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Difference between revisions of "2024 AMC 10B Problems/Problem 18"
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{{duplicate|[[2024 AMC 10B Problems/Problem 18|2024 AMC 10B #18]] and [[2024 AMC 12B Problems/Problem 14|2024 AMC 12B #14]]}}
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==Problem ==
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How many different remainders can result when the <math>100</math>th power of an integer is
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divided by <math>125</math>?
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<math>\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 125</math>
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==Solution 1==
 
==Solution 1==
 
First note that the totient function of <math>125</math> is <math>100</math>. We can set up two cases, which depend on whether a number is relatively prime to <math>125</math>.
 
First note that the totient function of <math>125</math> is <math>100</math>. We can set up two cases, which depend on whether a number is relatively prime to <math>125</math>.
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~lprado
 
~lprado
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==Solution 2==
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We split the cases into:
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1. If x is not a multiple of 5:
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we get x^100\equiv 1 (mod 125)
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2. If x is a multiple of 125:
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Clearly the only remainder provides 0
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Therefore, the remainders can only be 1 and 0, which gives the answer (B)2
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~mitsuihisashi14
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==See also==
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{{AMC10 box|year=2024|ab=B|num-b=17|num-a=19}}
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{{AMC12 box|year=2024|ab=B|num-b=13|num-a=15}}
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{{MAA Notice}}

Revision as of 01:24, 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.

Problem

How many different remainders can result when the $100$th power of an integer is divided by $125$?

$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 125$

Solution 1

First note that the totient function of $125$ is $100$. We can set up two cases, which depend on whether a number is relatively prime to $125$.

If $n$ is relatively prime to $125$, then $n^{100} \equiv 1 \pmod{125}$ because of Euler's Totient Theorem.

If $n$ is not relatively prime to $125$, it must be have a factor of $5$. Express $n$ as $5m$, where $m$ is some integer. Then $n^{100} \equiv (5m)^{100} \equiv 5^{100}\cdot m^{100} \equiv 125 \cdot 5^{97} \cdot m^{100} \equiv 0 \pmod{125}$.

Therefore, $n^{100}$ can only be congruent to $0$ or $1 \pmod{125}$. Our answer is $\boxed{2}$.

~lprado

Solution 2

We split the cases into:

1. If x is not a multiple of 5: we get x^100\equiv 1 (mod 125)

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 (B)2

~mitsuihisashi14

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. AMC logo.png

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