2009 UNCO Math Contest II Problems/Problem 4

Revision as of 23:01, 25 November 2016 by Mathisfun04 (talk | contribs) (Solution)


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 $4\cdot 3\cdot 2\cdot 1$)


We first factorize the product as $2^7\cdot3^2\cdot5\cdot7$. 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 four even powers of two that are less than or equal to $2^7$, namely $2^0$, $2^2$, $2^4$, $2^6$, as $0$ is even. Repeating this process with three, five, and seven, we find that three has two even powers, $3^0$, and $3^2$, and that five and seven have only one even power, $5^0$ and $7^0$ respectively. Multiplying this we have $4\cdot2\cdot1\cdot1 = 8$. Therefore, our answer is $\boxed8$.

See also

2009 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions
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