Question

Can you find the smallest `n` ? as `3^n-2^n` can be divided by `2015`

Answer 1

See below.

Explanation:

Let us prove by finite induction, that

`3^(2k)-2^(2k) equiv 0 mod 5`

1) for `k=1 rArr (3^2-2^2) = (3+2)(3-2) equiv 0 mod 5`

2) now supposing that `3^(2k)-2^(2k) equiv 0 mod 5` then

`3^(2(k+1))-2^(2(k+1))=9*3^(2k)+4*2^(2k) = 5*3^(2k)+4*(3^(2k)-2^(2k)) equiv 0 mod 5`

Once we now that `3^(2k)-2^(2k) equiv 0 mod 5` with some tabulations we obtain for `k=30`

`(3^60-2^60)/2015 = 21037795669510313652450415` so the solution is for `n = 2k=60`