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

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?

• It is 96. Brute-force enumeration says so. Mar 19, 2017 at 0:35
• Well obviously, but there has to be a better proof than brute force. Mar 19, 2017 at 0:39
• 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. Mar 19, 2017 at 0:46
• 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 Mar 21, 2017 at 22:03