Difference between revisions of "2024 AMC 10B Problems/Problem 7"
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+ | ==Solution 3== | ||
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+ | We start the same as solution 2, and find that: | ||
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+ | <cmath>7^1\equiv7\pmod{19}</cmath> | ||
+ | <cmath>7^2\equiv11\pmod{19}</cmath> | ||
+ | <cmath>7^3\equiv1\pmod{19}</cmath> | ||
+ | |||
+ | We know that for <math>2024</math>, <math>2025</math>, and <math>2026</math>, because there are three terms, we can just add them up. <math>1 + 7 + 11 = 19</math>, which is <math>0</math> mod <math>19</math>. | ||
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+ | ~ILoveMath31415926535 | ||
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== |
Revision as of 16:37, 14 November 2024
Contents
[hide]Problem
What is the remainder when is divided by ?
Solution 1
We can factor the expression as
Note that is divisible by , this expression is actually divisible by 19. The answer is .
Solution 2
If you failed to realize that the expression can be factored, you might also apply modular arithmetic to solve the problem.
Since , the powers of repeat every three terms:
The fact that , , and implies that .
Solution 3
We start the same as solution 2, and find that:
We know that for , , and , because there are three terms, we can just add them up. , which is mod .
~ILoveMath31415926535
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/QLziG_2e7CY?feature=shared
~ Pi Academy
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=24EZaeAThuE
See also
2024 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.