Find the number of factors of the product $5^8$ $7^5$ $2^3$ which are perfect squares.

Step 1 (Formulate Divisors): We are given the product $N=2^3\times 5^8\times 7^5$ (writing in increasing order of prime bases). Any factor of $N$ is of the form:

where $0\le a\le 3$, $0\le b\le 8$, and $0\le c\le 5$.

Step 2 (Perfect Square Condition): For $F$ to be a perfect square, all exponents in its prime factorization ($a,b,c$) must be even integers:

1. Choices for $a$ (even in $[0,3]$): $\{0,2\}$ (2 choices)

2. Choices for $b$ (even in $[0,8]$): $\{0,2,4,6,8\}$ (5 choices)

3. Choices for $c$ (even in $[0,5]$): $\{0,2,4\}$ (3 choices)

Step 3 (Apply Product Rule): The choices of exponents are independent, so the total number of perfect square factors is:

Step 4 (Conclusion): There are exactly 30 perfect square factors.