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Difference between revisions of "2005 AMC 10A Problems/Problem 15"
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and
 
and
  
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<math>(3*3)*(3*3)*(3*3)=9^3</math>
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<math>(3*4)*(3*4)*(2*6)=9^3</math>
  
 
So, we have 6 cubes total: <math>1^3 ,2^3, 3^3, 4^3, 6^3,</math> and <math>9^3</math> for a total of <math>6</math> cubes <math>\Rightarrow \mathrm{(E)}</math>
 
So, we have 6 cubes total: <math>1^3 ,2^3, 3^3, 4^3, 6^3,</math> and <math>9^3</math> for a total of <math>6</math> cubes <math>\Rightarrow \mathrm{(E)}</math>
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==See Also==
 
==See Also==

Revision as of 01:21, 15 January 2018

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:

$(3*2)*(3*2)*6=6^3$

and

$(3*4)*(3*4)*(2*6)=9^3$

So, we have 6 cubes total: $1^3 ,2^3, 3^3, 4^3, 6^3,$ and $9^3$ for a total of $6$ cubes $\Rightarrow \mathrm{(E)}$

See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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