"Just like in [problem 21](https://projecteuler.net/problem=21), we'll define an `aliquot_sum` function and use that to find all the [abundant numbers](https://en.wikipedia.org/wiki/Abundant_number) below 28,124 (as in problem 21, we could instead use a sieve to compute the divisor sums)."
"abundant_numbers = {k for k in range(1, 28124) if aliquot_sum(k) > k}"
]
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"Then we check every integer less than 28,124 to see if it's the sum of any two abundant numbers, and if it is, remove it from a set containing all those integers. Whatever's left in that set are the non-abundant sums."