170 HISTORY OF THE THEORY OF NUMBERS. [CHAP, vi
*A. J. M. Brogtrop73 treated periodic decimals.
G. Bellavitis74 noted that the use of base 2 renders much more compact and convenient Gauss'15 table and hence constructed such a table.
W. Shanks76 found that the period for I/p, where p = 487, is divisible by p, so that the period for l/p2 has p — 1 digits.
J, W. L. Glaisher76 formed the period 05263... for 1/19 as follows: List 5; divide it by 2 and list the quotient 2; since the remainder is 1, divide 12 by 2 and list the quotient 6; divide it by 2 and list the quotient, etc. To get the period for 1/199, start with 50. To get the period, apart from the prefixed zero, for 1/49, start with 20 and divide always by 5; for 1/499, start with 200.
Glaisher77 noted that, if we regard as the same periods those in which the digits and their cyclic order are the same, even if commencing at different places, a number q prime to 10 will have/periods each of a digits, where af=<t>(q). This was used to check Goodwyn's table.23 If # = 39, there are four periods each of six digits. If q — I belongs to the period for l/#, the two halves of every period are complementary; if not, the periods form pairs and the periods in each pair are complementary. For each prime J\T<1000, except 3 and 487, the period for 1/N* has uN*"1 digits if that for l/N has n digits.
Glaisher78 collected various known results on periodic decimals and gave an account of the tables relating thereto. If q is prime to 10 and if the period for 1/q has <£(<?) digits, the products of the period by the 4>(q) integers <q and prime to q have the same digits in the same cyclic order; for example, if 3=49. He gave (pp. 204-6) for each g<1024 and prime to 10 the number a of digits in the period for 1/q, the number n of periods of irreducible fractions p/q, not regarding as distinct two periods having the same digits in the same cyclic order, and, finally Euler's $(<?)• The values of a and n were obtained by mere counting from the entries in Good-wyn's23 "table of circles"; in every case, cm=<£(#). For the prime p = 487, he gave the full periods for l/p and l/p2, each of 486 digits, thus verifying Desmarest's38 statement of the exceptional character of this p [cf. Shanks76].
Glaisher79 again stated the chief rules for the lengths of periods.
The problem was proposed80 to find a number whose products by 2,..., 6 have the same digits, but in a new order.
Birger Hausted81 solved this problem. Start with any number a of one digit, multiply it by any number p and let 6 be the digit in the units
"Nieuw Archief voor Wiskunde, Amsterdam, 3,1877, 58-9.
74Atti Accad. Lincei, Mem. Sc. Pis. Mat., (3), 1, 1877, 778-800. Transunti, 206. See 62a
of Ch. VII.
76Proc. Roy. Soc. London, 25, 1877, 551-3. 76Messenger Math., 7, 1878, 190-1. Cf. Desmarest.38 "Report British Assoc., 1878, 471-3. 78Proc. Cambridge Phil. Soc., 3, 1878, 185-206.
"Solutions of the Cambridge Senate-House Problems and Riders for 1878, pp. 8-9. "Tidsskrift for Math., Kjobenhavn, 2, 1878, 28. "/bid, pp. 180-3. Jornal de Sc. Math, e Ast., 2, 1878, 154-6.