Find last digit of $7^{7^{7^7}}$

Question

Find last digit of $7^{7^{7^7}}$

I know that the last digit of $7^x$ depends on the remainder $x$ leaves when divided by $4$: ($x = 7^{7^7}$) $$7^{4k} \equiv 1 \bmod 10$$ $$7^{4k+1} \equiv 7 \bmod 10$$ $$7^{4k+2} \equiv 9 \bmod 10$$ $$7^{4k+3} \equiv 3 \bmod 10$$

And also that

$$7 \equiv -1 \bmod 4$$

So for all odd positive numbers, $7^{n} \equiv -1 \bmod 4$ and $7^{7}$ is odd. But which congruence relation should I use? $4k+3$ and $4k+1$ are both odd

Please explain, thanks.

Answer 1

(a)...$7^2\equiv 1$ mod $4\implies$ any odd power of $7$ is congruent to $7$ mod $ 4 \implies$ any odd power of $7$ is congruent to $3$ mod $4\implies 7^{7^7}\equiv 3$ mod $4$....(b)...$7^4\equiv 1$ mod $10\implies ( 7^n\equiv 7^3\equiv 3$ mod $10$ when $n\equiv 3$ mod $4)$...(c) Therefore $$7^{7^{7^7}}\equiv 3 \text { mod }10.$$

Answer 2

You're getting yourself confused. $7^7$ is odd, correct; but $7^7$ isn't the exponent you care about. To determine what form to use, the exponent you're looking at is $7^{7^7}$. By your observations, $7^7$ is odd, so $7^{7^7} \equiv -1 \mod 4$. $4n+1$ and $4n+3$ are both odd, yes, but only one of them is $-1\mod 4$: $4n+3$. So $7^{7^7} \equiv 3\mod 4$, and so $7^{7^{7^7}} \equiv 3\mod 10$ according to your table.