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> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } | + | <math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6 </math> |
==Solution== | ==Solution== |
Revision as of 18:51, 6 January 2017
Contents
[hide]Problem
How many positive cubes divide ?
Solution
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
Answer :
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.