Is there always a partition into distinct divisors of an abundant number involving the largest nontrivial divisor? With the obvious exception of weird numbers like $70$, for which such partitions are not possible at all.
For example, with $12$, we have $6 + 4 + 2 = 12$.
With a larger abundant number like $60$ there of course are partitions that don't use the largest nontrivial divisor, like $20 + 15 + 10 + 5 + 4 + 3 + 2 + 1 = 60$, but there is also at least one that does: $30 + 20 + 10$.
Does there exist an abundant number, but not weird, which can be partitioned into a selection of its distinct divisors, but none of which use the largest nontrivial divisor?
The answer to this question seems like it should obviously be "No, of course not." But how to prove it?
I've mostly only looked at even abundant numbers, but I've also looked at $945$. But of course there could exist some ridiculously large number meeting this criterion.
 A: I'm afraid I can't provide a full answer, but here's some results that I've managed to come up with.
Lets first agree on some terminology.
   A positive integer is called perfect if it equals the sum of its proper divisors.
   A positive integer is called deficient if it exceeds the sum of its proper divisors.
   A positive integer is called abundant if it is exceeded by the sum of its proper divisors.
   A positive integer is called pseudoperfect if it equals the sum of a subset of its proper divisors.
Clearly, every perfect number is also pseudoperfect, and no deficient number is pseudoperfect.
   A positive integer that is abundant, but not pseudoperfect is called weird.
So far, this is standard terminology; lets introduce a further concept:
   A positive integer is called strongly pseudoperfect if it equals the sum of a subset of its proper divisors containing the largest proper divisor.
It is clear that perfect numbers are strongly pseudoperfect and that strongly pseudoperfect numbers are pseudoperfect.
   Finally, we call an abundant number semi-weird if it is neither weird nor strongly pseudoperfect.
Then your question can be re-stated as: Do semi-weird numbers exist?
With the use of a computer, I have found that no semi-weird numbers less than one million exist.
To ease computation, we make use of the following lemmas:
(1) If $n$ is strongly pseudoperfect, then every multiple of $n$ is as well.
(2) If $n=2^m\cdot p$ for a positive integer $m$ and a prime $p<2^{m+1}$, then $n$ is strongly pseudoperfect.
Using these two facts, if $s$ is the smallest semi-weird number, then the following hold.


*

*$s$ is abundant.

*$s$ is not weird.

*All divisors of $s$ are either deficient or weird.

*$s$ is not of the form $2^m\cdot p$ for a prime $p<2^{m+1}$.


If you reduce the numbers below one million to just the candidates for minimal semi-weirdness that satisfy these four conditions, you get a much smaller list of 1795 numbers (starting at $350$ and ending with $999999$). It is then easy to check that all of these are in fact strongly pseudoperfect.
Therefore, if semi-weird numbers exist at all, they must be larger than a million.
