All numbers less than 100 with phi(n) = 64 So the question is to find all natural numbers less than 100 where $\phi(n) = 64$ (Euler totient function).
I read somewhere here on Mathematics SE, and tried something like below.
$64 = 2^6$
From
$$\prod p^{\alpha(p)-1}(p-1) = 64$$
I got {1, 2, 4, 16} as possible values for $(p-1)$, and from that: $p_i \in \{2, 3, 5, 17\}$.
From which I figured $n = 2^a\cdot3^b\cdot5^c\cdot17^d$
But from here onwards, I'm stuck. How do I go about figuring the a, b, c, and d values? And help in the right direction would be great! (please note that I'm kind of a beginner)
 A: As discussed in the comments:  
Simple divisibility considerations tell us that $$n=2^a3^b5^c17^d\quad 0≤a≤6\quad  0≤b,c,d≤1$$
We note that $$\varphi(3)=2\quad \varphi(5)=2^2\quad \varphi(17)=2^4$$
It follows that $\varphi(3^b5^c17^d)=2^{b+2c+4d}$
We work case by case, from values of $a$.
$a=0$ We want $b+2c+4d=6$  Easy to see the only case here is $(a,b,c,d)=(0,0,1,1)$  Thus we get $n=5\times 17=85$  Thus $\boxed {n=85}$ works
$a=1$ We want $b+2c+4d=6$  As $2\times 85>0$ we get no new cases.
$a=2$ We want $b+2c+4d=5$  Easy to see the only case here is $(a,b,c,d)=(2,1,0,1)$  Thus we get $n=2^2\times 3\times  17>100$
$a=3$  We want $b+2c+4d=4$  Easy to see the only case here is $(a,b,c,d)=(3,0,0,1)$  Thus we get $n=2^3\times  17>100$
$a=4$  We want $b+2c+4d=3$  Easy to see the only case here is $(a,b,c,d)=(4,1,1,0)$  Thus we get $n=2^4\times 3\times 5>100$
$a=5$  We want $b+2c+4d=2$  Easy to see the only case here is $(a,b,c,d)=(5,0,1,0)$  Thus we get $n=2^5\times  5>100$
$a=6$  We want $b+2c+4d=1$  Easy to see the only case here is $(a,b,c,d)=(6,1,0,0)$  Thus we get $n=2^6\times  3>100$
So, after all that work, the only solution less than $100$ is $85$.
A: You have to solve in natural numbers: $\; a+b+2c+4d=7$, with the constraints $b,c,d\le 1$ and of course, $a\le 7$.
Case 1: if $d=1$, you have to solve $a+b+2c=3$, whence
$$\begin{cases}c=1,\;b=1,\;a=0&(17\cdot5\cdot3)\\
c=1,\;b=0,\;a=1&(17\cdot5\cdot2)\\
c=0,\;b=1,\;a=2\quad&(17\cdot3\cdot 2^2)\\
c=0,\;b=0,\;a=3&(17\cdot 2^3)
\end{cases}$$
Case 2: if $d=0,\;c=1$, you have to solve $a+b=5$, whence
$$\begin{cases}
b=1,\;a=4\quad&(5\cdot3\cdot 2^4)\\
b=0,\;a=5&(5\cdot 2^5)
\end{cases}$$
Case 3: if $d=0,\;c=0$, you have to solve $a+b=7$, whence
$$\begin{cases}
b=1,\;a=6\quad&(3\cdot 2^6)\\
b=0,\;a=7&( 2^7)
\end{cases}$$
So the solutions are
$$n\in\{85,128,136,160,170,192,204,240\}.$$
