Difference between revisions of "2005 AMC 10A Problems/Problem 15"
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<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> | <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> | ||
Revision as of 21:58, 27 May 2013
Contents
[hide]Problem
How many positive cubes divide ?
Solution
Solution 1
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 :