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"
m (Improved formatting)
 
(38 intermediate revisions by 16 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
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>
+
We obtain <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</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 of 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 non-negative multiples 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:
+
This means
  
−
<math>a\in\{0,3,6\}</math> (<math>3</math> posibilities)
+
<cmath>\begin{align*}&a\in\{0,3,6\} \text{ (} 3 \text{ possibilities),} \\
 +
&b\in\{0,3\} \text{ (} 2 \text{ possibilities),} \\
 +
&c\in\{0\} \text{ (} 1 \text{ possibility), and} \\
 +
&d\in\{0\} \text{ (} 1 \text{ possibility),}\end{align*}</cmath>
  
−
<math>b\in\{0,3\}</math> (<math>2</math> posibilities)
+
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>.
  
−
<math>c\in\{0\}</math> (<math>1</math> posibility)
+
==Solution 2==
 +
As in Solution 1, we write <math>3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1)</math>, and now notice that there are a total of <math>3</math> <math>3</math>s, <math>3</math> <math>2</math>s, and <math>3</math> <math>1</math>s. This gives us our first <math>3</math> cubes: <math>3^3</math>, <math>2^3</math>, and <math>1^3</math>.
  
−
<math>d\in\{0\}</math>(<math>1</math> posibility)
+
However, we can also multiply smaller numbers in the expression to make bigger cubes. For example, <math>(2 \cdot 2) \cdot 4 \cdot 4 = 4 \cdot 4 \cdot 4 = 4^3</math> (where one 2 comes from the <math>3!</math>, and the other from the <math>5!</math>). Using this method, we can also make
  
 +
<cmath>(3 \cdot 2) \cdot (3 \cdot 2) \cdot 6 = 6^3</cmath>
  
−
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>
+
and
  
−
==Solution 2==
+
<cmath>(3 \cdot 4) \cdot (3 \cdot 4) \cdot (2 \cdot 6) = 12^3.</cmath>
  
−
If you factor 3!*5!*7! You get
+
So we have <math>6</math> possible cubes in total: <math>1^3</math>, <math>2^3</math>, <math>3^3</math>, <math>4^3</math>, <math>6^3</math>, and <math>12^3</math>, and the answer is <math>\boxed{\textbf{(E) } 6}</math>.
  
−
==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 17:52, 1 July 2025

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

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

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

This means

\begin{align*}&a\in\{0,3,6\} \text{ (} 3 \text{ possibilities),} \\ &b\in\{0,3\} \text{ (} 2 \text{ possibilities),} \\ &c\in\{0\} \text{ (} 1 \text{ possibility), and} \\ &d\in\{0\} \text{ (} 1 \text{ possibility),}\end{align*}

so the number of perfect cubes that divide $3! \cdot 5! \cdot 7!$ is $3\cdot2\cdot1\cdot1 = \boxed{\textbf{(E) } 6}$.

Solution 2

As in Solution 1, we write $3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1)$, and now notice that there are a total of $3$ $3$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 cubes. For example, $(2 \cdot 2) \cdot 4 \cdot 4 = 4 \cdot 4 \cdot 4 = 4^3$ (where one 2 comes from the $3!$, and the other from the $5!$). Using this method, we can also make

\[(3 \cdot 2) \cdot (3 \cdot 2) \cdot 6 = 6^3\]

and

\[(3 \cdot 4) \cdot (3 \cdot 4) \cdot (2 \cdot 6) = 12^3.\]

So we have $6$ possible cubes in total: $1^3$, $2^3$, $3^3$, $4^3$, $6^3$, and $12^3$, and the answer is $\boxed{\textbf{(E) } 6}$.

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_15&oldid=253082"