solution: First, we find $A$. The prime factorization of $500$ is $2^2 \cdot 5^3$. Therefore,  $$A=(1+2+2^2)(1+5+5^2+5^3)=(7)(156).$$To see why $(1+2+2^2)(1+5+5^2+5^3)$ equals the sum of the divisors of 500, note that if you distribute (without simplifying), you get 12 terms, with each divisor of $2^2\cdot 5^3$ appearing exactly once.
  
  Now we prime factorize $7 \cdot 156 = 7 \cdot 2^2 \cdot 3 \cdot 13$. The sum of the prime divisors of $A$ is $2+3+7+13=\boxed{25}$.
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  If $A$ is the sum of the positive divisors of $500$, what is the sum of the distinct prime divisors of $A$?
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 25
extra_info.question: If $A$ is the sum of the positive divisors of $500$, what is the sum of the distinct prime divisors of $A$?
extra_info.split: test