What is the last digit of this expression: $4^{2^1} + 4^{2^2} + 4^{2^3} + ...+ 4^{2^{2013}}$

Question

The expression:

$4^{2^1} + 4^{2^2} + 4^{2^3} + ...+ 4^{2^{2013}}$

I know that the last digit of each of the sumands is 6, but I have trouble proving that. I tried to prove it using induction, but then I realized that I don't have a clue how to write it properly.

Also, because of the fact that the last digit of every sumand is 6, the last digit of the given expression is the last digit of $2013 * 6$ which is 8.

Answer 1

You can write it as the following (following from Daniel's hint): $16x\equiv6\mod 10$ for any $x$ where $x\equiv6\mod10$. From there it is easy to show that $4^{2^k}\equiv6\mod10$ and you already have the rest.