2020 AMC 10A Problems/Problem 15
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 12! is 
* 
* 
* 
* 
.
This yields a total of * 
* 
* 
* 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 * 3 * 2 perfect squares. (For 2, you can have 0, 2, 4, 6, 8, or 10 2s, 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 = 23.
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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