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CHAP, vi]                    PEKIODIC DECIMAL FRACTIONS.                           173
p — I =sh and we have h sets of s fractions whose periods differ only by the cyclic permu nation of the digits. If p is a product of distinct prunes pi3 p2... and if the lengths of the periods of 1/p, l/p1} l/p2,... are s, slf s2,..., then s is the 1. c. m. of «j, s2, — If p=pi°p2/3- • •> and s, slt s' are the lengths of the periods of 1/p, I/pi, I/Pi", then $' is one of the numbers s^ SIP,..., Sipi""1 and hence divides (pi--l)pi0"1; and s is a divisor of <j>(p). Thus p divides 10*(p)-l.
C. A. Laisant95 used a lattice of points, whose abscissas are a+r, a+2r,..., a+pr and ordinates are their residues < p modulo p, to represent graphically periodic decimal fractions and to expand fractions into a difference of two series of ascending powers of fixed fractions.
*A. Rieke96 noted thaf a periodic decimal with a period of 2m digits equals (A+l)/(10m+l), where A is the first half of the period. He discussed the period length for any base.
W. E. Heal97 noted that, if B contains all the prime factors of N, the number of digits hi the fraction to the base B for M/N is the greatest integer in (n+ri — l)/n', where n—ri is the greatest difference found by subtracting the exponent of each prime factor of N from the exponent of the same prime factor of B. If B contains no prune factor of JV, the fraction for M/N is purely periodic, with a period of <KJV) digits. If B contains some, but not all, of the prime factors of AT, the number of digits preceding the period is the same as in the first theorem. The proofs are obscure. There is given the period for 1/p when p<100 and has 10 as a primitive root [the same p's as by Glaisher83]. Likewise for base 12, with p<50.
R. W. Genese98 noted that, if we multiply the period for 1/81 [Glaisher60] by m, where m<Sl and prime to it, we get a period containing the digits 0, 1,..., 9 except 9n—ra, where 9n is the multiple of 9 just exceeding m.
Jos. Mayer99 investigated the moduli with respect to which 10 belongs to a given exponent, and gave the factors of 10n—1, n< 12. He discussed the determination of the exponent to which 10 belongs for a given modulus by use of the theory of indices and by the methods of quadratic, cubic, biquadratic,... residues. He used also the fact that there are (a—a') 03 — /3')... divisors of pfpzpf... which divide no one of the fixed factors PiWps7 • • • > PiaP2bp3y> - ••> where a<a, 6</3,.. ., and p1} p2,. .. are distinct prunes. He gave the length of the period for 1/p, for each prune p:§2543 and 22 higher primes [Burckhardt20].
L. Contejean100 proved that, in the conversion of an irreducible fraction a/6 into a decimal fraction, if the remainders ar and am are congruent modulo 6, so that 10ra^lOma, then 10m""r—1 is divisible by the quotient b' of 6 by the highest factor 2'5a of 6. Thus the length of the period is
"Assoc. fran§. avanc. sc., 16, 1887, II, 228-235.
••Versuch iiber die periodischen Briiche, Progr., Kiga, 1887.
•'Annals of Math., 3, 1887, 97-103.
"Report British Assoc., 1888, 580-1.
"XJeber die Grosse der Periode eines unendlichen Dezimalbruches, oder die  Congruence
10*=1 (mod P).   Progr. K. Studienanstalt Burghausen, Mtinchen, 1888, 52 pp. 100BuU. soc. philomathique de Paris, (8), 4, 1891-2, 64-70.