solution: Prime factorize $284=2^2\cdot71$. The sum of the proper divisors of $284$ is
  \begin{align*}
  1+2+2^2+71+2 \cdot 71 &= (1+2+2^2)(1+71)-284 \\
  &= 220 \\
  &= 2^2\cdot5\cdot11.
  \end{align*}Here we have used the observation that multiplying out $(1+2+2^2)(1+71)$ by distributing yields an expression which is the sum of all $6$ factors of $284.$ Applying this observation again, we find that the sum of the proper divisors of $220$ is $$(1+2+2^2)(1+5)(1+11)-220=7\cdot 6\cdot 12-220=\boxed{284}.$$
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  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?
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 284
extra_info.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?
extra_info.split: test