2024 AMC 12B Problem 14

Attempt Problem 14 of the 2024 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 12B solutions, or check the answer key.

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14.

How many different remainders can result when the 100100th power of an integer is divided by 125?125?

11

22

55

2525

125125

Answer: B
Concepts:Euler’s Totient Functionmodular exponentiation
Difficulty rating: 1760
Solution:

If nn is coprime to 5,5, then since φ(125)=100,\varphi(125) = 100, Euler's theorem gives n1001(mod125).n^{100} \equiv 1 \pmod{125}. If nn is a multiple of 5,5, then n100n^{100} is divisible by 5100,5^{100}, hence by 125,125, leaving remainder 0.0.

So the only possible remainders are 00 and 1,1, which is 22 distinct values.

Thus, the correct answer is B.

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