Difference between revisions of "2005 AMC 10A Problems/Problem 15"
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<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> | ||
− | Therefore, a perfect cube that divides <math> 3! \cdot 5! \cdot 7! </math> must be in the form <math>2^{a}\cdot3^{b}\cdot5^{c}\cdot7^{d}</math> where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are non-negative multiples of <math>3</math> that are less than | + | Therefore, a perfect cube that divides <math> 3! \cdot 5! \cdot 7! </math> must be in the form <math>2^{a}\cdot3^{b}\cdot5^{c}\cdot7^{d}</math> where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are non-negative multiples of <math>3</math> that are less than or equal to <math>8</math>, <math>4</math>, <math>2</math>, and <math>1</math> respectively. |
So: | So: |
Revision as of 20:06, 1 August 2006
Problem
How many positive cubes divide ?
Solution
Therefore, a perfect cube that divides must be in the form where , , , and are non-negative 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