$\displaystyle 9^{100}-1$ is divisible by 10.
How can I do this problem? My book doesn't give adequate info on how to do it.
Use Euler's theorem. Coincidentally, I just mentioned this in the other thread you started not long ago.
$\displaystyle \varphi(10)=4$
$\displaystyle \gcd(9,10)=1$
$\displaystyle 9^{100}-1\equiv 9^{25\cdot4}-1\equiv \left(9^4\right)^{25}-1\equiv1^{25}-1\equiv1-1\equiv0\ (\text{mod}\ 10)$
Using the the following rule: If $\displaystyle a\equiv b (mod\ n)$, then $\displaystyle a^k\equiv b^k(mod\ n)$.
Now, notice $\displaystyle 9\equiv -1(mod\ 10)$, then $\displaystyle 9^{100}\equiv (-1)^{100}\equiv 1(mod\ 10)$.
Then $\displaystyle 9^{100}-1\equiv 0 (mod\ 10)$.