Find n if $ϕ(n) = 2^{6} \times 17$ and $τ(n) = 12$. Find n if $ϕ(n) = 2^{6} \times 17$ and $τ(n) = 12$.
 A: Note that for any prime $p$, $\phi(p^k)=p^k-p^{k-1}$ and $\tau(p^k)=k+1$, so $n$ is not a power of a single prime.
Now, writing $n=p_1^{k_1}\cdots p_m^{k_m}$, where the $p_j$ are distinct primes indexed in increasing order, and where the $k_j$ are positive integers, we have $$\tau(n)=(k_1+1)\cdots(k_m+1).$$ Hence, there are only a few possibilities for the form of the decomposition of $n$ into prime powers:

(1) $n=p_1^1p_2^5,$
(2) $n=p_1^5p_2^1,$
(3) $n=p_1^2p_2^3,$
(4) $n=p_1^3p_2^2,$
(5) $n=p_1^1p_2^1p_3^2,$
(6) $n=p_1^1p_2^2p_3^1,$
(7) $n=p_1^2p_2^1p_3^1.$

(Why are there no other possibilities?)
In cases (1) through (4) (with $n=p_1^{k_1}p_2^{k_2}$), we would have $$\phi(n)=(p_1^{k_1}-p_1^{k_1-1})(p_2^{k_2}-p_2^{k_2-1}),$$ and in cases (5) through (7) (with $n=p_1^{k_1}p_2^{k_2}p_3^{k_3}$), we would have $$\phi(n)=(p_1^{k_1}-p_1^{k_1-1})(p_2^{k_2}-p_2^{k_2-1})(p_3^{k_3}-p_3^{k_3-1}).$$ Note that each $p_j-1$ will necessarily divide $\phi(n)$, regardless of which of the cases holds, and when $k_j>1$ we have that $p_j^{k_j-1}$ is a (nontrivial) factor of $\phi(n)$, as well. Since the primes are indexed in increasing order, this alone should let you rule out several cases quickly. Can you take it from here?
