Find the last two digits of the number $7^{100}-8^{100}$. So I tried and found that $$7^{100} \equiv 1 \pmod{100}$$ but I got stuck with $8^{100}$. Help me out please. 

First, note that $7^2\equiv8^2\pmod{5}$. So $$7^{100}-8^{100}=(7^2-8^2)\left((7^2)^{49}+(7^2)^{48}(8^2)+\cdots+(8^2)^{49}\right)\equiv 0\pmod{25},$$
since the second term consists of $50$ summands of the same modulo $5$. 
Second $7^{100}-8^{100}\equiv 7^{100}\pmod{4}\equiv 1\pmod{4}$.
Chinese remainder theorem then implies that the answer is $25$.
