342 History of the Theory of Numbers. [Chap, xii 
primes 6<200 and certain larger primes, from which are easily deduced 
tests for the divisor 6. 
Several"" noted that if 10 belongs to the exponent n modulo d, and if 
Si, Si, ■ ■ .denote the sums of every nth digit of N beginning with the first, 
second, ... at the right, the remainder on the division of A^ by d is that of 
S1 + IOS2+IO-S3+ . . . 
J. Fontes^^ would find the least residue of A^ modulo M. If 10" has the 
residue q modulo M, we do not change the least residue of N if we multiply 
a set of n digits of A^ by the same power of q as of 10". Thus for M = 19, 
iV=10433 = 10'+4-10H33, 10" has the residue 5 modulo 19 and we may 
replace N by 5- +4 ',5+ 33. The method is applied to each prime M^ 149. 
Fontes^^ gave a history of the tests for divisibility, and an "extension 
of the method of Pascal," similar to that in his preceding paper. 
P. Valerio*° would test the divisibility of N by 39, for example, by sub- 
tracting from N a multiple of 39 with the same ending as N. 
F. Belohldvek^^ noted that 10A-\-B is divisible by 10p±l if A=FpB is. 
C. Borgen'^^ ^oted that Z = a„-10"+ . . . +ai-10+ao is divisible by A^ if 
"T' (a_„+rlO''-'+ . . . +a,)(10"-A^)''/'' 
..=0 
is divisible by N. For A'' = 7, take a = 1 ; then 10"— AT = 3 and Z is divisible 
by 7 if ao+3ai+2a2 — ^3 — 3a4 — 2a5+ ... is divisible by 7. 
J. J. Sylvester^^" noted that, if the r digits of A'^, 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, 10^, . . ., modulo 13], the sum 
of the products is divisible by 13 if and only if N is divisible by 13. 
C. L. Dodgson^^** discussed the quotient and remainder on division by 
9 or 11. 
L. T. Riess^' noted that, if p is not divisible by 2 or 5, 106+a(a<10) 
is divisible by p if b — xa is divisible by p, where mp = 10x-\-a (a< 10) and 
m = l, 7, 3, 9 according as p=l, 3, 7, 9 (mod 10), respectively. 
A. Loir^^ gave tests for prime divisors < 100 by uniting them by twos 
or threes so that the product P ends in 1 , as 7 -43 = 30 1 . To test N, multiply 
the number formed of the last two digits of A'^ by the number preceding 01 in 
P, subtract the product from A^, 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. 
Plakhowo"*^ gave the test by Bougaief, but without using congruences. 
'"'Math. Quest. Educ. Times, 57, 1892, 111. 
"Assoc, frang. avanc. sc, 22, 1893, II, 240-254. 
»M4m. ac. sc. Toulouse, (9), 5, 1893, 459-475. 
"La Revue Scientifique de France, (3), 52, 1893, 765 
"Casopis, Prag, 23, 1894, 59. "Mature, 57, 1897-8, 54. 
«MEducat. Times, March, 1897. Proofs. Math. Quest. Educ. Times, 66, 1897, 108. Cf. W. E. 
Heal, Amer. Math. Monthly, 4, 1897, 171-2. 
"'^Nature, 56, 1897, 565-6. 
«Russ. Nat., 1898, 329. Cf. Jahrb. Fortschritte Math., 29, 1898, 137. 
"Assoc, frang. avanc. sc, 27, 1898, II, 144-6. 
«»Bull. des sc. math, et phys. 61(§mentaires, 4, 1898-9, 241-3, 
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