How many perfect squares are divisors of the product 1! ∙ 2! ∙ 3! ∙ 4! ∙ 5! ∙ 6! ∙ 7! ?

]]>Whenever tackling a divisors problem, oftentimes it's a great strategy to first consider the prime factorization.

The prime factorization of the product is \(2^{16}\cdot3^{7}\cdot5^{3}\cdot7\). Since we're dealing with perfect squares, let's think about what the prime factorization of a perfect square looks like-- in a perfect square, all of the exponents are even! For example, \(2^4\cdot3^2\) is a perfect square, so is \(2^2 \cdot 5^2\), so is \(3^6\), but not \(2^3\cdot3^2\) or \(3^4\cdot 5\).

So, using \(2^{16}\cdot3^{7}\cdot5^{3}\cdot7\), we have to find all of the possibilities to make a perfect square using each of these factors. Let's tackle each of the factors individually!

Every perfect square that is a divisor of that product is either going to have \(2^0,2^2,2^4,2^6,2^8,2^{10},2^{12},2^{14},\) or \(2^{16}\) in its prime factorization. This gives us \(9\) possibilities for the power of \(2\).

Then, using the same logic, every perfect square that is a divisor of that product has a power of \(3\) that's either \(3^0,3^2,3^4\) or \(3^{6}\) in its prime factorization. This gives us \(4\) possibilities for the power of \(2\).

Finally, there are only \(2\) possibilities for a perfect square power of \(5\): \(5^0\) and \(5^2\). We can stop here since we have no other options for any factors greater than 5.

So, multiplying our possibilities together, we have \(9\cdot4\cdot2=\boxed{72}\) in total. Let me know if this made sense and if you still have any questions!

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