Difference between revisions of "2005 AMC 10A Problems/Problem 15"
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Mr doudou14 (talk | contribs) |
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<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6 </math> | <math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6 </math> | ||
− | + | == Solution 1 == | |
<math> 3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1) = 2^{8}\cdot3^{4}\cdot5^{2}\cdot7^{1}</math> | <math> 3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1) = 2^{8}\cdot3^{4}\cdot5^{2}\cdot7^{1}</math> | ||
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==Solution 2== | ==Solution 2== | ||
− | + | <math> 3! \cdot 5! \cdot 7! = </math>3*2*5*4*3*2*7*6*5*4*3*2 | |
Revision as of 01:05, 15 January 2018
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
3*2*5*4*3*2*7*6*5*4*3*2
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.