2005 AMC 10A Problems/Problem 15: Difference between revisions
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<math>3 \cdot 2 = 6</math> | <math>3 \cdot 2 = 6</math> | ||
Answer : \boxed{E} | Answer : <math>\boxed{E}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 21:58, 27 May 2013
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 : 
