# 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 $8$, $5$, $2$ 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 $3\cdot2\cdot1\cdot1 = 6 \Rightarrow \mathrm{(E)}$

## 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 also 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 find: $(3*2)*(3*2)*6=6^3$

and $(3*4)*(3*4)*(2*6)=12^3$

So, we have 6 cubes total: $1^3 ,2^3, 3^3, 4^3, 6^3,$ and $12^3$ for a total of $6$ cubes $\Rightarrow \mathrm{(E)}$

## See also

 2005 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 14 Followed byProblem 16 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

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