2020 AMC 10A Problems/Problem 15
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 
0 s, etc.)
The probability that the divisor chosen is a perfect square is ![\[\frac{6\cdot 3\cdot 2}{11\cdot 6\cdot 3\cdot 2\cdot 2}=\frac{1}{22} \implies \frac{m}{n}=\frac{1}{22} \implies m + n = 1 + 22 = 23 \implies \boxed{\textbf{(E) } 23 }\]](http://latex.artofproblemsolving.com/a/6/7/a6710cee723ef2f940e036e5f72551c4920b2450.png)
~mshell214
Video Solution
~IceMatrix
See Also
| 2020 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
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| All AMC 10 Problems and Solutions | ||
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