Difference between revisions of "2024 AMC 10B Problems/Problem 7"
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+ | ==Problem== | ||
+ | What is the remainder when <math>7^{2024}+7^{2025}+7^{2026}</math> is divided by <math>19</math>? | ||
+ | <math>\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 7 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 18</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | We can factor the expression as | ||
+ | |||
+ | <cmath>7^{2024} (1 + 7 + 7^2) = 7^{2024} (57).</cmath> | ||
+ | |||
+ | Note that <math>57=19\cdot3</math>, this expression is actually divisible by 19. The answer is <math>\boxed{\textbf{(A) } 0}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | If you failed to realize that the expression can be factored, you might also apply modular arithmetic to solve the problem. | ||
+ | |||
+ | Since <math>7^3\equiv1\pmod{19}</math>, the powers of <math>7</math> repeat every three terms: | ||
+ | |||
+ | <cmath>7^1\equiv7\pmod{19}</cmath> | ||
+ | <cmath>7^2\equiv11\pmod{19}</cmath> | ||
+ | <cmath>7^3\equiv1\pmod{19}</cmath> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ~[[User:Bloggish|Bloggish]] | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | We start the same as solution 2, and find that: | ||
+ | |||
+ | <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>. | ||
+ | |||
+ | |||
+ | ==Solution 4 (Given more advanced knowledge)== | ||
+ | |||
+ | By Fermat's Little Theorem (FLT), we know that <cmath>7^{18}\equiv1\pmod{19}</cmath> Then its order must divide <math>18</math>. Trying simple values we try and succeed: <cmath>7^3\equiv1\pmod{19}</cmath> | ||
+ | |||
+ | |||
+ | So the expression is equivalent to <math>1+7+49\pmod{19}</math>, which gives <math>\boxed{\textbf{(A) } 0}</math> when divided by 19. | ||
+ | |||
+ | |||
+ | ~xHypotenuse | ||
+ | |||
+ | ==🎥✨ Video Solution by Scholars Foundation ➡️ (Easy-to-Understand 💡✔️)== | ||
+ | |||
+ | https://youtu.be/T_QESWAKUUk?si=5euBbKNMaYBROuTV&t=100 | ||
+ | |||
+ | ==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== | ||
+ | {{AMC10 box|year=2024|ab=B|num-b=6|num-a=8}} | ||
+ | {{MAA Notice}} |
Latest revision as of 00:52, 21 November 2024
Contents
[hide]Problem
What is the remainder when is divided by ?
Solution 1
We can factor the expression as
Note that , 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 .
Solution 4 (Given more advanced knowledge)
By Fermat's Little Theorem (FLT), we know that Then its order must divide . Trying simple values we try and succeed:
So the expression is equivalent to , which gives when divided by 19.
~xHypotenuse
🎥✨ Video Solution by Scholars Foundation ➡️ (Easy-to-Understand 💡✔️)
https://youtu.be/T_QESWAKUUk?si=5euBbKNMaYBROuTV&t=100
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.