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?
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You have a nice answer below for your “twice” case in your question, but your actual question posits any multiple, not just a number’s double. Do you also want any multiple or just simply twice? – Eleven-Eleven May 28 '19 at 15:13
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@Eleven-Eleven any multiple – user662650 May 28 '19 at 15:20
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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.
Dietrich Burde
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So did you add all the divisors manually or is there a short cut to find the sum? – user662650 May 29 '19 at 04:27
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No, there is a recursion formula and several special cases, where I do not need to find the divisors myself. See the link I have given above. For more information see also this post. – Dietrich Burde May 29 '19 at 08:07