Difference between revisions of "2024 AMC 10B Problems/Problem 7"
Xhypotenuse (talk | contribs) |
m |
||
| (3 intermediate revisions by 2 users not shown) | |||
| Line 46: | Line 46: | ||
~xHypotenuse | ~xHypotenuse | ||
| + | |||
| + | ==Solution 5 (Basic Mod Almost same as sol one)== | ||
| + | |||
| + | <math>7^{2024} * (1+7+49) == 0 (mod 19)</math> | ||
| + | <math>57 == 0 mod 19</math> | ||
| + | |||
| + | so it is just <math>\boxed{\textbf{(A) } 0}</math>. | ||
| + | |||
| + | ~Cheerfulfrog | ||
==🎥✨ Video Solution by Scholars Foundation ➡️ (Easy-to-Understand 💡✔️)== | ==🎥✨ Video Solution by Scholars Foundation ➡️ (Easy-to-Understand 💡✔️)== | ||
| Line 59: | Line 68: | ||
==Video Solution 2 by SpreadTheMathLove== | ==Video Solution 2 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=24EZaeAThuE | https://www.youtube.com/watch?v=24EZaeAThuE | ||
| + | |||
| + | ==Video Solution by TheBeautyofMath== | ||
| + | https://youtu.be/ZaHv4UkXcbs?t=208 | ||
| + | |||
| + | ~IceMatrix | ||
==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}} | ||
Latest revision as of 21:19, 6 October 2025
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4 (Given more advanced knowledge)
- 6 Solution 5 (Basic Mod Almost same as sol one)
- 7 🎥✨ Video Solution by Scholars Foundation ➡️ (Easy-to-Understand 💡✔️)
- 8 Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
- 9 Video Solution 2 by SpreadTheMathLove
- 10 Video Solution by TheBeautyofMath
- 11 See also
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 
, 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 .
Solution 4 (Given more advanced knowledge)
By Fermat's Little Theorem (FLT), we know that ![\[7^{18}\equiv1\pmod{19}\]](http://latex.artofproblemsolving.com/5/2/7/52712c18689479933820da01da7f52a9a365ba8b.png)
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
Solution 5 (Basic Mod Almost same as sol one)
so it is just .
~Cheerfulfrog
🎥✨ 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
Video Solution by TheBeautyofMath
https://youtu.be/ZaHv4UkXcbs?t=208
~IceMatrix
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. 
