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Difference between revisions of "2021 GMC 10B Problems/Problem 10"

(Created page with "==Problem== What is the remainder when <math>88!^{{{{(88!-1)}^{(88!-2)}}^{(88!-3)}}^{.....1}}\cdot 1^{2^{3^{4^{.....88!}}}}</math> is divided by <math>89</math>? <math>\textb...")
 
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Preface: there is a 89% chance this is wrong.
 
Preface: there is a 89% chance this is wrong.
  
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Note that by Wilson’s Theorem,
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Note that by [[Wilson's Theorem]],
 
<cmath>88! \equiv -1 \pmod{89}.</cmath>
 
<cmath>88! \equiv -1 \pmod{89}.</cmath>
  

Latest revision as of 19:35, 7 March 2022

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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