Okay, let's tackle this problem step by step. The question is: If A is the sum of the positive divisors of 500, what is the sum of the distinct prime divisors of A? Hmm, alright. First, I need to find the sum of all positive divisors of 500, which is called A. Then, once I have A, I need to factorize it into its prime factors and sum up the distinct ones. Let me break this down.

Starting with the first part: finding the sum of divisors of 500. I remember that there's a formula for the sum of divisors of a number based on its prime factorization. The formula is if a number n can be expressed as \( n = p_1^{a_1} \times p_2^{a_2} \times \dots \times p_k^{a_k} \), then the sum of its divisors is \( (1 + p_1 + p_1^2 + \dots + p_1^{a_1}) \times (1 + p_2 + p_2^2 + \dots + p_2^{a_2}) \times \dots \times (1 + p_k + p_k^2 + \dots + p_k^{a_k}) \). So, first, I need to find the prime factorization of 500.

500. Let's factorize it. 500 divided by 2 is 250, divided by 2 again is 125. 125 is 5 cubed. So, 500 is \( 2^2 \times 5^3 \). Let me check that: 2 squared is 4, 4 times 5 cubed (125) is 4*125=500. Yep, that's correct.

So, applying the sum of divisors formula: for the prime factors 2 and 5. The exponents are 2 and 3 respectively. So, the sum of divisors would be (1 + 2 + 2^2) multiplied by (1 + 5 + 5^2 + 5^3). Let me compute each part.

First part: 1 + 2 + 4 = 7. Second part: 1 + 5 + 25 + 125. Let's compute that: 1 + 5 is 6, 6 + 25 is 31, 31 + 125 is 156. So, multiplying 7 by 156. Let me do that: 7*156. 7*150 is 1050, and 7*6 is 42, so 1050 + 42 = 1092. So, A is 1092. Wait, is that right? Let me verify.

Alternatively, maybe I made a mistake in the calculation. Let me double-check. The sum of divisors for 2^2 is 1 + 2 + 4 = 7, that's correct. For 5^3, the sum is 1 + 5 + 25 + 125. Let's add them again: 1 + 5 = 6, 6 +25=31, 31 +125=156. Then 7*156. Let me compute 7*156: 150*7=1050, 6*7=42; 1050+42=1092. Okay, that seems right. So A is 1092.

Now, the next part is to find the sum of the distinct prime divisors of A, which is 1092. So, I need to factorize 1092 into its prime factors and then sum the unique primes. Let's do that.

First, let's factorize 1092. Let's start by dividing by small primes. 1092 is even, so divide by 2: 1092 ÷ 2 = 546. 546 is also even, divide by 2 again: 546 ÷ 2 = 273. Now, 273: check divisibility by 3. 2 + 7 + 3 = 12, which is divisible by 3. So, 273 ÷ 3 = 91. Now, 91: check divisibility. 91 divided by 7 is 13. So, 91 = 7 * 13. Therefore, the prime factors of 1092 are 2, 2, 3, 7, 13. But we need the distinct primes, so they are 2, 3, 7, 13. Sum them up: 2 + 3 + 7 + 13. Let's compute that. 2 + 3 = 5, 5 +7=12, 12 +13=25. So, the sum is 25.

Wait, but let me confirm that 1092 is indeed 2^2 * 3 * 7 *13. Let's check 2^2 is 4, 4*3=12, 12*7=84, 84*13=1092. Yes, that's correct. So the prime factors are 2, 3, 7, 13. Sum is 2+3+7+13=25. Therefore, the answer should be 25.

But let me double-check each step to make sure I didn't skip anything or make a mistake. First, the sum of divisors of 500: 500 is 2^2 *5^3. Sum of divisors formula gives (1+2+4)*(1+5+25+125)=7*156=1092. Correct.

Then, factorizing 1092: divide by 2 twice to get 273, then divide by 3 to get 91, then 7*13. So primes are 2,3,7,13. Sum is 25. All steps look correct.

So, the final answer is 25. Therefore, the sum of the distinct prime divisors of A is 25.
Thus, the sum of the distinct prime divisors of $A$ is $\boxed{25}$.
</think>