Difference between revisions of "2005 AMC 10A Problems/Problem 15"
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So, we have 6 cubes total:<math>1^3 ,2^3, 3^3, 4^3, 6^3,</math> and <math>12^3</math> for a total of <math>6</math> cubes <math>\Rightarrow \mathrm{(E)}</math> | So, we have 6 cubes total:<math>1^3 ,2^3, 3^3, 4^3, 6^3,</math> and <math>12^3</math> for a total of <math>6</math> cubes <math>\Rightarrow \mathrm{(E)}</math> | ||
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− | + | [[Category:Introductory Combinatorics Problems]] | |
− | + | [[Category:Introductory Number Theory Problems]] | |
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{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:43, 13 August 2019
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==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 |
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