How many even perfect square factors does $2^4 \cdot 7^9$ have?
A positive integer is a factor of $2^4\cdot7^9$ if and only if its prime factorization is of the form $2^a\cdot 7^b$ for exponents $a$ and $b$ satisfying $0\leq a \leq 4$ and $0\leq b\leq 9$.  A positive integer is a perfect square if and only if the exponents in its prime factorization are even.  A positive integer is even if and only if the exponent of 2 in its prime factorization is at least 1.  Therefore, we may choose $a=2$ or $4$ and $b=0,$ $2,$ $4,$ $6,$ or $8.$  Since we have 2 choices for $a$ and 5 choices for $b$, there are $2\times5=\boxed{10}$ ways to make these two decisions.
10
Level 5
Number Theory
Problem: How many even perfect square factors does $2^4 \cdot 7^9$ have?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the even perfect square factors of the number \(2^4 \cdot 7^9\), we first identify the general form of a factor. A factor of \(2^4 \cdot 7^9\) can be expressed as \(2^a \cdot 7^b\), where \(0 \leq a \leq 4\) and \(0 \leq b \leq 9\).

Next, we focus on the conditions for the factor to be a perfect square. A number is a perfect square if all the exponents in its prime factorization are even. Therefore, \(a\) must be even, and \(b\) must also be even.

The possible even values for \(a\) (from \(0\) to \(4\)) are \(0, 2, 4\). This gives us \(3\) choices for \(a\).

The possible even values for \(b\) (from \(0\) to \(9\)) are \(0, 2, 4, 6, 8\). This gives us \(5\) choices for \(b\).

Now, we need to consider the requirement for the factor to be even. For the factor \(2^a \cdot 7^b\) to be even, \(a\) must be at least \(1\). Therefore, the valid even values for \(a\) that satisfy this condition are \(2\) and \(4\). This gives us \(2\) choices for \(a\).

Since \(b\) can still take any of the \(5\) even values, we can now calculate the total number of even perfect square factors. The total number of combinations is given by multiplying the number of choices for \(a\) and \(b\):

Total even perfect square factors = Number of choices for \(a\) × Number of choices for \(b\) = \(2 \times 5 = 10\).

Thus, the final answer is \(10\).<|END_OF_LATENT|> The final answer is $10$.