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 [[nonnegative]] [[multiple]]s of <math>3</math> that are less than or equal to <math>8</math>, <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 [[nonnegative]] [[multiple]]s of <math>3</math> that are less than or equal to <math>8</math>, <math>5</math>, <math>2</math> and <math>1</math>, respectively. |
So: | So: | ||
− | <math>a\in\{0,3, | + | <math>a\in\{0,3,5\}</math> (<math>3</math> possibilities) |
<math>b\in\{0,3\}</math> (<math>2</math> possibilities) | <math>b\in\{0,3\}</math> (<math>2</math> possibilities) |
Revision as of 11:06, 15 January 2018
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
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.