Find what values of $n$ give $\varphi(n) = 10$ 
For what values of $n$ do we get $\varphi(n) = 10$?

Here, $\varphi$ is the Euler Totient Function. Now, just by looking at it, I can see that this happens when $n = 11$. Also, my friend told me that it happens when $n = 22$, but both of these are lucky guesses, or educated guesses. I haven't actually worked it out, and I don't know if there are any more.
How would I go about answering this question?

Suppose $\varphi(n)=10$.  If $p \mid n$ is prime then $p-1$ divides $10$.  Thus $p$ is one of $2,3,11$.
If $3 \mid n$, it does so with multiplicity $1$.  But then there would exist $m \in \mathbb{N}$ such that $\varphi(m)=5$, and this quickly leads to a contradiction (e.g. note that such values are always even).
Thus $n$ fits the form $2^a\cdot 11^b$, and we claim that $b=1$.  If $b>1$ we have $11 \mid \varphi(n)$, a contradiction.  As well, $b=0$ gives $\varphi(n)$ a power of $2$.  Thus $n=2^a \cdot 11$, and it's easy to see from here that $n=11,22$ are the only solutions.
