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Difference between revisions of "2005 AMC 10A Problems/Problem 15"

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==Problem==
 
==Problem==
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How many positive cubes divide <math> 3! \cdot 5! \cdot 7! </math> ?
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How many positive cubes divide <math>3! \cdot 5! \cdot 7!</math>?
  
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<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6 </math>
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<math>
 +
\textbf{(A) } 2\qquad \textbf{(B) } 3\qquad \textbf{(C) } 4\qquad \textbf{(D) } 5\qquad \textbf{(E) } 6
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</math>
  
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==Solution==
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== Solution 1 ==
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=== Solution 1 ===
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We obtain <math> 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</math>.
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<math> 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}</math>
 
  
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Therefore, a [[perfect cube]] that divides <math> 3! \cdot 5! \cdot 7! </math> must be in the form <math>2^{a}\cdot3^{b}\cdot5^{c}\cdot7^{d}</math> where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are [[nonnegative]] [[multiple]]s of <math>3</math> that are less than or equal to <math>8</math>, <math>4</math>, <math>2</math> and <math>1</math>, respectively.  
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Therefore, a perfect cube that divides <math>3! \cdot 5! \cdot 7! </math> must be of the form <math>2^{a}\cdot3^{b}\cdot5^{c}\cdot7^{d}</math>, where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are non-negative multiples of <math>3</math> that are less than or equal to <math>8</math>, <math>5</math>, <math>2</math> and <math>1</math> respectively.  
  
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So:
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This means
  
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<math>a\in\{0,3,6\}</math> (<math>3</math> possibilities)
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<cmath>\begin{align*}&a\in\{0,3,6\} \text{ (} 3 \text{ possibilities),} \\
 +
&b\in\{0,3\} \text{ (} 2 \text{ possibilities),} \\
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&c\in\{0\} \text{ (} 1 \text{ possibility), and} \\
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&d\in\{0\} \text{ (} 1 \text{ possibility),}\end{align*}</cmath>
  
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<math>b\in\{0,3\}</math> (<math>2</math> possibilities)
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so the number of perfect cubes that divide <math>3! \cdot 5! \cdot 7!</math> is <math>3\cdot2\cdot1\cdot1 = \boxed{\textbf{(E) } 6}</math>.
  
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<math>c\in\{0\}</math> (<math>1</math> possibility)
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==Solution 2==
 +
As in Solution 1, we write <math>3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1)</math>, and now notice that there are a total of <math>3</math> <math>3</math>s, <math>3</math> <math>2</math>s, and <math>3</math> <math>1</math>s. This gives us our first <math>3</math> cubes: <math>3^3</math>, <math>2^3</math>, and <math>1^3</math>.
  
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<math>d\in\{0\}</math>(<math>1</math> possibility)
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However, we can also multiply smaller numbers in the expression to make bigger cubes. For example, <math>(2 \cdot 2) \cdot 4 \cdot 4 = 4 \cdot 4 \cdot 4 = 4^3</math> (where one 2 comes from the <math>3!</math>, and the other from the <math>5!</math>). Using this method, we can also make
  
 +
<cmath>(3 \cdot 2) \cdot (3 \cdot 2) \cdot 6 = 6^3</cmath>
  
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So the number of perfect cubes that divide <math> 3! \cdot 5! \cdot 7! </math> is <math>3\cdot2\cdot1\cdot1 = 6 \Rightarrow \mathrm{(E)}</math>
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and
  
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===Solution 2===
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<cmath>(3 \cdot 4) \cdot (3 \cdot 4) \cdot (2 \cdot 6) = 12^3.</cmath>
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Answer : <math>\boxed{D}</math>
 
  
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==See Also==
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So we have <math>6</math> possible cubes in total: <math>1^3</math>, <math>2^3</math>, <math>3^3</math>, <math>4^3</math>, <math>6^3</math>, and <math>12^3</math>, and the answer is <math>\boxed{\textbf{(E) } 6}</math>.
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*[[2005 AMC 10A Problems]]
 
  
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*[[2005 AMC 10A Problems/Problem 14|Previous Problem]]
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==See also==
 +
{{AMC10 box|year=2005|ab=A|num-b=14|num-a=16}}
  
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*[[2005 AMC 10A Problems/Problem 16|Next Problem]]
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[[Category:Introductory Number Theory Problems]]
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[[Category:Introductory Combinatorics Problems]]  
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*[[Factorial]]
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[[Category:Introductory Number Theory Problems]]
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*[[Prime factorization]]
 
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[[Category:Introductory Combinatorics Problems]]  
 
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[[Category:Introductory Number Theory Problems]]
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:52, 1 July 2025

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
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All AMC 10 Problems and Solutions

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