# Special property of multiples of 6

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 $$n$$?

• 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? – Arthur Nov 9 at 9:46
• The question is not clear to me. – Shobhit Nov 9 at 9:47
• 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. – Vivek Dubey Nov 9 at 9:48
• 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. – Vivek Dubey Nov 9 at 9:50
• Are you just asking for $N + 2N + 3N = 6N$? – Brian Moehring Nov 9 at 10:04

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