Constant Product?

The answer is simply the $5$ -fold Dirichlet convolution $N = (1 \star 1 \star 1 \star 1 \star 1)(1458000)$ . Since $1$ is multiplicative, so is its $5$ -fold convolution. Moreover $(1 \star 1 \star 1 \star 1 \star 1)(p^n) \; = \; \tfrac{1}{24}(n+1)(n+2)(n+3)(n+4)$ for any positive integer $n$ and prime $p$ . Thus $N \; = \; (1 \star 1 \star 1 \star 1 \star 1)(2^4) \times (1 \star 1 \star 1 \star 1 \star 1)(3^6) \times (1 \star 1 \star 1 \star 1 \star 1)(5^3) \; =\; 70 \times 210 \times 35 \; = \; 514560$ and hence $N = 2^2 \times 3 \times 5^3 \times 7^3$ , making the answer $2+1+3+3=\boxed{9}$ .

Given number 1458000 = 2^4 * 3^6 * 5^3.

So, now the problem becomes finding the number of ways of partitioning a group of four 2s, six 3s and three 5s into five partitions/groups. Thereby, we get the product of numbers in each partition/group to be one value of the ordered quintuplet.

Thus, the total number of ways of making that partition is equal to the number of ordered quintuplets.

Number of partitions

= number of ways of ordering four 2s, six 3s and three 5s into five partitions/groups.

One intuitive approach for this is: Imagine four 2s, six 3s and three 5s as string of literals. Now add four '|' symbols to the original string. The | symbol acts as a separator of groups.

Now the total number of combinations turn into an elementary combinatorics problem like: "In how many ways can you re-arrange

17 characters(4 |s + 4 2s + 6 3s + 3 5s) of which 4 are |s, 4 are 2s and so on

The result is 17!/(4! 4! 6!*3!) = 142942800

This number can be factorized as 142942800 = 2^4 * 3^1 * 5^2 * 7^2 * 2431

Hence, the required form has been achieved and the answer becomes sum of 4+1+2+2 = 9