Edit: My attempt below has a mistake. As anankElpis pointed out in the comments, I should not be searching modulo 9 and 11, not 3 and 11. That way we get that $n$ must be a multiple of both 5 and 6, which gives us 30.
Original post: Because $99=3*3*11$, we have $$5^n\equiv 1\mod 99\qquad\Leftrightarrow\qquad 5^n\equiv 1\mod 3\quad\wedge\quad 5^n\equiv 1\mod 11.$$ Now because \begin{align*} 5^1=5&=2\mod 3\\ 5^2=2\cdot5=10&=1\mod 3 \end{align*} and so on, we know that $n$ must be even, and because \begin{align*} 5^1&=5\mod 11\\ 5^2=5\cdot 5=25&=3\mod 11\\ 5^3=3\cdot 5=15&=4\mod 11\\ 5^4=4\cdot 5=20&=9\mod 11\\ 5^5=9\cdot 5=45&=1\mod 11, \end{align*} we know that $n$ must be a multiple of 5. So now we've narrowed $n$ down to $10,20,30,...$. But I don't know how to continue from here.
Another thing I tried: We also know that $5^{\phi(99)}\equiv 1\mod 99$, where $\phi$ is Euler's totient function. I calculated $\phi(99)=60$ by removing all multiples of 3 (33 numbers) and then the remaining multiples of 11 (6 numbers), so $99-33-6=60$. So we know that $5^{60}\equiv 1\mod 99$. However, when I checked with WolframAlpha, I found out that $5^{30}\equiv 1\mod 99$, which makes $n=30$ the correct answer.
Can anyone give me a hint about how to find the answer? Thank you for your time.
