Abelian groups of order n

Question

I'm having some trouble understanding answers to a question I saw posted here: HERE

The explanation posted is that to find the number of abelian groups of order $n$ I must:

1. Calculate the prime factorisation of $n$.
2. Multiply the partition number of each power together.

eg: If $n = 69,984 = 2^5\cdot 3^7$, then there are $7⋅15=105$ different abelian groups, since $5$ has $7$ partitions and $7$ has $15$.

However, the explanation goes on to calculate $n = 1,000,000$. I would have thought it would go something like this:

$$1,000,000 = 2^6\cdot 5^6$$

partitions$(6)=11$

Therefore, there are $11 \cdot 11 = 121$ abelian groups.

I'm obviously quite far off as the solution is apparently: $n=(2^4)\cdot(3^4)\cdot(5^4)\cdot(7^4)\cdot(11^4)\cdot(13^4)\cdot(17^2)\cdot(19^2)\cdot(23^2)\cdot(29^2)\cdot(31^2)\cdot(37^2) = 4,965,978,981,753,783,895,734,117,534,249,000$

Can someone please try and explain what I'm not understanding/missing?

Thanks

Answer 1

The problem considered in the linked question is not asking for the number of abelian groups of order $10^6$. It's asking you to find $n$ such that the number of abelian groups of order $n$ is exactly $10^6$. To rephrase, the question asks if there is a positive integer $n=p_1^{k_1}\cdots p_m^{k_m}$, such that $$P(k_1)\cdots P(k_m)=10^6,$$ where $P$ is the number of partitions.

Because $10^6=2^65^6=P(2)^6P(4)^6$, the required $n$ must have $6+6$ prime factors, 6 of them to power 2, and the remaining 6 to power 4.

I'm obviously quite far off as the solution is apparently: $n=4,965,978,981,753,783,895,734,117,534,249,000$

Note that this is not the solution, it is a solution. However, it is the minimal solution.