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May 14th, 2014, 06:04 AM   #1
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Summation of all proper divisor

What is the maximum 4 digit integer n for which the sum of its proper divisors is (n-1)?

Thanks in advance
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May 14th, 2014, 07:12 AM   #2
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u may notice that for the sum of divisors of an integer n to be n-1 it should be of the form 2^k
to find the maximum integer u have to find k to do that u may solve the following:
2^k=9999(where 9999 is the maximum four digit integer)
then k=ln9999/ln2 =13.2.....
then k=13
and hence n=2^13=8192
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May 14th, 2014, 07:27 AM   #3
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Oops! I used the wrong definition of proper divisors!

Last edited by v8archie; May 14th, 2014 at 07:32 AM.
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May 14th, 2014, 07:42 AM   #4
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Hi Islam

Thanks for your answer. But How did you get this formula like:

If the sum of all proper divisor of n to be (n-1) then n=2^k?

Why did we take ln? Why cant we take log?

Thanks in advance and Regards
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May 14th, 2014, 07:45 AM   #5
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Hi Islam

It gives you the same result using ln and log. So one question is solved. But what is the formula for this?

Thanks
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May 14th, 2014, 08:48 AM   #6
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Almost perfect number – Wikipedia, the free encyclopedia
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