let n a Natural number whose decimal expansion consists of 3^n identical digits, sshow by induction 3^n divide

Question

what are the possible reminders when 100th power of an integer is divided by 125.

Answer 1

First one.

The n+1 st number will be the nth times a number that you can quickly prove is divisible by 3 (because the remainder of any power of 10, when divided by 3, is exactly 1). That's the core of the inductive step. You can surely construct a proof from there.

For the second one. Hmm. Case 1: The integer is divisible by 5. Then you know the remainder. :)

Case 2: It isn't. Then it's 4th power has remainder 1 when divided by 5. Call it 5k+1. (5k+1)^25, when expanded out, would seem to have every term divisible by 125 except for the last one, which is just 1.

So the answer is just "0 and 1"