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2005 AMC 10A Problems/Problem 15

Problem

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

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

Solution 1

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

Therefore, a perfect cube that divides $3! \cdot 5! \cdot 7!$ must be of the form $2^{a}\cdot3^{b}\cdot5^{c}\cdot7^{d}$, where $a$, $b$, $c$, and $d$ are non-negative multiples of $3$ that are less than or equal to $8$, $5$, $2$ and $1$ respectively.

This means

\begin{align*}&a\in\{0,3,6\} \text{ (} 3 \text{ possibilities),} \\ &b\in\{0,3\} \text{ (} 2 \text{ possibilities),} \\ &c\in\{0\} \text{ (} 1 \text{ possibility), and} \\ &d\in\{0\} \text{ (} 1 \text{ possibility),}\end{align*}

so the number of perfect cubes that divide $3! \cdot 5! \cdot 7!$ is $3\cdot2\cdot1\cdot1 = \boxed{\textbf{(E) } 6}$.

Solution 2

As in Solution 1, we write $3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1)$, and now notice that there are a total of $3$ $3$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 cubes. For example, $(2 \cdot 2) \cdot 4 \cdot 4 = 4 \cdot 4 \cdot 4 = 4^3$ (where one 2 comes from the $3!$, and the other from the $5!$). Using this method, we can also make

\[(3 \cdot 2) \cdot (3 \cdot 2) \cdot 6 = 6^3\]

and

\[(3 \cdot 4) \cdot (3 \cdot 4) \cdot (2 \cdot 6) = 12^3.\]

So we have $6$ possible cubes in total: $1^3$, $2^3$, $3^3$, $4^3$, $6^3$, and $12^3$, and the answer is $\boxed{\textbf{(E) } 6}$.

See also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 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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