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2021 WSMO Speed Round Problems/Problem 10

Revision as of 11:13, 23 December 2021 by Pinkpig (talk | contribs) (Created page with "==Problem== Find the remainder when <math>\underbrace{2021^{2022^{\ldots^{2022^{2021}}}}}_{2021\text{ } 2021\text{'}s}\cdot\underbrace{2022^{2021^{\ldots^{2021^{2022}}}}}_{202...")
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Problem

Find the remainder when $\underbrace{2021^{2022^{\ldots^{2022^{2021}}}}}_{2021\text{ } 2021\text{'}s}\cdot\underbrace{2022^{2021^{\ldots^{2021^{2022}}}}}_{2022\text{ }2022\text{'}s}$ is divided by 11.

Solution

First, note that $\underbrace{2021^{2022^{\ldots^{2022^{2021}}}}}_{2021\text{ } 2021\text{'}s}\cdot\underbrace{2022^{2021^{\ldots^{2021^{2022}}}}}_{2022\text{ }2022\text{'}s}\equiv\underbrace{8^{2022^{\ldots^{2022^{2021}}}}}_{2021\text{ } 2021\text{'}s}\cdot\underbrace{9^{2021^{\ldots^{2021^{2022}}}}}_{2022\text{ }2022\text{'}s}.$ Now, note that $a^b\equiv a^{b+10}\pmod{11}$ for all $a$ and $b.$ This means that we can take $\pmod{10}$ on $\underbrace{2022^{2021^{\ldots^{2021^{2022}}}}}_{2020\text{ }2021\text{'}s}\text{ and }\underbrace{2021^{2022^{\ldots^{2022^{2021}}}}}_{2021\text{ }2022\text{'}s}.$ We can easily find that $\underbrace{2021^{2022^{\ldots^{2022^{2021}}}}}_{2021\text{ }2022\text{'}s}\equiv1^{\text{something}}\equiv1\pmod{10}.$ In addition, $\underbrace{2022^{2021^{\ldots^{2021^{2022}}}}}_{2020\text{ }2021\text{'}s}\equiv2^{\text{something}\equiv1\pmod{4}}\equiv2\pmod{10}.$ Thus, $\underbrace{2021^{2022^{\ldots^{2022^{2021}}}}}_{2021\text{ } 2021\text{'}s}\cdot\underbrace{2022^{2021^{\ldots^{2021^{2022}}}}}_{2022\text{ }2022\text{'}s}\equiv8^2\cdot9\equiv\boxed{4}\pmod{11}.$

~pinkpig

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