Difference between revisions of "2005 AMC 10A Problems/Problem 15"
(→Solution 1) |
Supermath123 (talk | contribs) m (→Solution 2) |
||
| Line 35: | Line 35: | ||
<math>(3*4)*(3*4)*(2*6)=12^3</math> | <math>(3*4)*(3*4)*(2*6)=12^3</math> | ||
| − | So, we have 6 cubes total: <math>1^3 ,2^3, 3^3, 4^3, 6^3,</math> and <math>12^3</math> for a total of <math>6</math> cubes <math>\Rightarrow \mathrm{(E)}</math> | + | So, we have 6 cubes total:<math>1^3 ,2^3, 3^3, 4^3, 6^3,</math> and <math>12^3</math> for a total of <math>6</math> cubes <math>\Rightarrow \mathrm{(E)}</math> |
==See Also== | ==See Also== | ||
Revision as of 18:55, 10 June 2019
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
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
