Difference between revisions of "2005 AMC 10A Problems/Problem 15"
(→See Also) |
m (→See==See also) |
||
| Line 37: | Line 37: | ||
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== | |
{{AMC10 box|year=2005|ab=A|num-b=14|num-a=16}} | {{AMC10 box|year=2005|ab=A|num-b=14|num-a=16}} | ||
Revision as of 13:52, 13 August 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
| 2005 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
