FoolForMath Group Title Just another cute number theory problem: $$(1)$$ Find the remainder when $$100!^{100!}+99$$ is divided by $$101$$. PS: This problem is posted on request of @Limitless 2 years ago 2 years ago

1. .Sam.

Wilson's Theorem?

2. FoolForMath

Bingo! ^^

3. mathmagician

first, you should see that 100^k mod 101 is equal to 100, if k is odd and 1 if k is even. The same happens with 100!^k. So, 100!^100!=1mod 101 and 100!^100!+99=100 mod 101

4. FoolForMath

By Wilson's theorem, $100! \equiv -1 \pmod {101}$ Since $$100!$$ is even: $100!^ {100!} \equiv 1 \pmod {101} \implies 100!^ {100!} +99 \equiv 100 \pmod {101}$ Hence, $$100$$ is the remainder. Ref:http://en.wikipedia.org/wiki/Wilson's_theorem

5. FoolForMath

EDIT: $100!^ {100!} \equiv 1 \pmod {101}$$\implies 100!^ {100!} +99 \equiv 100 \pmod {101}$

6. KingGeorge

You don't even need Wilson's Theorem for this. Since $$100!$$ is a multiple of $$100$$, and coprime to 101, we have that $$100!^{100!}\equiv1\pmod{101}$$ by Fermat's Little Theorem. Hence, $$100!^{100!}+99\equiv100\pmod{101}$$.