Difference between revisions of "2024 AMC 10B Problems/Problem 7"
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| + | ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ||
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| + | https://youtu.be/QLziG_2e7CY?feature=shared | ||
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| + | ~ Pi Academy | ||
==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 10:00, 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 
.
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/QLziG_2e7CY?feature=shared
~ Pi Academy
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
