What is the largest number smaller than 100 such that the sum of its divisors is larger than twice the number...
4
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
edited Nov 22 '18 at 12:33
amWhy
192k28225439
asked Mar 19 '17 at 0:29
Gerard L.
1,372726
add a comment |
4
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
edited Nov 22 '18 at 12:33
amWhy
192k28225439
asked Mar 19 '17 at 0:29
Gerard L.
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
add a comment |
4
4
4
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
edited Nov 22 '18 at 12:33
amWhy
192k28225439
asked Mar 19 '17 at 0:29
Gerard L.
1,372726
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
prime-factorization divisor-sum
edited Nov 22 '18 at 12:33
amWhy
192k28225439
asked Mar 19 '17 at 0:29
Gerard L.
1,372726
edited Nov 22 '18 at 12:33
amWhy
192k28225439
asked Mar 19 '17 at 0:29
Gerard L.
1,372726
edited Nov 22 '18 at 12:33
amWhy
192k28225439
edited Nov 22 '18 at 12:33
amWhy
192k28225439
edited Nov 22 '18 at 12:33
amWhy
192k28225439
192k28225439
asked Mar 19 '17 at 0:29
Gerard L.
1,372726
asked Mar 19 '17 at 0:29
Gerard L.
1,372726
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
add a comment |
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
add a comment |
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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