2005 AMC 10A Problems/Problem 15

Problem

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

$\mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6$

Solution 1

$3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1) = 2^{8}\cdot3^{4}\cdot5^{2}\cdot7^{1}$

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 $8$ , $4$ , $2$ 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 $3\cdot2\cdot1\cdot1 = 6 \Rightarrow \mathrm{(E)}$

Solution 2

$3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1)$

In the expression, we notice that there are 3 $3's$ , 3 $2's$ , and 3 $1's$ . This gives us our first 3 cubes: $3^3$ , $2^3$ , and $1^3$ .

However, we can multiply smaller numbers in the expression to make bigger expressions. For example, $(2*2)*4*4=4*4*4=4^3$ (one 2 comes from the $3!$ , and the other from the $5!$ ). Using this method, we also find: