A positive integer $n$ is a Zumkeller number iff its divisors can be partitioned into two sets with equal sum.
If $\sigma(n)$ denotes the divisor-sum-function , this means that there are distinct divisors of $n$ ($1$ and $n$ are also allowed) with sum $S:=\frac{\sigma(n)}{2}$. Of course , $\sigma(n)$ must be even for this purpose , so $n$ can neither be a perfect square nor twice a perfect square.
Every perfect number ($\sigma(n)=2n$) is a Zumkeller number , the other Zumkeller numbers must be abundant ($\sigma(n)>2n$). Also, the OEIS-links shows some useful sufficient conditions , the most powerful one being that $n$ is divisble by $6$ and the $3$-adic valuation is odd.
If we assume that we know the prime factorization of $n$ , is there an efficient way to determine whether $n$ is a Zumkeller number ?
Checking all subsets is of course not efficient (the subset sum problem probably can in general only be solved this way).
A method often working is the greedy algorithm : We choose always the largest possible divisor so that the target sum is not exceeded. I have not found a number yet where this method fails. Is there one ?
