Correct Answer: C) $36$ To determine the number of factors of $N$ that are perfect squares, we first consider the general form of any factor of $N$. Given the prime factorization $N = 2^{5} \times 3^{4} \times 5^{3} \times 7^{2}$, any factor $F$ of $N$ must be of the form: $$F = 2^a \times 3^b \times 5^c \times 7^d$$ where $a$, $b$, $c$, and $d$ are non-negative integers satisfying the following conditions based on the exponents in $N$: $$0 \le a \le 5$$ $$0 \le b \le 4$$ $$0 \le c \le 3$$ $$0 \le d \le 2$$ For $F$ to be a perfect square, all the exponents in its prime factorization ($a, b, c, d$) must be even integers. Let's find the possible even values for each exponent within its allowed range: 1. **For the exponent $a$ (corresponding to the prime factor $2$):** $a$ must be an even integer and $0 \le a \le 5$. The possible values for $a$ are $\{0, 2, 4\}$. The number of choices for $a$ is $3$. 2. **For the exponent $b$ (corresponding to the prime factor $3$):** $b$ must be an even integer and $0 \le b \le 4$. The possible values for $b$ are $\{0, 2, 4\}$. The number of choices for $b$ is $3$. 3. **For the exponent $c$ (corresponding to the prime factor $5$):** $c$ must be an even integer and $0 \le c \le 3$. The possible values for $c$ are $\{0, 2\}$. The number of choices for $c$ is $2$. 4. **For the exponent $d$ (corresponding to the prime factor $7$):** $d$ must be an even integer and $0 \le d \le 2$. The possible values for $d$ are $\{0, 2\}$. The number of choices for $d$ is $2$. To find the total number of factors of $N$ that are perfect squares, we multiply the number of choices for each exponent (since the choices for each exponent are independent): $$\text{Total number of perfect square factors} = (\text{Choices for } a) \times (\text{Choices for } b) \times (\text{Choices for } c) \times (\text{Choices for } d)$$ $$\text{Total number of perfect square factors} = 3 \times 3 \times 2 \times 2 = 36$$ Thus, there are $36$ factors of $N$ that are perfect squares.