2024 AMC 12B Problems/Problem 14
Revision as of 01:10, 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 2, which gives the answer (B)2 ~mitsuihisashi14
