Difference between revisions of "2020 AMC 10A Problems/Problem 15"

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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 <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.)
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

Revision as of 17:27, 1 February 2020

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 $7$ and $11$ can not be in the prime factorization of a perfect square 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 $1$0 $2$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

https://youtu.be/ZGwAasE32Y4

~IceMatrix

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
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

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