Comparison of no. of abelian groups of order $p^r$ with that of order $q^r$ 
[J.B. Fraleigh, Exercises 11, problem 37] Let $p$ and $q$ be distinct prime numbers. How does the number of abelian groups of order $p^r$ compare with the number of abelian groups of order $q^r$ ?

For instance, I take $p = 2$, $q = 3$ and $r=3$. Then, 
for $p = 2, r = 3$, we have $\mathbb Z_8$, $\mathbb Z_4 × \mathbb Z_2$ and $\mathbb Z_2 × \mathbb Z_2 × \mathbb Z_2$. Similarly for $q = 3, r = 3$, we have $\mathbb Z_{27}$, $\mathbb Z_9 × \mathbb Z_3$ and $\mathbb Z_3 × \mathbb Z_3 × \mathbb Z_3$.
It seems that they have same no. of abelian groups, since it depends on $r$. But I'm not sure if I'm correct. Any help or hint would be appreciated.

According to the Fundamental Theorem of Finite Abelian Groups, these must be equal. 
See 
https://proofwiki.org/wiki/Fundamental_Theorem_of_Finite_Abelian_Groups
