors of 10n — 1 f or n = 6-10, 12-16, 18, 21, 22, 28, 33, 35, 41, 44, 46, 58, 60, 96.
E. Midy27 noted that, if an, a"1, . . . are the least powers of a which, diminished by unity, give remainders divisible by qh, q^1, . . . , respectively (q, qi,. . . being distinct primes), and if the quotients are not divisible by q, ql} . . .Respectively, and if t is the 1. c. m. of n, n1} . . . , then a belongs to the exponent t modulo p = qhqihl . . . , and a* — 1 is divisible by q only h times.
Let the period of the pure decimal fraction for a/b have 2n digits. If b is prime to 10n — 1, the sum of corresponding digits hi the half periods is always 9, and the sum of corresponding remainders is 6. Next, let b and 10n — 1 have d>l as their g. c. d. and set b' = b/d. Let an be the nth remainder in finding the decimal fraction. Then a+an = b'k, ai+an-M==^/^ij etc. The sums q+qn, gi+<?ŧ+i,... of corresponding digits hi the half periods equal
Similar results hold when the period of mn digits is divided into n parts of m digits each.   For example, in the period
for 1/403, the two halves are not complementary (1015 — 1 being divisible
by 31); for i = l, 2, 3, the sum of the digits of rank i, i+3, i+6, . . ., i+27
is always 45, while the corresponding sums of the remainders are 2015.
N. Druckenmuller27a noted that any fraction can be expressed as a/
J. Westerberg28 gave in 1838 factors of 10nąl for n^ 15.
G. R. Perkins29 considered the remainder rx when Nx is divided by P, and the quotient q in Nrx-i=Pqx+rx. If rk = P— 1, there are 2k terms in the period of remainders, and
rk+z+rx = P,     .     (fcH-x+^W-l-[These results relate to l/P written to the base N.]
25Polytechnisches Journal (ed., J. G. Dingier), Stuttgart, 34, 1829, 68; extract from Mechanics'
Magazine, N. 313, p. 411.
26Von den Perioden der Decimalbriiche, Progr., Oels, 1834.
"De quelquea propri6t6a dcs nombres et des fractions d<5cimale3 pe*riodiques, Nantes, 1836,21pp. ""Theorie der Kettenreihen . . ., Trier, 1837. 28See Chapter on Perfect Numbers.104 "Arner. Jour. Sc. Arts, 40, 1841, 112-7.