2005 AMC 10A Problems/Problem 15: Difference between revisions
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So the number of perfect cubes that divide <math> 3! \cdot 5! \cdot 7! </math> is <math>3\cdot2\cdot1\cdot1 = 6 \Rightarrow \mathrm{(E)}</math> | So the number of perfect cubes that divide <math> 3! \cdot 5! \cdot 7! </math> is <math>3\cdot2\cdot1\cdot1 = 6 \Rightarrow \mathrm{(E)}</math> | ||
==Solution 2== | |||
If you factor 3!*5!*7! You get | |||
==See Also== | ==See Also== | ||
Revision as of 17:25, 27 May 2013
Problem
How many positive cubes divide 
?
Solution
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:
( posibilities)
( posibilities)
( posibility)
( posibility)
So the number of perfect cubes that divide is 
Solution 2
If you factor 3!*5!*7! You get
