What is remainder if a long integer(X) is divided by 3^n. What if divided by 4^n...?

Question

OK, i know that for dividing X by 2^n, i shud actually find remainder of last n digits of dividend divided by 2^n. Is there any similar strategy for other divisors like 3^n, 4^n, 5^n, etc? -thanks

sorry, but i wrote 4^n by mistake. i want to know strategies(if any) for 3^n, 5^n, etc..

Answer 1

I think this only works for powers of 2. So4^n = 2^2n and should work. I don't think it works for anything but 2^kn.

Answer 2

it fairly is the dazzling answer on account that those on the numerator are addition and subtraction you could divide them via the denominator one after the different and once you do you get 6m^2n-2mn^2+m