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

Revision as of 17:28, 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!*5!*7! You get

2^7 * 3^4 * 5^2

A cube need three same factors for it to be a cube.

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 * 2 = 6 Answer : E

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

Revision as of 17:28, 27 May 2013

Contents

Problem

Solution

Solution 2

See Also

@@ Line 24: / Line 24: @@
 ==Solution 2==
 If you factor 3!*5!*7! You get
+^7 * 3^4  *  5^2
+A cube need three same factors for it to be a cube.
+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.
+* 2 = 6
+Answer : E
 ==See Also==