Difference between revisions of "2021 GMC 10B Problems/Problem 10"
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Preface: there is a 89% chance this is wrong. | Preface: there is a 89% chance this is wrong. | ||
| − | Note that by | + | Note that by [[Wilson's Theorem]], |
<cmath>88! \equiv -1 \pmod{89}.</cmath> | <cmath>88! \equiv -1 \pmod{89}.</cmath> | ||
Latest revision as of 18:35, 7 March 2022
Problem
What is the remainder when 
is divided by 
?
Solution
Preface: there is a 89% chance this is wrong.
Note that by Wilson's Theorem,
![\[88! \equiv -1 \pmod{89}.\]](http://latex.artofproblemsolving.com/9/6/7/96723d6a90eac4d5d766472c3f471d38b959d7a8.png)
We can substitute this in for 
to have
![\[88!^{{{{(88!-1)}^{(88!-2)}}^{(88!-3)}}^{.....1}} \equiv (-1)^{{{{(88!-1)}^{(88!-2)}}^{(88!-3)}}^{.....1}} \pmod{89}\]](http://latex.artofproblemsolving.com/6/c/6/6c67336fb284a5d1020debb1a96b3aaa6e7c466e.png)
Note that the parity of is even. This means that 
is odd. Since 
is odd, 
is consequently odd. Applying this to our congruence, we have
![\[88!^{{{{(88!-1)}^{(88!-2)}}^{(88!-3)}}^{.....1}} \equiv -1 \equiv \boxed{\textbf{(E)}~88} \pmod{89}.\]](http://latex.artofproblemsolving.com/a/c/0/ac0651e66915bfae785e64067c352c7a53373265.png)
~pineconee
