2003 AMC 12A Problem 23

Problem 23 of 25HarderNumber Theory

How many perfect squares are divisors of the product

1! \cdot 2! \cdot 3! \cdots 9!\,?

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Solution

The product is

1! \cdot 2! \cdots 9! = 2^{30}\cdot3^{13}\cdot5^{5}\cdot7^{3}.

A perfect-square divisor has the form

2^{2a}3^{2b}5^{2c}7^{2d}

with

0\le a\le15,

0\le b\le6,

0\le c\le2,

and

0\le d\le1.

The number of choices is

16\cdot7\cdot3\cdot2=672.

Thus, the correct answer is B.

Tagged: prime factorization · perfect square · factor counting