2024 AMC 10B Problems/Problem 18

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 (Euler Totient)

We split the cases into:

1. If x is not a multiple of 5: we get $x^{100} \equiv 1 \pmod{125}$

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

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

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2024 AMC 10B Problems/Problem 18

Contents

Problem

Solution 1

Solution 2 (Euler Totient)

See also