Sum of powers of the sum of all the square numbers in a factorial

Question

Here's the exact question:

The sum of all positive integers $p$ such that $13! \div p$ is a perfect square, can be written as $2^a . 3^b . 5^c . 7^d . 11^e . 13^f$ where $a, b, c, d, e$ and $f$ are positive integers. Find $(a + b + c + d + e + f)$

Can someone please help me how to start with this question?

Here is my approach:
I tried to factorise $13!$ as $2^{10} . 3^5 . 5^2 . 7^1 . 11^1 . 13^1$
Then I knew that $p$ must have factors of $2^{even} . 3^{odd} . 5^{even} . 7^1 . 11^1 . 13^1$
But now I cannot proceed further on thinking the sum of all such possible $p$'s.....

Any hint/suggestion will work! Also, anyone can also suggest formatting issues/edit the question without changing its actual meaning, thank you!!!

PS: I donot think this is a duplicate, I did not find similar questions to this....

Answer 1

OP has stated, with a slight correction, that

$p$ must be of the form $2^{even} . 3^{odd} . 5^{even} . 7^1 . 11^1 . 13^1$

Let's clarify that further (and it's implied in the comments that OP knows this too):

If $p = 2^a \times 3^b \times 5^c \times 7^d \times 11^e \times 13^f$, then
$a \in \{ 0, 2, 4, 6, 8, 10 \}, b \in \{ 1, 3, 5 \}, c \in \{0, 2 \}, d = 1 , e = 1 , f = 1 $.
Furthermore, this represents the complete classification of all such $p$.

How can we find the sum of all of these?
As lulu mentioned in the comments, we could just sum up the 36 values.
OP is asking in the comments if there is a more efficient method?

Hint: Find an expression such that in the expanded form, each of the terms correspond to a possible value of $p$.
Then, evaluate that expression.

$(2^0 + 2^2 + 2^4 + 2^6 + 2^8 + 2^{10} ) \times ( 3^1 + 3^3 + 3^5) \times ( 5^0 + 5^2 ) \times 7 \times 11 \times 13$.

Note: This approach is very similar to finding the sum of divisors of an integer. Lulu mentioned that "your divisor must be divisible by 3 and what's left over is the square of a divisor of $ 2^5 \times 3^2 \times 5$", which explains the similarities.