Abelian groups of order n. 
Is there a number $n$ such that there are exactly 1 million abelian groups of order $n$?

Can anyone please explain. I would yes because numbers are infinitive, and so any number n can be expressed as a direct product of cyclic groups of order n.
Can anyone please help  me understand. 
Thank you.

Given any number $n$, the number of different (up to isomorphism) abelian groups of order $n$ is dependant on the prime factorization of $n$. If $n$ is a prime power $p^k, k\in \Bbb N$, then the number of abelian groups of order $n$ is exactly the number of distinct partitions of $k$. Example: For $p = 3, k = 4$ and thus $n = 81$, there are $5$ different abelian groups:
$$
\Bbb Z_{81} \quad \Bbb Z_{27}\times \Bbb Z_3 \quad \Bbb Z_{9} \times \Bbb Z_9 \quad \Bbb Z_9 \times \Bbb Z_3 \times \Bbb Z_3 \quad (\Bbb Z_3)^4
$$
Now, if there are multiple primes in the factorization of $n$, the number of partitions allowed by each one are just multiplied together. Example: If $n = 2^53^7 = 69,984$, then there are $7\cdot 15 = 105$ different abelian groups, since $5$ has $7$ partitions and $7$ has $15$.
Now, to construct a million, we need $2^65^6$ different partitions total. The easiest way (maybe the only way) is to let $n$ have be $6$ primes to the power $2$ ($2$ partitions) and $6$ primes to the power $4$ (five partitions). Thus, I believe the smallest such $n$ is given by
$$
n = 2^43^45^47^411^413^417^219^223^229^231^237^2 \\= 49,659,789,817,537,838,957,341,175,342,490,000
$$
