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>. | ||
| + | |||
| + | ~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
Problem
What is the remainder when 
is divided by 
?
Solution 1
We can factor the expression as
![\[7^{2024} (1 + 7 + 7^2) = 7^{2024} (57).\]](http://latex.artofproblemsolving.com/0/5/f/05f12ad7ed22ebbea9c9e4437c5ba95e02476353.png)
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:
![\[7^1\equiv7\pmod{19}\]](http://latex.artofproblemsolving.com/c/2/4/c241c520618add32d93f1ceec343a869bff401e0.png)
![\[7^2\equiv11\pmod{19}\]](http://latex.artofproblemsolving.com/7/8/8/7884bae9a7c83119054fe26642be6f54fa5d3558.png)
![\[7^3\equiv1\pmod{19}\]](http://latex.artofproblemsolving.com/9/6/a/96a87b5ffd4ca057d6e94eef31d6fb7de25d590c.png)
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. 
