Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
(→‎Solution 2)
 
(36 intermediate revisions by 16 users not shown)
Line 2: Line 2:
 
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> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6 </math>
+
<math> \textbf{(A) } 2\qquad \textbf{(B) } 3\qquad \textbf{(C) } 4\qquad \textbf{(D) } 5\qquad \textbf{(E) } 6 </math>
  
−
==Solution==
+
== 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</math>, <math>4</math>, <math>2</math> and <math>1</math>, respectively.  
+
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:  
  
−
<math>a\in\{0,3,6\}</math> (<math>3</math> posibilities)
+
<math>a\in\{0,3,6\}</math> (<math>3</math> possibilities)
  
−
<math>b\in\{0,3\}</math> (<math>2</math> posibilities)
+
<math>b\in\{0,3\}</math> (<math>2</math> possibilities)
  
−
<math>c\in\{0\}</math> (<math>1</math> posibility)
+
<math>c\in\{0\}</math> (<math>1</math> possibility)
  
−
<math>d\in\{0\}</math>(<math>1</math> posibility)
+
<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 = 6 \Rightarrow \mathrm{(E)}</math>
+
==Solution 2==
 +
<math> 3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1)</math>
  
−
==Solution 2==
+
In the expression, we notice that there are 3 <math>3's</math>, 3 <math>2's</math>, and 3 <math>1's</math>. This gives us our first 3 cubes: <math>3^3</math>, <math>2^3</math>, and <math>1^3</math>.
  
−
If you factor 3!*5!*7! You get
+
However, we can also multiply smaller numbers in the expression to make bigger expressions. For example, <math>(2 \cdot 2) \cdot 4 \cdot 4=4 \cdot 4 \cdot 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:
  
−
2^7 * 3^4  *  5^2
+
<math>(3 \cdot 2) \cdot (3 \cdot 2) \cdot 6 = 6^3</math>
  
−
A cube need three same factors for it to be a cube.
+
and
  
−
There are 3 ways for the first factor of a cube:  2^0, 2^3, and 2^6. And the second ways are:  3^0, and 3^3.
+
<math>(3 \cdot 4) \cdot (3 \cdot 4) \cdot (2 \cdot 6) = 12^3</math>
  
−
3 * 2 = 6  
+
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
−
Answer : E
 
  
−
==See Also==
+
==See also==
−
*[[2005 AMC 10A Problems]]
+
{{AMC10 box|year=2005|ab=A|num-b=14|num-a=16}}
  
−
*[[2005 AMC 10A Problems/Problem 14|Previous Problem]]
+
[[Category:Introductory Number Theory Problems]]
−
 
+
[[Category:Introductory Combinatorics Problems]]  
−
*[[2005 AMC 10A Problems/Problem 16|Next Problem]]
+
[[Category:Introductory Number Theory Problems]]
−
 
+
{{MAA Notice}}
−
*[[Factorial]]
 
−
 
 
−
*[[Prime factorization]]
 
−
 
 
−
 
 
−
[[Category:Introductory Combinatorics Problems]]  
 
−
[[Category:Introductory Number Theory Problems]]
 

Latest revision as of 18:29, 13 December 2021

Problem

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

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

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 $8, 5, 2$ 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 $3\cdot2\cdot1\cdot1 = \boxed{\textbf{(E) }6}$

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 $3's$, 3 $2's$, and 3 $1's$. This gives us our first 3 cubes: $3^3$, $2^3$, and $1^3$.

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

$(3 \cdot 2) \cdot (3 \cdot 2) \cdot 6 = 6^3$

and

$(3 \cdot 4) \cdot (3 \cdot 4) \cdot (2 \cdot 6) = 12^3$

So, we have 6 cubes total: $1^3 ,2^3, 3^3, 4^3, 6^3,$ and $12^3$ for a total of $\boxed{\textbf{(E) }6}$ 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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_15&oldid=167661"