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Difference between revisions of "2005 AMC 10A Problems/Problem 15"

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In the expression, we notice that there are 3 <math>3's</math> and 2 <math>2's</math>. This gives us our first 2 cubes: <math>3^3</math> and <math>2^2</math>.  
 
In the expression, we notice that there are 3 <math>3's</math> and 2 <math>2's</math>. This gives us our first 2 cubes: <math>3^3</math> and <math>2^2</math>.  
  
−
However, we can multiply smaller numbers in the expression to make bigger expressions. For example, <math>(2*2)*4*4=4*4*4=4^3</math> (one 2 comes from the <math>3!</math>, and the other from the <math>5!</math>).
+
However, we can multiply smaller numbers in the expression to make bigger expressions. For example, <math>(2*2)*4*4=4*4*4=4^3</math> (one 2 comes from the <math>3!</math>, and the other from the <math>5!</math>). Using this method, we find:
 +
 
 +
<math>(3*2)*(3*2)*6=6^3</math>
  
 
==See Also==
 
==See Also==

Revision as of 01:12, 15 January 2018

Problem

How many positive cubes divide $3! \cdot 5! \cdot 7!$ ?

$\mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6$

Solution 1

$3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1) = 2^{8}\cdot3^{4}\cdot5^{2}\cdot7^{1}$

Therefore, a perfect cube that divides $3! \cdot 5! \cdot 7!$ must be in the form $2^{a}\cdot3^{b}\cdot5^{c}\cdot7^{d}$ where $a$, $b$, $c$, and $d$ are nonnegative multiples of $3$ that are less than or equal to $8$, $4$, $2$ and $1$, respectively.

So:

$a\in\{0,3,6\}$ ($3$ possibilities)

$b\in\{0,3\}$ ($2$ possibilities)

$c\in\{0\}$ ($1$ possibility)

$d\in\{0\}$($1$ possibility)


So the number of perfect cubes that divide $3! \cdot 5! \cdot 7!$ is $3\cdot2\cdot1\cdot1 = 6 \Rightarrow \mathrm{(E)}$

Solution 2

$3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1)$

In the expression, we notice that there are 3 $3's$ and 2 $2's$. This gives us our first 2 cubes: $3^3$ and $2^2$.

However, we can multiply smaller numbers in the expression to make bigger expressions. For example, $(2*2)*4*4=4*4*4=4^3$ (one 2 comes from the $3!$, and the other from the $5!$). Using this method, we find:

$(3*2)*(3*2)*6=6^3$

See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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