<Question>
The proper divisors of 12 are 1, 2, 3, 4 and 6. A proper divisor of an integer $N$ is a positive divisor of $N$ that is less than $N$. What is the sum of the proper divisors of the sum of the proper divisors of 284?
</Question>

<think>
<Plan_1>
Okay, let's tackle this problem step by step. The question is asking for the sum of the proper divisors of the sum of the proper divisors of 284. That sounds a bit complicated, but if I break it down, it should be manageable.

First, let me make sure I understand what a proper divisor is. The problem states that a proper divisor of an integer N is a positive divisor of N that is less than N. So, for example, the proper divisors of 12 are 1, 2, 3, 4, and 6. That makes sense because each of those numbers divides 12 without leaving a remainder, and none of them are equal to 12.

The task is to find the sum of the proper divisors of the sum of the proper divisors of 284. Let's parse that again. We need to first find all the proper divisors of 284, add them up to get a sum, then find all the proper divisors of that sum, and finally add those proper divisors to get the final result.

So, step 1: Find all proper divisors of 284.

<prompt_1.1> Follow the above plan to find all proper divisors of 284 and calculate their sum. </prompt_1.1>
-----

Alternatively, maybe there's a formula or a smarter way to compute the sum of divisors. For example, the sum of divisors function is multiplicative, so if we can factorize the number into primes, we can compute the sum. Let's try that for 284.

<prompt_1.2> Use the above plan to factorize 284 and compute the sum of its proper divisors using the sum of divisors function. </prompt_1.2>

Then for 1 to 283, we can use the same method. Let's try to find the proper divisors of 1, 2, 4, 71, 142, 283.

<prompt_1.3> Follow the above plan to find the proper divisors of 1, 2, 4, 71, 142, and 283, and calculate their sum. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
Then, step 2: Sum those proper divisors. Let's call this sum S. Then, step 3: Find all proper divisors of S. Finally, step 4: Sum those proper divisors of S. That final sum is the answer we need.

Okay, starting with step 1: Find the proper divisors of 284. Let's list all the numbers less than 284 that divide 284 without leaving a remainder.

First, since 284 is even, 2 is a divisor. Let's divide 284 by 2: 284 ÷ 2 = 142. So, 2 and 142 are divisors. Then, check if 4 is a divisor: 284 ÷ 4 = 71. So, 4 and 71 are divisors. Wait, 71 is a prime number, right? Let me confirm: 71 is only divisible by 1 and 71. So, yes, 71 is prime.

So far, the divisors we have are 1, 2, 4, 71, 142, and 284. But since we're looking for proper divisors, we exclude 284. So the proper divisors of 284 are 1, 2, 4, 71, and 142.
</execution_1.1>

<execution_1.2>
First, prime factorization of 284. Let's see: 284 divided by 2 is 142. 142 divided by 2 is 71. 71 is a prime number (since it's not divisible by 2, 3, 5, 7; 7*10 is 70, 71 is prime). So 284 = 2^2 * 71^1. Then, the sum of divisors function σ(n) for n = p^a * q^b is (1 + p + p^2 + ... + p^a)(1 + q + ... + q^b). So σ(284) = (1 + 2 + 4)(1 + 71) = 7 * 72 = 504. Therefore, the sum of all divisors of 284 is 504. But we need the sum of proper divisors, which is σ(284) - 284 - 1? Wait, no. Wait, the sum of all divisors includes 1, 2, 4, 71, 142, 284. So proper divisors exclude 284 itself. Therefore, the sum of proper divisors is σ(284) - 284. Since σ(284) is 504, then the sum of proper divisors is 504 - 284 = 220. Wait, but let's check that with the actual divisors. Let's list them again: 1, 2, 4, 71, 142. Adding them up: 1 + 2 = 3, 3 + 4 = 7, 7 + 71 = 78, 78 + 142 = 220. Yep, that's correct. So the sum of proper divisors of 284 is 220.
</execution_1.2>

<execution_1.3>
Wait, 284 is 4 times 71. 71 is a prime number, right? Because 71 divided by primes up to its square root (which is about 8.4) doesn't divide evenly. Let's check: 2, 3, 5, 7. 7*10 is 70, so 7*10+1 is 71. 71 divided by 2 is 35.5, not integer. 71 divided by 3 is around 23.666...., not integer. Divided by 5 is 14.2, nope. Divided by 7 is about 10.142...., nope. So 71 is prime. Therefore, 284 factors into 2^2 * 71^1.

The sum of divisors function for 284 would be (1 + 2 + 4) * (1 + 71) = 7 * 72 = 504. But that's the sum of all divisors, including 284. So the sum of proper divisors would be 504 - 284 = 220. Wait, that's interesting. So the sum of proper divisors of 284 is 220. So that's a quicker way. But since I already found the proper divisors manually and confirmed that their sum is 220, this matches. So that's good.

So maybe using the sum of divisors function would have been faster. But since 284 is a knownamicable number with 220, right? Because the sum of proper divisors of 284 is 220, and the sum of proper divisors of 220 is 284. That's the classic amicable pair. So maybe I could have just used that knowledge, but since the problem is about the sum of proper divisors of the sum of proper divisors of 284, which would be the sum of proper divisors of 220. So then, we need to compute the sum of proper divisors of 220, which is 284. Therefore, the answer is 284.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let me double-check that. Is there any number between 2 and 142 that divides 284? Let's see.

<prompt_2.1> Follow the above plan to verify the list of proper divisors of 284 and their sum. </prompt_2.1>
-----

Based on execution_1.3:
But let's go through the steps again to make sure.

First, find the proper divisors of 284: 1, 2, 4, 71, 142. Sum is 1 + 2 + 4 + 71 + 142 = 220.

<prompt_2.2> Follow the above plan to verify the solution using the known amicable pair property and confirm the result. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Let's check 3: 284 ÷ 3 is approximately 94.666...., which is not an integer. 5: 284 ends with a 4, so not divisible by 5. 6: 284 ÷ 6 is about 47.333...., not integer. 7: 284 ÷ 7 is around 40.571...., nope. 8: 284 ÷ 8 = 35.5, not integer. 9: 284 ÷ 9 ≈ 31.555...., no. 10: 284 ÷ 10 = 28.4, nope. Then 11: Let's do the divisibility test for 11. Alternating sum: (2 + 4) - 8 = 6 - 8 = -2, which is not divisible by 11, so 11 doesn't divide 284. 12: 284 ÷ 12 ≈ 23.666...., nope. 13: 13*21 = 273, 284 - 273 = 11, so remainder 11. Not divisible. 14: 14*20 = 280, 284 - 280 = 4, so remainder 4. Not divisible. 15: 15*18 = 270, 284 - 270 = 14, remainder. 16: 16*17 = 272, 284 - 272 = 12, remainder. 17: 17*16 = 272, same as above. 18: 18*15 = 270, remainder 14. 19: 19*14 = 266, 284 - 266 = 18, remainder. 20: 20*14 = 280, remainder 4. So up to 20, none divide. Then 22: 22*12 = 264, 284 - 264 = 20, remainder. 23: 23*12 = 276, 284 - 276 = 8, remainder. 24: 24*11 = 264, same as above. 25: 25*11 = 275, 284 - 275 = 9, remainder. 26: 26*10 = 260, 284 - 260 = 24, remainder. 27: 27*10 = 270, remainder 14. 28: 28*10 = 280, remainder 4. 29: 29*9 = 261, 284 - 261 = 23, remainder. 30: 30*9 = 270, remainder 14. So on and so forth. So there's no divisors between 2 and 142. So the proper divisors are indeed 1, 2, 4, 71, 142. Sum is 220. So that seems correct.
</execution_2.1>

<execution_2.2>
Then, the sum of proper divisors of 220. Let's compute that. Let's list the proper divisors of 220.

First, prime factorization of 220. 220 divided by 2 is 110. Divided by 2 again is 55. 55 is 5*11. So prime factors are 2^2 * 5^1 * 11^1. Therefore, the proper divisors of 220 are all the divisors except 1 and 220. Wait, no. Wait, proper divisors are all divisors less than the number, so 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110. Let me check that.

Let's list them systematically. Starting from 1: 1 is a divisor. 220 ÷ 1 = 220. 2: 220 ÷ 2 = 110. 4: 220 ÷ 4 = 55. 5: 220 ÷ 5 = 44. 10: 220 ÷ 10 = 22. 11: 220 ÷ 11 = 20. Then, 20: 220 ÷ 20 = 11. 22: 220 ÷ 22 = 10. 44: 220 ÷ 44 = 5. 55: 220 ÷ 55 = 4. 110: 220 ÷ 110 = 2. So the divisors are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110, 220. Therefore, proper divisors are all except 220. So proper divisors are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110. Now, let's sum these up.

Calculating the sum step by step:

1 + 2 = 3

3 + 4 = 7

7 + 5 = 12

12 + 10 = 22

22 + 11 = 33

33 + 20 = 53

53 + 22 = 75

75 + 44 = 119

119 + 55 = 174

174 + 110 = 284

So the sum of proper divisors of 220 is 284. Therefore, the answer is 284. That's interesting because 284 and 220 form an amicable pair. Each is the sum of the proper divisors of the other. So this problem is essentially asking for the sum of the proper divisors of the sum of the proper divisors of 284, which loops back to 284 itself. Therefore, the answer is 284.

But let me confirm once more to ensure there are no mistakes in my calculations.

First proper divisors of 284:

1, 2, 4, 71, 142. Sum is 1 + 2 + 4 + 71 + 142 = 220. Correct.

Proper divisors of 220:

1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110. Sum is 1+2=3, 3+4=7, 7+5=12, 12+10=22, 22+11=33, 33+20=53, 53+22=75, 75+44=119, 119+55=174, 174+110=284. Correct.

Therefore, the final answer is 284.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.2:
Wait, but the problem says "the sum of the proper divisors of the sum of the proper divisors of 284". So step by step:

<prompt_3.1> Follow the above plan to confirm the solution and ensure it matches the problem statement. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
1. Sum of proper divisors of 284: 220.

2. Then sum of proper divisors of 220: 284.

Therefore, the answer is 284.

But let me check if there's any alternative interpretation of the problem. For instance, does "sum of the proper divisors" include the number itself? But no, the problem specifies "proper divisors" which exclude the number. So, for example, when we sum the proper divisors of 284, we exclude 284, getting 220. Then, when summing the proper divisors of 220, we exclude 220, getting 284. Hence, the result is 284.

Therefore, the answer is 284.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
**Final Answer**
\boxed{284}
</Plan_4>

<Execution_4>
No prompt to execute.
</Execution_4>

<Final_answer>
Therefore, the sum of the proper divisors of the sum of the proper divisors of 284 is $\boxed{284}$.
</Final_answer>
</think>