2005 AMC 10A Problems/Problem 15
Contents
Problem
How many positive cubes divide ?
Solution 1
We obtain .
Therefore, a perfect cube that divides must be of the form
, where
,
,
, and
are non-negative multiples of
that are less than or equal to
,
,
and
respectively.
This means
so the number of perfect cubes that divide is
.
Solution 2
As in Solution 1, we write , and now notice that there are a total of
s,
s, and
s. This gives us our first
cubes:
,
, and
.
However, we can also multiply smaller numbers in the expression to make bigger cubes. For example, (where one 2 comes from the
, and the other from the
). Using this method, we can also make
and
So we have possible cubes in total:
,
,
,
,
, and
, and the answer is
.
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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