2024 AMC 12B Problems/Problem 14

Revision as of 01:11, 14 November 2024 by Mitsuihisashi14 (talk | contribs)

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