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> | ||
| − | 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> | + | 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>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 = \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> | ||
Revision as of 20:32, 8 June 2020
Contents
Problem
A positive integer divisor of 
is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as 
, where 
and 
are relatively prime positive integers. What is 
?
Solution
The prime factorization of is 
.
This yields a total of 
divisors of 
In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that 
and 
can not be in the prime factorization of a perfect square because there is only one of each in Thus, there are 
perfect squares. (For 
, you can have 
, , 
, 
, 
, or 
s, etc.)
The probability that the divisor chosen is a perfect square is ![\[\frac{6\cdot 3\cdot 2}{11\cdot 6\cdot 3\cdot 2\cdot 2}=\frac{1}{22} \implies \frac{m}{n}=\frac{1}{22} \implies m\ +\ n = 1\ +\ 22 = \boxed{\textbf{(E) } 23 }\]](http://latex.artofproblemsolving.com/d/b/4/db4113777e6a63ed02b4da07eadda444458a7ab4.png)
~mshell214, edited by Rzhpamath
Video Solution
~IceMatrix
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. 
