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

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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>
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In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Thus, there are <math>5</math> * <math>3</math> * <math>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. 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>.)
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In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Thus, there are <math>5 \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. 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>.)
 
The probability that the divisor chosen is a perfect square is 1/22. m + n = 1 + 22 = 23 <math>\implies \boxed{\textbf{(E) } 23 }</math>
 
The probability that the divisor chosen is a perfect square is 1/22. m + n = 1 + 22 = 23 <math>\implies \boxed{\textbf{(E) } 23 }</math>
  

Revision as of 22:07, 31 January 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. Thus, there are $5 \cdot 3 \cdot 2$ perfect squares. (For $2$, you can have $0$, $2$, $4$, $6$, $8$, or $1$0 $2$s, etc. Note that $7$ and $11$ can not be in the prime factorization of a perfect square because there is only one of each in $12!$.) The probability that the divisor chosen is a perfect square is 1/22. m + n = 1 + 22 = 23 $\implies \boxed{\textbf{(E) } 23 }$

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

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