Okay, let's tackle this math 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 into smaller parts, it should be manageable.

First, I need to find the proper divisors of 284. Proper divisors are numbers that divide 284 evenly without leaving a remainder, but they have to be less than 284 itself. So, starting with 1, since 1 is a divisor of every number. Then, checking 2: 284 divided by 2 is 142, which is an integer, so 2 is a proper divisor. Next, 3: 284 divided by 3 is approximately 94.666..., which isn't an integer, so 3 isn't a divisor. Then 4: 284 divided by 4 is 71, which is an integer, so 4 is a proper divisor. Let's check 5: 284 divided by 5 is 56.8, which isn't an integer. 6: 284 divided by 6 is about 47.333..., not an integer. 7: 284 divided by 7 is around 40.571..., nope. 8: 284 divided by 8 is 35.5, still not an integer. 9: 284 divided by 9 is approximately 31.555..., nope. 10: 284 divided by 10 is 28.4, not an integer. Hmm, maybe I need a better approach than trial division here.

Wait, maybe 284 is a known number with a specific property. Let me recall. Oh, right! 284 is part of an amicable pair with 220. Amicable numbers are pairs where each number is the sum of the proper divisors of the other. So if 284 and 220 are amicable, then the sum of the proper divisors of 284 should be 220. Let me verify that to be sure.

So, if the sum of proper divisors of 284 is 220, then the next step is to find the sum of the proper divisors of 220. Then, the final answer would be the sum of the proper divisors of that sum. Wait, no. Wait, the problem says: sum of the proper divisors of the sum of the proper divisors of 284. So first, compute the sum of proper divisors of 284, which is S1. Then, compute the sum of proper divisors of S1, which is S2. Then, the answer is S2.

But if 284 and 220 are amicable, then S1 = 220. Then, I need to find the sum of proper divisors of 220, which would be S2. Then, the answer is S2. So, maybe the answer is 284? Because in amicable pairs, the sum of proper divisors of each is the other number. So if S1 = 220, then S2 would be 284. But let me check that.

Wait, but amicable numbers are pairs where each is the sum of the proper divisors of the other. So if 220 and 284 are amicable, then the sum of proper divisors of 220 is 284, and the sum of proper divisors of 284 is 220. Therefore, the sum of proper divisors of 284 is 220, then the sum of proper divisors of 220 is 284. Therefore, the answer should be 284. But let me verify this by actually calculating the proper divisors of 284 and 220 to confirm.

Starting with 284. Let's find all proper divisors. Prime factorization might help. Let's factor 284. 284 divided by 2 is 142, so 2 is a factor. Then 142 divided by 2 is 71. So 284 = 2^2 * 71. Therefore, the divisors of 284 are 1, 2, 4, 71, 142, and 284. The proper divisors are 1, 2, 4, 71, 142. Let's sum those: 1 + 2 + 4 + 71 + 142. Calculating step by step: 1+2=3; 3+4=7; 7+71=78; 78+142=220. Yes, so the sum of proper divisors of 284 is indeed 220. That confirms the amicable pair. Then, the sum of proper divisors of 220 would be the next step.

Now, let's find the sum of proper divisors of 220. First, factorize 220. 220 divided by 2 is 110, then 110 divided by 2 is 55. 55 is 5*11. So the prime factors of 220 are 2^2 * 5 * 11. The divisors are generated by exponents of prime factors: exponents for 2: 0,1,2; for 5: 0,1; for 11: 0,1. So all combinations:

2^0 *5^0 *11^0 =1

2^0 *5^0 *11^1=11

2^0 *5^1 *11^0=5

2^0 *5^1 *11^1=5*11=55

2^1 *5^0 *11^0=2

2^1 *5^0 *11^1=2*11=22

2^1 *5^1 *11^0=2*5=10

2^1 *5^1 *11^1=2*5*11=110

2^2 *5^0 *11^0=4

2^2 *5^0 *11^1=4*11=44

2^2 *5^1 *11^0=4*5=20

2^2 *5^1 *11^1=4*5*11=220

So the divisors are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110, 220. The proper divisors are all except 220. So sum them up: 1 + 2 + 4 + 5 + 10 + 11 + 20 + 22 + 44 + 55 + 110. Let's add 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

Yes, so the sum of proper divisors of 220 is 284. Therefore, the sum of proper divisors of the sum of proper divisors of 284 is 284. So the answer is 284. But let me just cross-verify once more to be sure.

First, sum proper divisors of 284: 1+2+4+71+142=220. Then sum proper divisors of 220: 1+2+4+5+10+11+20+22+44+55+110=284. So yes, the answer is 284. Therefore, the final answer is 284.
Therefore, the sum of the proper divisors of the sum of the proper divisors of 284 is $\boxed{284}$.
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