107(mod77)
I tried repeated squaring, which worked but took many computations. I also tried Fermat's little theorem, but since 7<77 I didn't know how to use it.
Any simpler way to do this?
Answer
Not much effort saved, but work separately mod 11 (easy, since 10≡−1(mod11)) and mod 7 (easy since by Fermat 107≡10(mod7)). Then stitch the answers together using the Chinese Remainder Theorem. In this case that can be done by inspection: the answer is 10.
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