Difference between revisions of "2005 AMC 10A Problems/Problem 15"
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− | <math>c\in\{0\}</math> (<math>1</math> | + | <math>c\in\{0\}</math> (<math>1</math> possibility) |
− | <math>d\in\{0\}</math>(<math>1</math> | + | <math>d\in\{0\}</math>(<math>1</math> possibility) |
Revision as of 20:38, 10 May 2015
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:
(
possibilities)
(
possibilities)
(
possibility)
(
possibility)
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 :
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.