Is there a genral formula to find if a number X has it's factor such that they add upto a muliple of X?

Question

I was solving problems while I frequently encountered problems such as "Check if the positive factors of 496 add up to twice of 496", etc. So instead of finding all it's factors which is tiresome, I wanted to know is there a general formula?

Answer 1

Yes, there is a formula for $\sigma(n)$, the sum of positive divisors of $n$, see here. According to the formula $\sigma(496)=992=2\cdot 496$, so yes, "it adds up to twice of $496$". A positive integer $n$ is called perfect if $\sigma(n)=2n$. For references see here. The first perfect numbers are $$ n=6, 28, 496, 8128, 33550336, 8589869056, $$

$$ 137438691328, 2305843008139952128, $$

$$ 2658455991569831744654692615953842176, $$

$$ 191561942608236107294793378084303638130997321548169216 $$

More generally, $n$ is called $m$-multiple-perfect, if $\sigma(n)=m\cdot n$. For example $\sigma(120)=3\cdot 120$. See also this MO-question.