Difference between revisions of "2005 AMC 10A Problems/Problem 15"
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In the expression, we notice that there are 3 <math>3's</math> and 2 <math>2's</math>. This gives us our first 2 cubes: <math>3^3</math> and <math>2^2</math>. | In the expression, we notice that there are 3 <math>3's</math> and 2 <math>2's</math>. This gives us our first 2 cubes: <math>3^3</math> and <math>2^2</math>. | ||
− | However, we can multiply smaller numbers in the expression to make bigger expressions. For example, <math>(2*2)*4*4=4*4*4=4^3</math> (one 2 comes from the <math>3!</math>, and the other from the <math>5!</math>). | + | However, we can multiply smaller numbers in the expression to make bigger expressions. For example, <math>(2*2)*4*4=4*4*4=4^3</math> (one 2 comes from the <math>3!</math>, and the other from the <math>5!</math>). Using this method, we find: |
+ | |||
+ | <math>(3*2)*(3*2)*6=6^3</math> | ||
==See Also== | ==See Also== |
Revision as of 02:12, 15 January 2018
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 and 2
. This gives us our first 2 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 find:
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.