Use strong induction to prove the ones digit of 4^k.

Question

The question is to prove the following using strong induction:

"For 4^k, where k is a nonnegative integer, if k is even then the ones digit of 4^k is 6 and if k is odd then the ones digit of 4^k is 4."

The problem also hints at using the quotient-remainder theorem (n = dq + r). By setting d = 10, r would be the ones digit of n. I don't know how to connect this with the problem.

I've been thinking about this problem for literally hours and have no idea what to do. Any help would be appreciated.

Answer 1

What happens if you multiply a number that ends in a 6 by 4? What about if the number ends in a 4 instead?

Answer 2

This can be solved with a relatively simple induction – the easiest way would probably be to note that $4^3 \equiv 4 \pmod{10}$, and therefore, for all $n\geq 1$, $4^{2n+1} \equiv 4 \pmod{10}$ and $4^{2n+2} = (4^{2n+1})\cdot4 \equiv 16 \equiv 6 \pmod{10}$, the first identity being proved by induction.

I'll leave you to fill in the specific details for the induction hypothesis and induction step.