Find the smallest positive odd integer n such that φ(n)/n = 7680/12121 In a previous problem, I was able to deduce that if you have φ(n)/n = a/b where gcd(a,b) = 1 then the largest prime factor of b must also be the largest prime factor of n. 
I found the prime factorization of both integers:
7680 = 2^9*3*5
12121 = 17*23*31
I believe that in the prime factorization of n we must have the following numbers: 31, 4, 2 since 
φ(n)/n = {(p1-1)(p2-1)...(pr-1)}/p1*p2*...*pr
I'm not sure where to go from this. Thanks for your help!

Let $F$ be a finite subset of prime numbers.   Write $n$ as $\prod_{p \in F} p^{\alpha(p)}$.  Then $\phi(n)/n = \prod_{p \in F} (1-1/p)$.  From OP's factorisation, $17,23,31\in F$.  Observe that the factor $11$ should not be present in $\frac{16×22×30}{17×23×31}$, so we add $11$ to $F$.  Repeat the same step to see that $n=5×11×17×23×31$.
