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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.

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  • $\begingroup$ Can you somehow prove that the last digit of $4^{2k} = (4^2)^k$ is $6$ for all $k \geqslant 1$? $\endgroup$ Feb 20, 2014 at 21:35
  • $\begingroup$ I have a little trouble when writing that. I don't know how I should write the hypothesis of the induction. Should I just write it like this: "Let's assume that the last digit of $4^{2k}$ is 6."? $\endgroup$
    – BowTie
    Feb 20, 2014 at 21:38
  • $\begingroup$ Are you familiar with the notation $4^2 \equiv 6 \pmod{10}$? $\endgroup$ Feb 20, 2014 at 21:40
  • $\begingroup$ Uh, cannot believe how I couldn't think of that. Thanks a bunch :D $\endgroup$
    – BowTie
    Feb 20, 2014 at 21:41

2 Answers 2

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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.

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$4^2\equiv 6 \mod 10$ and $6^2\equiv 6\mod 10$, which is equivalent to $4(^2)^2=4^4$.

With the $(6 \mod 10)$ relation, we can continue doubling the exponent of $4$ with ease.

$(4^4)^2=4^8\equiv 6^2=6 \mod 10, \dots$

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