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

Revision as of 15:01, 1 January 2017

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:

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

Answer : $\boxed{D}$

@@ Line 24: / Line 24: @@
 ===Solution 2===
+Answer : <math>\boxed{D}</math>
-If you factor <math>3! \cdot5 ! \cdot 7!</math> You get
-<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>.
-<math>3 \cdot 2 = 6</math>
-Answer : <math>\boxed{E}</math>
 ==See Also==

Page

Toolbox

Search