Difference between revisions of "2024 AMC 10B Problems/Problem 18"
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so <math>(5+n)^{100} \equiv n^{100}</math>. Substituting <math>-n</math> for <math>n</math>, we get <math>(5-n)^{100} \equiv n^{100}</math>. Therefore, the remainders when divided by <math>125</math> repeat every <math>5</math> integers, so we only need to check the <math>100</math>th powers of <math>0, 1, 2, 3, 4</math>. But we have that <math>(5-1)^{100} \equiv 1^{100}</math> and <math>(5-2)^{100} \equiv 2^{100}</math>, so we really only need to check <math>0, 1, 2</math>. We know that <math>0, 1</math> produce different remainders, so the answer to the problem is either <math>2</math> or <math>3</math>. But <math>3</math> is not an answer choice, so the answer is <math>\boxed{\textbf{(B) } 2}</math>. | so <math>(5+n)^{100} \equiv n^{100}</math>. Substituting <math>-n</math> for <math>n</math>, we get <math>(5-n)^{100} \equiv n^{100}</math>. Therefore, the remainders when divided by <math>125</math> repeat every <math>5</math> integers, so we only need to check the <math>100</math>th powers of <math>0, 1, 2, 3, 4</math>. But we have that <math>(5-1)^{100} \equiv 1^{100}</math> and <math>(5-2)^{100} \equiv 2^{100}</math>, so we really only need to check <math>0, 1, 2</math>. We know that <math>0, 1</math> produce different remainders, so the answer to the problem is either <math>2</math> or <math>3</math>. But <math>3</math> is not an answer choice, so the answer is <math>\boxed{\textbf{(B) } 2}</math>. | ||
+ | |||
+ | ==Solution 4 (Totient)== | ||
+ | Euler's Totient Function, <math>\phi(n)</math> returns <math>n\cdot\left(1-\dfrac{1}{x}\right)</math> as a product of each prime divisor of <math>n</math>. | ||
+ | |||
+ | Euler's Totient Theorem states that if <math>{a}</math> is an integer and <math>p</math> is a positive integer relatively prime to <math>a</math>, then <math>{a}^{\phi (p)}\equiv 1 \pmod {p}</math>. | ||
+ | |||
+ | In this case, <math>p=125</math>, which is convenient because <math>125</math> only has one prime factor, <math>5</math>, therefore <math>\phi(125)=100</math>, so <cmath>a^{100}\equiv1\pmod{125}</cmath>where <math>\gcd(a,125)=1</math>. Every single number that isn't a multiple of <math>5</math> is relatively prime to <math>125</math>, therefore we have two cases: | ||
+ | |||
+ | 1) <math>a\%5\neq0\implies a^{100}\equiv1\pmod{125}</math> | ||
+ | 2) <math>a\%5=0\implies a^{100}=5^3\cdot x\implies a^{100}\equiv0\pmod{125}</math> | ||
+ | |||
+ | The answer is <math>\boxed{\text{(B) }2}</math> ~Tacos_are_yummy_1 | ||
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== |
Revision as of 12:05, 14 November 2024
- The following problem is from both the 2024 AMC 10B #18 and 2024 AMC 12B #14, so both problems redirect to this page.
Contents
[hide]Problem
How many different remainders can result when the th power of an integer is divided by ?
Fast Solution
https://www.youtube.com/watch?v=S7l_Yv2Sd7E
Solution 1
First note that the totient function of is . We can set up two cases, which depend on whether a number is relatively prime to .
If is relatively prime to , then because of Euler's Totient Theorem.
If is not relatively prime to , it must be have a factor of . Express as , where is some integer. Then .
Therefore, can only be congruent to or . Our answer is .
~lprado
Solution 2 (Euler Totient)
We split the cases into:
1. If x is not a multiple of 5: we get
2. If x is a multiple of 125: Clearly the only remainder provides 0
Therefore, the remainders can only be 1 and 0, which gives the answer .
~mitsuihisashi14
Solution 3
Note that
Taking this mod , we can ignore most of the terms except the for the last :
so . Substituting for , we get . Therefore, the remainders when divided by repeat every integers, so we only need to check the th powers of . But we have that and , so we really only need to check . We know that produce different remainders, so the answer to the problem is either or . But is not an answer choice, so the answer is .
Solution 4 (Totient)
Euler's Totient Function, returns as a product of each prime divisor of .
Euler's Totient Theorem states that if is an integer and is a positive integer relatively prime to , then .
In this case, , which is convenient because only has one prime factor, , therefore , so where . Every single number that isn't a multiple of is relatively prime to , therefore we have two cases:
1) 2)
The answer is ~Tacos_are_yummy_1
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/c6nhclB5V1w?feature=shared
~ Pi Academy
See also
2024 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 |
2024 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.