### Instruction: How many even perfect square factors does $2^4 \cdot 7^9$ have?### Response: I need to find the number of factors of $2^4 \cdot 7^9$ that are both even and perfect squares.
A factor of $2^4 \cdot 7^9$ must be of the form $2^a \cdot 7^b$, where $0 \leq a \leq 4$ and $0 \leq b \leq 9$.
To be even, a factor must have $a > 0$, since $2^0 = 1$ is odd.
To be a perfect square, a factor must have both $a$ and $b$ even, since an odd power of a prime is not a perfect square.
So I need to count how many ways I can choose $a$ and $b$ to be even and positive.
For $a$, I have two choices: $2$ or $4$.
For $b$, I have five choices: $0, 2, 4, 6, 8$.