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 "2020 AMC 10A Problems/Problem 15"

(→‎Solution)
(→‎Video Solution)
 
(7 intermediate revisions by 7 users not shown)
Line 8: Line 8:
 
The prime factorization of <math>12!</math> is <math>2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11</math>.  
 
The prime factorization of <math>12!</math> is <math>2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11</math>.  
 
This yields a total of <math>11 \cdot 6 \cdot 3 \cdot 2 \cdot 2</math> divisors of <math>12!.</math>
 
This yields a total of <math>11 \cdot 6 \cdot 3 \cdot 2 \cdot 2</math> divisors of <math>12!.</math>
−
In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that <math>7</math> and <math>11</math> can not be in the prime factorization of a perfect square because there is only one of each in <math>12!.</math> Thus, there are <math>6 \cdot 3 \cdot 2</math> perfect squares. (For <math>2</math>, you can have <math>0</math>, <math>2</math>, <math>4</math>, <math>6</math>, <math>8</math>, or <math>1</math>0 <math>2</math>s, etc.)
+
In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that the divisor can't have any factors of <math>7</math> and <math>11</math> in the prime factorization because there is only one of each in <math>12!.</math> Thus, there are <math>6 \cdot 3 \cdot 2</math> perfect squares. (For <math>2</math>, you can have <math>0</math>, <math>2</math>, <math>4</math>, <math>6</math>, <math>8</math>, or <math>10</math> <math>2</math>s, etc.)
−
The probability that the divisor chosen is a perfect square is <cmath>\frac{6\cdot 3\cdot 2}{11\cdot 6\cdot 3\cdot 2\cdot 2}=\frac{1}{22} \implies \frac{m}{n}=\frac{1}{22} \implies m + n = 1 + 22 = 23 \implies \boxed{\textbf{(E) } 23 }</cmath>
+
The probability that the divisor chosen is a perfect square is <cmath>\frac{6\cdot 3\cdot 2}{11\cdot 6\cdot 3\cdot 2\cdot 2}=\frac{1}{22} \implies \frac{m}{n}=\frac{1}{22} \implies m\ +\ n = 1\ +\ 22 = \boxed{\textbf{(E) } 23 }</cmath>
  
 
~mshell214, edited by Rzhpamath
 
~mshell214, edited by Rzhpamath
  
 
==Video Solution==
 
==Video Solution==
 +
 +
Education, The Study of Everything
 +
 +
https://youtu.be/ipZV6QfN3iU
 +
 +
The Beauty of Math
 +
 
https://youtu.be/ZGwAasE32Y4
 
https://youtu.be/ZGwAasE32Y4
  
 
~IceMatrix
 
~IceMatrix
 +
 +
https://youtu.be/XVbBKfbvELw
 +
 +
~savannahsolver
 +
 +
== Video Solution ==
 +
https://youtu.be/wopflrvUN2c?t=407
 +
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Latest revision as of 19:04, 3 December 2023

Problem

A positive integer divisor of $12!$ is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23$

Solution

The prime factorization of $12!$ is $2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11$. This yields a total of $11 \cdot 6 \cdot 3 \cdot 2 \cdot 2$ divisors of $12!.$ In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that the divisor can't have any factors of $7$ and $11$ in the prime factorization because there is only one of each in $12!.$ Thus, there are $6 \cdot 3 \cdot 2$ perfect squares. (For $2$, you can have $0$, $2$, $4$, $6$, $8$, or $10$ $2$s, etc.) The probability that the divisor chosen is a perfect square is \[\frac{6\cdot 3\cdot 2}{11\cdot 6\cdot 3\cdot 2\cdot 2}=\frac{1}{22} \implies \frac{m}{n}=\frac{1}{22} \implies m\ +\ n = 1\ +\ 22 = \boxed{\textbf{(E) } 23 }\]

~mshell214, edited by Rzhpamath

Video Solution

Education, The Study of Everything

https://youtu.be/ipZV6QfN3iU

The Beauty of Math

https://youtu.be/ZGwAasE32Y4

~IceMatrix

https://youtu.be/XVbBKfbvELw

~savannahsolver

Video Solution

https://youtu.be/wopflrvUN2c?t=407

~ pi_is_3.14

See Also

2020 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=2020_AMC_10A_Problems/Problem_15&oldid=206435"