Difference between revisions of "2024 AMC 10B Problems/Problem 7"
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Note that <math>57</math> is divisible by <math>19</math>, this expression is actually divisible by 19. The answer is <math>\boxed{\textbf{(A) } 0}</math>. | Note that <math>57</math> is divisible by <math>19</math>, this expression is actually divisible by 19. The answer is <math>\boxed{\textbf{(A) } 0}</math>. | ||
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+ | ==Solution 2== | ||
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+ | If you failed to realize that the expression can be factored, you might also apply modular arithmetic to solve the problem. | ||
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+ | Since <math>7^3\equiv1\pmod{19}</math>, the powers of <math>7</math> repeat every three terms: | ||
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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> | ||
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+ | The fact that <math>2024\equiv2\pmod3</math>, <math>2025\equiv0\pmod3</math>, and <math>2026\equiv1\pmod3</math> implies that <math>7^{2024}+7^{2025}+7^{2026}\equiv11+1+7\equiv19 \equiv0\pmod{19}</math>. | ||
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+ | ~[[User:Bloggish|Bloggish]] | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=B|num-b=6|num-a=8}} | {{AMC10 box|year=2024|ab=B|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 04:39, 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 .
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 |
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