2021 GMC 10B Problems/Problem 10

Problem

What is the remainder when $88!^{{{{(88!-1)}^{(88!-2)}}^{(88!-3)}}^{.....1}}\cdot 1^{2^{3^{4^{.....88!}}}}$ is divided by $89$ ?

$\textbf{(A)} ~0 \qquad\textbf{(B)} ~1 \qquad\textbf{(C)} ~44 \qquad\textbf{(D)} ~59 \qquad\textbf{(E)} ~88$

Solution

Preface: there is a 89% chance this is wrong.

Note that by Wilson's Theorem, $88! \equiv -1 \pmod{89}.$

We can substitute this in for $88!$ to have $88!^{{{{(88!-1)}^{(88!-2)}}^{(88!-3)}}^{.....1}} \equiv (-1)^{{{{(88!-1)}^{(88!-2)}}^{(88!-3)}}^{.....1}} \pmod{89}$

Note that the parity of $88!$ is even. This means that $88! - 1$ is odd. Since $88!-1$ is odd, ${{{{(88!-1)}^{(88!-2)}}^{(88!-3)}}^{.....1}}$ is consequently odd. Applying this to our congruence, we have $88!^{{{{(88!-1)}^{(88!-2)}}^{(88!-3)}}^{.....1}} \equiv -1 \equiv \boxed{\textbf{(E)}~88} \pmod{89}.$ ~pineconee

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2021 GMC 10B Problems/Problem 10

Problem

Solution