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

Revision as of 21:58, 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

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:

$a\in\{0,3,6\}$ ( $3$ 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 : $\boxed{E}$

@@ Line 5: / Line 5: @@
 ==Solution==
+=== Solution 1 ===
 <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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