AMC10 2005 10A, Problem 15

How many positive cubes divide $3! \cdot 5! \cdot 7!$ ?

Solution: $3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1) = 2^{8}\cdot3^{4}\cdot5^{2}\cdot7^{1}$ Therefore, a perfect cube that divides $3! \cdot 5! \cdot 7!$ must be in the form $2^{a}\cdot3^{b}\cdot5^{c}\cdot7^{d}$ where $a$, $b$, $c$, and $d$ are nonnegative multiples of $3$ that are less than or equal to $8, 5, 2$ and $1,$ respectively. So: $a\in\{0,3,6\}$ ($3$ possibilities) $b\in\{0,3\}$ ($2$ possibilities) $c\in\{0\}$ ($1$ possibility) $d\in\{0\}$($1$ possibility) So the number of perfect cubes that divide $3! \cdot 5! \cdot 7!$ is $3\cdot2\cdot1\cdot1 = \fbox{6}$

Answer: 6