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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+ | CHECK OUT Video Solution: https://youtu.be/cIgUSeRl5Io | ||
==See also== | ==See also== |
Revision as of 19:46, 30 October 2020
Contents
[hide]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
CHECK OUT Video Solution: https://youtu.be/cIgUSeRl5Io
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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