2005 AMC 10A Problems/Problem 15
Contents
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 
, 3 
, and 3 
. This gives us our first 3 cubes: 
, 
, and 
.
However, we can also multiply smaller numbers in the expression to make bigger expressions. For example, 
(one 2 comes from the 
, and the other from the 
). Using this method, we find:
and
So, we have 6 cubes total:
and 
for a total of 
cubes 
See==See also
| 2005 AMC 10A (Problems • Answer Key • Resources) | ||
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Followed by Problem 16 | |
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