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 also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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