Difference between revisions of "2005 AMC 10A Problems/Problem 15"
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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 | ||
− | + | <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
Contents
[hide]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}