Difference between revisions of "2024 AMC 12B Problems/Problem 14"
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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 (B)2 | ||
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~mitsuihisashi14 | ~mitsuihisashi14 | ||
Revision as of 01:12, 14 November 2024
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
