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2005 AMC 10A Problems/Problem 15: Difference between revisions

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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>
<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>


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 non-negative multiples 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.  
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.  


So:  
So:  
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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 E</math>
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>


==See Also==
==See Also==
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*[[2005 AMC 10A Problems/Problem 16|Next Problem]]
*[[2005 AMC 10A Problems/Problem 16|Next Problem]]
*[[Factorial]]
*[[Prime factorization]]
[[Category:Introductory Combinatorics Problems]]
[[Category:Introductory Number Theory Problems]]

Revision as of 10:04, 2 August 2006

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

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

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:

$a\in\{0,3,6\}$ ($3$ posibilities)

$b\in\{0,3\}$ ($2$ posibilities)

$c\in\{0\}$ ($1$ posibility)

$d\in\{0\}$($1$ posibility)


So the number of perfect cubes that divide $3! \cdot 5! \cdot 7!$ is $3\cdot2\cdot1\cdot1 = 6 \Rightarrow \mathrm{(E)}$

See Also

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