Difference between revisions of "2005 AMC 10A Problems/Problem 15"
Mr doudou14 (talk | contribs) (→Solution 2) |
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<math>(3*3)*(3*3)*(3*3)=9^3</math> | <math>(3*3)*(3*3)*(3*3)=9^3</math> | ||
− | So, we have 6 cubes total: <math>1^3 ,2^3, 3^3, 4^3, 6^3, and 9^3</math> for a total of <math>6</math> cubes | + | So, we have 6 cubes total: <math>1^3 ,2^3, 3^3, 4^3, 6^3,</math> and <math>9^3</math> for a total of <math>6</math> cubes |
==See Also== | ==See Also== |
Revision as of 01:18, 15 January 2018
Contents
[hide]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 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 also find:
and
So, we have 6 cubes total: and
for a total of
cubes
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.