Stephen Froggatt indicates in Math Forum an equation, it can help in finding remainders Divisibility Rules to 50 and Beyond.
His explanation is as follow (divisibility by 7 in decimal):
Assume, that 10b+a = 7k.
Then b+5a = b+5(7k-10b) = 35k-49b = 7(5k-7b).
But where are 5 get from?
There exists a numbers M, K and formula
M*P - K*D = 1
for base P and divisor D. Let M is as small positive as possible.
Then the number N is split into two parts, b is N with cutting least significant digit off, and this digit a.
This equation is equivalent to congruence
M*P = 1 (mod D)
Such case
N*M = (b*P+a)M = b(P*M)+a*M = b*1 + a*M = b+aM (mod D)
In mentioned example we check: 10=3 (7), 20=-1 (7), 30=2 (7), 40=5 (7), 50=1(7) it is this, M=5.
Number N is divisible by D iff b+M*a is divisible by D.
Congruence allows the second formulae, when M is bigger than half D. Simply decrease M by D:
Number N is divisible by D iff b+(M-D)*a is divisible by D.
Upgrade [SF-]
We can use congruence
M*P = -1 (mod D)
Such case there is
N*M = -b+aM = -(b-aM) (mod D)
This method doesn't detect quotient N / D.
