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
Best Answer
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 $$ \frac1a+\frac1b+\frac1c=1.$$