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

If $$N=p^a\times q^b\times r^c$$, then the number of factors is given by $$(a+1)(b+1)(c+1)$$

Now, we are given $$N=3^5 \times 5^8 \times 7^2$$, and we need to find the factors which are perfect squares.

A number is a perfect square if all exponents in its prime factorization are even. So, we need to find the even exponent possible values for each prime number in N

Exponent of 3 in N is 5, thus, the even values which are possible = 0, 2, 4 {3 possibilities}

Exponent of 5 in N is 8, thus, the even values which are possible = 0, 2, 4, 6, 8 {5 possibilities}

Exponent of 7 in N is 2, thus, the even values which are possible = 0, 2 {2 possibilities}

Thus, the total number of even factors for N will be $$=3\times5\times2=30$$