Difference between revisions of "2005 AMC 10A Problems/Problem 15"
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How many positive cubes divide <math> 3! \cdot 5! \cdot 7! </math> ? | How many positive cubes divide <math> 3! \cdot 5! \cdot 7! </math> ? | ||
− | <math> \ | + | <math> \textbf{(A) } 2\qquad \textbf{(B) } 3\qquad \textbf{(C) } 4\qquad \textbf{(D) } 5\qquad \textbf{(E) } 6 </math> |
== Solution 1 == | == 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> | ||
− | 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 | + | 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, 5, 2</math> and <math>1,</math> respectively. |
So: | So: | ||
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<math>d\in\{0\}</math>(<math>1</math> possibility) | <math>d\in\{0\}</math>(<math>1</math> possibility) | ||
− | + | So the number of perfect cubes that divide <math> 3! \cdot 5! \cdot 7! </math> is <math>3\cdot2\cdot1\cdot1 = \boxed{\textbf{(E) }6}</math> | |
− | So the number of perfect cubes that divide <math> 3! \cdot 5! \cdot 7! </math> is <math>3\cdot2\cdot1\cdot1 = | ||
==Solution 2== | ==Solution 2== | ||
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<math>(3 \cdot 4) \cdot (3 \cdot 4) \cdot (2 \cdot 6) = 12^3</math> | <math>(3 \cdot 4) \cdot (3 \cdot 4) \cdot (2 \cdot 6) = 12^3</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 | + | 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>\boxed{\textbf{(E) }6}</math> cubes |
==See also== | ==See also== |
Latest revision as of 19:29, 13 December 2021
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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.