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.  Note that 7 and 11 can not be in the prime factorization of a perfect square because there is only one of each in $12 !$. Thus, there are $6 \cdot 3 \cdot 2$ perfect squares. (For $2$ , you can choose $0,2,4,6,8,$ or $10$, etc.  The probability that the divisor chosen is a perfect square is $$ \frac{6 \cdot 3 \cdot 2}{11 \cdot 6 \cdot 3 \cdot 2 \cdot 2}=\frac{1}{22} \Longrightarrow \frac{m}{n}=\frac{1}{22} \Longrightarrow m+n=1+22=\text { (E) } 23 $$. 
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 $m+n$?~ (2020 AMC 10A Problem, Question \#15) 
Options:
- $$3$$ 
- $$5$$ 
- $$12$$ 
- $$18$$ 
- $$23$$ 