Question

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

Hint:

To find the number of factors that are perfect cubes, you would count the number of exponents divisible by 3. For factors that are \(k^{th}\) powers, count exponents divisible by k. The principle remains the same.

Answer 1

Correct Answer: 30

Step 1: Understanding the Concept:
A number is a perfect square if all the exponents in its prime factorization are even. We need to find how many factors of the given number satisfy this condition.
Step 2: Key Formula or Approach:
Let the given number be \( N = p_1^{e_1} \times p_2^{e_2} \times \dots \times p_k^{e_k} \). A factor of N will be of the form \( f = p_1^{a_1} \times p_2^{a_2} \times \dots \times p_k^{a_k} \), where \( 0 \le a_i \le e_i \). For 'f' to be a perfect square, each exponent \(a_i\) must be an even number. The total number of such factors is the product of the number of possible even choices for each exponent.
Step 3: Detailed Explanation:
The given number is \( N = 3^5 \times 5^8 \times 7^2 \). Let a factor be \( f = 3^x \times 5^y \times 7^z \). For f to be a factor of N, the exponents must satisfy: \( 0 \le x \le 5 \) \( 0 \le y \le 8 \) \( 0 \le z \le 2 \)
For f to be a perfect square, the exponents x, y, and z must be even. Let's find the number of choices for each exponent:

Possible even values for x in the range [0, 5]: \{0, 2, 4\}. There are 3 choices.
Possible even values for y in the range [0, 8]: \{0, 2, 4, 6, 8\}. There are 5 choices.
Possible even values for z in the range [0, 2]: \{0, 2\}. There are 2 choices.
The total number of factors that are perfect squares is the product of the number of choices for each exponent. \[ \text{Number of perfect square factors} = (\text{choices for x}) \times (\text{choices for y}) \times (\text{choices for z}) \] \[ = 3 \times 5 \times 2 = 30 \] Step 4: Final Answer:
The number of factors that are perfect squares is 30.