The number of factors of $3^5 \times 5^8 \times 7^2$ that are perfect squares is

A perfect square factor will have the form $3^a \times 5^b \times 7^c$ where each exponent must be even.

This is because a perfect square is a number that can be expressed as $n^2$ for some integer $n$.

For our number $3^5 \times 5^8 \times 7^2$, we need to find all possible even exponents that don't exceed the original exponents:

• For $3^5$: The exponent $a$ can be 0, 2, 4 (3 possibilities)
  • For $5^8$: The exponent $b$ can be 0, 2, 4, 6, 8 (5 possibilities)
  • For $7^2$: The exponent $c$ can be 0, 2 (2 possibilities)

Using the multiplication principle, the total number of perfect square factors is:

$3 \times 5 \times 2 = 30$

Therefore, the number of factors of $3^5 \times 5^8 \times 7^2$ that are perfect squares is 30.