Finding the Final Digits in a Stacked Exponents Problem

Question

How to find the the last two digits of the number $$\underbrace{\huge 7^{7^{7^{...}}}}_{1+n}$$ When there are $n$ sevens in the ascending exponents.

I have no idea how to condense this or where to begin.

Answer 1

We are interested only in the last two digits, so we care about the remainder of your power tower, modulo $100$.

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My hint is this: $$7^4 \equiv 2401 \equiv 1 \pmod{100}$$ $$7^2 \equiv 1 \pmod{4}$$

Answer 2

By Eulero-Fermat theorem, the tetration $\text{ }^n7$ verifies for all $n\ge 3$: $$ \text{ }^n7\bmod{100} \equiv 7^{(\text{ }^{n-1}7 \bmod{\varphi(100)})} \equiv 7^{7^{(\text{ }^{n-2}7 \bmod{\varphi(\varphi(100))})}} \equiv \text{ }^37\bmod{100}. $$

Indeed $\varphi(\varphi(100))=8$, and $\text{ }^{n-2}7 \bmod{8}=7\bmod{8}$. Notice that the same method, together with Chinese remainder theorem, works for the reduction of any congruence of the type $\text{ }^na\bmod{b}$.