Difference between revisions of "2005 AMC 10A Problems/Problem 15"
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How many positive cubes divide <math> 3! \cdot 5! \cdot 7! </math> ? | How many positive cubes divide <math> 3! \cdot 5! \cdot 7! </math> ? | ||
| − | <math> \ | + | <math> \textbf{(A) } 2\qquad \textbf{(B) } 3\qquad \textbf{(C) } 4\qquad \textbf{(D) } 5\qquad \textbf{(E) } 6 </math> |
== Solution 1 == | == Solution 1 == | ||
Revision as of 18:26, 13 December 2021
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. 
