Difference between revisions of "2021 WSMO Speed Round Problems/Problem 10"
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==Solution== | ==Solution== | ||
| − | First, note that <math>\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}.</math> Now, note that <math>a^b\equiv a^{b+10}\pmod{11}</math> for all <math>a</math> and <math>b.</math> This means that we can take <math>\pmod{10}</math> on <math>\underbrace{2022^{2021^{\ldots^{2021^{2022}}}}}_{2020\text{ }2021\text{'}s}\text{ and }\underbrace{2021^{2022^{\ldots^{2022^{2021}}}}}_{2021\text{ }2022\text{'}s}.</math> We can easily find that <math>\underbrace{2021^{2022^{\ldots^{2022^{2021}}}}}_{2021\text{ }2022\text{'}s}\equiv1^{\text{something}}\equiv1\pmod{10}.</math> In addition, <math>\underbrace{2022^{2021^{\ldots^{2021^{2022}}}}}_{2020\text{ }2021\text{'}s}\equiv2^{\text{something}\equiv1\pmod{4}}\equiv2\pmod{10}.</math> Thus, <math>\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}.</math> | + | First, note that <math>\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}.</math> Now, note that <math>a^b\equiv a^{b+10}\pmod{11}</math> for all <math>a</math> and <math>b.</math> This means that we can take <math>\pmod{10}</math> on <math>\underbrace{2022^{2021^{\ldots^{2021^{2022}}}}}_{2020\text{ }2021\text{'}s}\text{ and }\underbrace{2021^{2022^{\ldots^{2022^{2021}}}}}_{2021\text{ }2022\text{'}s}.</math> We can easily find that <math>\underbrace{2021^{2022^{\ldots^{2022^{2021}}}}}_{2021\text{ }2022\text{'}s}\equiv1^{\text{something}}\equiv1\pmod{10}.</math> In addition, <math>\underbrace{2022^{2021^{\ldots^{2021^{2022}}}}}_{2020\text{ }2021\text{'}s}\equiv2^{\text{something }\equiv1\pmod{4}}\equiv2\pmod{10}.</math> Thus, <math>\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}.</math> |
~pinkpig | ~pinkpig | ||
Latest revision as of 11:14, 23 December 2021
Problem
Find the remainder when 
is divided by 11.
Solution
First, note that 
Now, note that 
for all 
and 
This means that we can take 
on 
We can easily find that 
In addition, 
Thus, 
~pinkpig
