342 HISTORY or THE THEORY OF NUMBERS. [CHAP, xn
primes 6<200 and certain larger primes, from which are easily deduced tests for the divisor b.
Several370 noted that if 10 belongs to the exponent n modulo d, and if Si, 82, . . . denote the sums of every nth digit of N beginning with the first, second, . . .at the right, the remainder on the division of N by d is that of
J. Fontes38 would find the least residue of N modulo M. If 10n has the residue q modulo M, we do not change the least residue of N if we multiply a set of n digits of N by the same power of q as of 10n. Thus for M = 19, AT= 10433 = 104+4-102+33, io2 has the residue 5 modulo 19 and we may replace N by 52-f4-_5-f33. The method is applied to each prime M^ 149,
Fontes39 gave a history of the tests for divisibility, and an "extension of the method of Pascal," similar to that in his preceding paper.
P. Valeric40 would test the divisibility of N by 39, for example, by subtracting from N a multiple of 39 with the same ending as N.
F. Belohlavek41 noted that 10A+J5 is divisible by 10p=±= 1 if A^pB is.
C. Borgen42 noted that Z = <vlOn-f . . . +a140+a0 is divisible by N if
is divisible by A^ For JV = 7, take a= 1 ; then 10a AT = 3 and Z is divisible by 7 if a0+3a14-2a2 a3 3a4 2as+. . . is divisible by 7.
J. J. Sylvester420 noted that, if the r digits of N, read from left to right, be multiplied by the first r terms of the recurring series 1, 4, 3, 1, 4, 3; 1, 4, ... [the residues, in reverse order, of 10, IO2, . . ., modulo 13], the sum of the products is divisible by 13 if and only if N is divisible by 13.
C. L. Dodgson426 discussed the quotient and remainder on division by 9 or 11.
L. T. Riess43 noted that, if p is not divisible by 2 or 5, 10&+a(a<10) is divisible by p if bxa is divisible by p, where mp = Wx+a (a< 10) and m = l, 7, 3, 9 according as p=l, 3, 7, 9 (mod 10), respectively.
A. Loir44 gave tests for prime divisors < 100 by uniting them by twos or threes so that the product P ends in 0 1 , as 7 -43 = 30 1 . To test N, multiply the number formed of the last two digits of N by the number preceding 01 in P, subtract the product from N, and proceed in the same manner with the difference. Then P is a factor if we finally get a difference which is zero. If a difference is a multiple of a prime factor p of P, then N is divisible by p.
Plakhowo45 gave the test by Bougaief, but without using congruences.
""Math. Quest. Educ. Times, 57, 1892, 111.
38Assoc. franc.. avanc. sc., 22, 1893, II, 240-254.
39Me"rn. ac. sc. Toulouse, (9), 5, 1893, 459-475.
"La Revue Scientifique de France, (3), 52, 1893, 765.
"Casopis, Prag, 23, 1894, 59. 42Nature, 57, 1897-8, 54.
4saEducat. Times, March, 1897. Proofs, Math. Quest. Educ. Times, 66, 1897, 108. Cf. W. E.