How many factors of $2^4 \times 3^5 \times 10^4$ are perfect squares which are greater than $1?$

Let's find how many factors of $2^4 \times 3^5 \times 10^4$ are perfect squares greater than 1.

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First, we need to express everything in terms of prime factors.

Given: $2^4 \times 3^5 \times 10^4$

Since $10 = 2 \times 5$, we have:

$10^4 = (2 \times 5)^4 = 2^4 \times 5^4$

So our expression becomes:

$2^4 \times 3^5 \times 10^4 = 2^4 \times 3^5 \times 2^4 \times 5^4 = 2^8 \times 3^5 \times 5^4$

Working with prime factorization makes it easier to identify perfect squares and count factors systematically.

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A number is a perfect square if and only if all exponents in its prime factorization are even.

For example:

$36 = 2^2 \times 3^2$ is a perfect square (both exponents are even)

$72 = 2^3 \times 3^2$ is NOT a perfect square (exponent of 2 is odd)

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Any factor of $2^8 \times 3^5 \times 5^4$ has the form $2^a \times 3^b \times 5^c$ where:

$0 \leq a \leq 8$

$0 \leq b \leq 5$

$0 \leq c \leq 4$

For this factor to be a perfect square, we need $a$, $b$, and $c$ to all be even.

Possible values:

For $a$: $0, 2, 4, 6, 8$ → 5 choices

For $b$: $0, 2, 4$ → 3 choices (since $b \leq 5$, the largest even number is 4)

For $c$: $0, 2, 4$ → 3 choices

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Using the multiplication principle:

Total perfect square factors = $5 \times 3 \times 3 = 45$

We need factors greater than 1, so we must exclude the case where $a = b = c = 0$ (which gives us $2^0 \times 3^0 \times 5^0 = 1$).

Therefore: Perfect square factors greater than 1 = $45 - 1 = 44$

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We can also write: $2^8 \times 3^5 \times 5^4 = (2^4 \times 3^2 \times 5^2)^2 \times 3$

The perfect square factors come from: $(2^4 \times 3^2 \times 5^2)^2 \times 3^{even}$

Since $3^{even}$ can be $3^0, 3^2, 3^4$, and we can choose any factor of $(2^4 \times 3^2 \times 5^2)^2$, we get:

$(4+1)(2+1)(2+1) = 45$ total perfect square factors

Excluding 1: $45 - 1 = 44$

Answer: 44