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What is the largest number smaller than 100 such that the sum of its divisors is larger than twice the number itself?

After doing some guess and check, I found that $36$ had quite a few factors, and proceeded to use the largest multiple of $6$ less than $100$, using $96$ as my answer.

Is there a more solid proof?

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  • $\begingroup$ It is 96. Brute-force enumeration says so. $\endgroup$ Mar 19, 2017 at 0:35
  • $\begingroup$ Well obviously, but there has to be a better proof than brute force. $\endgroup$
    – Gerard L.
    Mar 19, 2017 at 0:39
  • $\begingroup$ I hope so. Meanwhile, the largest numbers less than 1000, 10000, 100000, are 996, 9996, 99996, respectively. But then at 1000000, the result is 999999. The series of numbers ending in 6 then resumes until we get to 999999999999. $\endgroup$ Mar 19, 2017 at 0:46
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    $\begingroup$ What you are looking for is abundant numbers. $10^n-4$ will always be abundant because it is a multiple of 6, so it can be written as itself plus half of itself plus a third of itself. Similarly, $10^n$ will always be abundant because it is a multiple of 20. The question is why there are often no abundant numbers between $10^n-4$ and $10^n$. $10^n-3$ will usually not be abundant because it is 1 mod 6; a list of consecutive abundant numbers can be found here: oeis.org/A096399/b096399.txt $\endgroup$ Mar 21, 2017 at 22:03

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