2009 UNCO Math Contest II Problems/Problem 4

Problem

How many perfect squares are divisors of the product $1!\cdot 2!\cdot 3!\cdot 4!\cdot 5!\cdot 6!\cdot 7!\cdot 8!$ ? (Here, for example, $4!$ means $4\cdot 3\cdot 2\cdot 1$ )

Solution

We first factorize the product as $2^{7}\cdot3^{6}\cdot4^{5}\cdot5^{4}\cdot6^{3}\cdot7^{2}\cdot8 = 2^{23}\cdot3^{9}\cdot5^{4}\cdot7^{2}$ . Since we want only perfect squares, we are looking for even powers in the prime factorization of the divisors. Working with each term in the prime factorization, we find that there are twelve even powers of two that are less than or equal to $2^{23}$ , namely $2^0$ , $2^2$ , $2^4$ , $2^6$ , $2^8$ , $2^{10}$ , $2^{12}$ , $2^{14}$ , $2^{16}$ , $2^{18}$ , $2^{20}$ , $2^{22}$ , as $0$ is even. Repeating this process with three, five, and seven, we find that three has five even powers, $3^0$ , $3^2$ , $3^4$ , $3^6$ and $3^8$ , and that five has 3 and and seven has two even powers. Multiplying this we have $12\cdot{5}\cdot{3}\cdot{2} = 360$ . Therefore, our answer is $\boxed{360}$ .

2009 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2009 UNCO Math Contest II Problems/Problem 4

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