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Can it be proved that if $n$ is a multiple of $6$, then there exists a subset of the proper divisors of $n$ that add up to $n$?

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  • $\begingroup$ Do you mean to ask whether all perfect numbers are multiples of $6$? Or that multiples of $6$ are the only numbers with enough proper divisors to even hope to be perfect? $\endgroup$ – Arthur Nov 9 at 9:46
  • $\begingroup$ The question is not clear to me. $\endgroup$ – Shobhit Nov 9 at 9:47
  • $\begingroup$ No I stumbled upon this problem from here codechef.com/problems/MCHEF001 In point 2, it is clear that the sum of subsets of factor of that number should not be equal to itself. $\endgroup$ – Vivek Dubey Nov 9 at 9:48
  • $\begingroup$ After going through various contestants, I saw one has just checked for multiples of 6 for satisfying point 2 of the question i.e if a number is a multiple of 6 then subset of its proper divisors add up to that number. Same is not with other number. $\endgroup$ – Vivek Dubey Nov 9 at 9:50
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    $\begingroup$ Are you just asking for $N + 2N + 3N = 6N$? $\endgroup$ – Brian Moehring Nov 9 at 10:04
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If $n$ is a multiple of $6$. Then the set of its divisors contains $\frac{n}{2}$, $\frac{n}{3}$, and $\frac{n}{6}$ which sum to $n$.

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Here is a proof that if $n$ is a multiple of $6$, then there exists a subset of factors to $n$ that adds up to $n$. If $n = 6k$, then the subset $\{k,2k,3k\} = \{\frac n6, \frac n3, \frac n2\}$ adds up to $n$.

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Sure, but it also follows from definition, that any multiple of a semiperfect number works.

It's literally, that $n$ divided by each of divisors of the semiperfect numbers $y$, that sum to $y$, will then sum to $n$.

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