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)</math> | <math> 3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1)</math> | ||
− | + | In the expression, we notice that there are 3 <math>3's</math> and 2 <math>2's</math>. | |
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==See Also== | ==See Also== |
Revision as of 01:07, 15 January 2018
Contents
[hide]Problem
How many positive cubes divide ?
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:
(
possibilities)
(
possibilities)
(
possibility)
(
possibility)
So the number of perfect cubes that divide is
Solution 2
In the expression, we notice that there are 3 and 2
.
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.