The definition: we define number as semi perfect , if the number equals to the sum of exactly k of its divisors.

the question: prove that for every n (n>0 | n belong to N) n is semi perfect order 3 if and only if n is divided by 6.

my attempt:


n is divided by 6 because of that, its divisors are $$\frac{n}{6} ,\frac{n}{2} ,\frac{n}{3} $$

we will sum those 3 divisors - $$\frac{n}{6} + \frac{n}{2} + \frac{n}{3} = n$$ and we finished the proof for one side


this is the side I have a bit more problem with I know n uphold the definition above but Im having trouble proving that n is divided by 6 with no remainder.

I thought maybe trying with contradiction ( assuming it doesnt divided by 6) but still stuck with this question

would love to please have some help with it


1 Answer 1


The proof that multiples of 6 work already contains a good hint for the other direction: It uses the fact that $$ \frac12+\frac13+\frac16=1.$$ Perhaps try to find (or rather: show that one cannot find) any other integer triple $(a,b,c)$ with $1<a<b<c$ and $$ \frac1a+\frac1b+\frac1c=1.$$

From $$1=\frac1a+\frac1b+\frac1c<\frac1a+\frac1a+\frac1a,$$ we obtain $a<3$, i.e., necessarily $a=2$. After this, from $$ \frac12=1-\frac1a=\frac1b+\frac1c<\frac1b+\frac1b, $$ we obtain $b<4$, so necessarily $b=3$. Finally, $\frac1c=1-\frac12-\frac13$ implies $c=6$.


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