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

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If you factor <math>3! \cdot5 ! \cdot 7!</math> You get
If you factor <math>3! \cdot5 ! \cdot 7!</math> You get


<center><math>2^7 \cdot 3^4  \cdot  5^2</math><\center>
<math>2^7 \cdot 3^4  \cdot  5^2</math>


There are 3 ways for the first factor of a cube:  <math>2^0</math>, <math>2^3</math>, and <math>2^6</math>. And the second ways are:  <math>3^0</math>, and <math>3^3</math>.
There are 3 ways for the first factor of a cube:  <math>2^0</math>, <math>2^3</math>, and <math>2^6</math>. And the second ways are:  <math>3^0</math>, and <math>3^3</math>.

Revision as of 21:57, 27 May 2013

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 $3\cdot2\cdot1\cdot1 = 6 \Rightarrow \mathrm{(E)}$

Solution 2

If you factor $3! \cdot5 ! \cdot 7!$ You get

$2^7 \cdot 3^4 \cdot 5^2$

There are 3 ways for the first factor of a cube: $2^0$, $2^3$, and $2^6$. And the second ways are: $3^0$, and $3^3$.

$3 \cdot 2 = 6$ Answer : \boxed{E}

See Also

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