Difference between revisions of "2005 AMC 10A Problems/Problem 15"
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So the number of perfect cubes that divide <math> 3! \cdot 5! \cdot 7! </math> is <math>3\cdot2\cdot1\cdot1 = 6 \Rightarrow \mathrm{(E)}</math> | So the number of perfect cubes that divide <math> 3! \cdot 5! \cdot 7! </math> is <math>3\cdot2\cdot1\cdot1 = 6 \Rightarrow \mathrm{(E)}</math> | ||
− | ==Solution 2== | + | ===Solution 2=== |
− | If you factor 3! | + | If you factor <math>3! \cdot5 ! \cdot 7!</math> You get |
− | 2^7 | + | <center><math>2^7 \cdot 3^4 \cdot 5^2</math><\center> |
− | 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. | + | 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>. |
− | 3 | + | <math>3 \cdot 2 = 6</math> |
− | Answer : E | + | Answer : \boxed{E} |
==See Also== | ==See Also== |
Revision as of 22:57, 27 May 2013
Contents
Problem
How many positive cubes divide ?
Solution
Therefore, a perfect cube that divides must be in the form where , , , and are nonnegative multiples of that are less than or equal to , , and , respectively.
So:
( posibilities)
( posibilities)
( posibility)
( posibility)
So the number of perfect cubes that divide is
Solution 2
If you factor You get
There are 3 ways for the first factor of a cube: , , and . And the second ways are: , and .
Answer : \boxed{E}