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?

The second is not a lucky guess. If $n$ is odd, then $\varphi(2n)=\varphi(2)\varphi(n)=\varphi(n)$.
There is clearly only one prime $n$ such that $\varphi(n)=10$. And it has the automatic companion $22$. To search for other non-primes, recall that $\varphi$ is multiplicative, meaning that if $a$ and $b$ are relatively prime, we have $\varphi(ab)=\varphi(a)\varphi(b)$.
Let us look at the possibilities $\varphi(a)=2$, $\varphi(b)=5$. The equation $\varphi(b)=5$ has no solutions, since by symmetry $\varphi(n)$ is even if $n\gt 2$. 
Remark: Consider $n=2k$ where $k$ is odd. If $k=1$, the equation $\varphi(n)=n$ has the solutions $3$, $6$, and $4$.
If $k\gt 1$, and $2k+1$ is not prime, then we cannot find an $x$ such that $\varphi(a)=2k$. If $2k+1$ is prime, the equation $\varphi(x)=2k$ has precisely two solutions, $2k+1$ and $4k+2$. The argument is basically the same as in the case $k=5$ above. 
