2024 AMC 10B Problems/Problem 7: Difference between revisions
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<cmath>7^{2024} (1 + 7 + 7^2) = 7^{2024} (57).</cmath> | <cmath>7^{2024} (1 + 7 + 7^2) = 7^{2024} (57).</cmath> | ||
Note that <math>57 | 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== | ==Solution 2== | ||
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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>. | 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 | |||
==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 💡✔️)== | |||
https://youtu.be/T_QESWAKUUk?si=5euBbKNMaYBROuTV&t=100 | |||
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ||
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==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
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 | ||
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