2024 AMC 12B Problem 14

Below is the professionally curated solution for Problem 14 of the 2024 AMC 12B, 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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Concepts:Euler’s Totient Functionmodular exponentiation

Difficulty rating: 1760

14.

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

11

22

55

2525

125125

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.

Problem 14 in Other Years