Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2024 AMC 10B Problems/Problem 18"

(→‎Solution 2)
(→‎Solution 2 (Euler Totient))
Line 27: Line 27:
 
Clearly the only remainder provides 0
 
Clearly the only remainder provides 0
  
−
Therefore, the remainders can only be 1 and 0, which gives the answer (B)2
+
Therefore, the remainders can only be 1 and 0, which gives the answer <math>\(B)boxed{2}</math>.
  
 
~mitsuihisashi14
 
~mitsuihisashi14

Revision as of 01:40, 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 (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)boxed{2}$ (Error compiling LaTeX. Unknown error_msg).

~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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10B_Problems/Problem_18&oldid=233644"