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 the divisor | + | 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 = \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 14:16, 5 November 2022
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 the divisor can't have any factors of 
and 
in the prime factorization 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
Education, The Study of Everything
~IceMatrix
~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. 
