Difference between revisions of "2024 AMC 12B Problems/Problem 14"
| Line 7: | Line 7: | ||
Clearly the only remainder provides 0 | Clearly the only remainder provides 0 | ||
| − | Therefore, the remainders can only be 1 and | + | Therefore, the remainders can only be 1 and 0, which gives the answer (B)2 |
~mitsuihisashi14 | ~mitsuihisashi14 | ||
Revision as of 01:11, 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
