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











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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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  • It is 96. Brute-force enumeration says so.
    – Fabio Somenzi
    Mar 19 '17 at 0:35










  • Well obviously, but there has to be a better proof than brute force.
    – Gerard L.
    Mar 19 '17 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.
    – Fabio Somenzi
    Mar 19 '17 at 0:46








  • 1




    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
    – alphacapture
    Mar 21 '17 at 22:03















up vote
4
down vote

favorite












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?










share|cite|improve this question
























  • It is 96. Brute-force enumeration says so.
    – Fabio Somenzi
    Mar 19 '17 at 0:35










  • Well obviously, but there has to be a better proof than brute force.
    – Gerard L.
    Mar 19 '17 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.
    – Fabio Somenzi
    Mar 19 '17 at 0:46








  • 1




    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
    – alphacapture
    Mar 21 '17 at 22:03













up vote
4
down vote

favorite









up vote
4
down vote

favorite











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?










share|cite|improve this question















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?







prime-factorization divisor-sum






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 12:33









amWhy

191k28224439




191k28224439










asked Mar 19 '17 at 0:29









Gerard L.

1,372726




1,372726












  • It is 96. Brute-force enumeration says so.
    – Fabio Somenzi
    Mar 19 '17 at 0:35










  • Well obviously, but there has to be a better proof than brute force.
    – Gerard L.
    Mar 19 '17 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.
    – Fabio Somenzi
    Mar 19 '17 at 0:46








  • 1




    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
    – alphacapture
    Mar 21 '17 at 22:03


















  • It is 96. Brute-force enumeration says so.
    – Fabio Somenzi
    Mar 19 '17 at 0:35










  • Well obviously, but there has to be a better proof than brute force.
    – Gerard L.
    Mar 19 '17 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.
    – Fabio Somenzi
    Mar 19 '17 at 0:46








  • 1




    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
    – alphacapture
    Mar 21 '17 at 22:03
















It is 96. Brute-force enumeration says so.
– Fabio Somenzi
Mar 19 '17 at 0:35




It is 96. Brute-force enumeration says so.
– Fabio Somenzi
Mar 19 '17 at 0:35












Well obviously, but there has to be a better proof than brute force.
– Gerard L.
Mar 19 '17 at 0:39




Well obviously, but there has to be a better proof than brute force.
– Gerard L.
Mar 19 '17 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.
– Fabio Somenzi
Mar 19 '17 at 0:46






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.
– Fabio Somenzi
Mar 19 '17 at 0:46






1




1




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
– alphacapture
Mar 21 '17 at 22:03




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
– alphacapture
Mar 21 '17 at 22:03















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