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May 14th, 2014, 06:04 AM  #1 
Member Joined: Apr 2014 From: Missouri Posts: 39 Thanks: 1  Summation of all proper divisor
What is the maximum 4 digit integer n for which the sum of its proper divisors is (n1)? Thanks in advance 
May 14th, 2014, 07:12 AM  #2 
Senior Member Joined: Nov 2010 Posts: 288 Thanks: 1 
u may notice that for the sum of divisors of an integer n to be n1 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 
May 14th, 2014, 07:27 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,622 Thanks: 2611 Math Focus: Mainly analysis and algebra 
Oops! I used the wrong definition of proper divisors!
Last edited by v8archie; May 14th, 2014 at 07:32 AM. 
May 14th, 2014, 07:42 AM  #4 
Member Joined: Apr 2014 From: Missouri Posts: 39 Thanks: 1 
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 (n1) then n=2^k? Why did we take ln? Why cant we take log? Thanks in advance and Regards 
May 14th, 2014, 07:45 AM  #5 
Member Joined: Apr 2014 From: Missouri Posts: 39 Thanks: 1 
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 
May 14th, 2014, 08:48 AM  #6 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 155 Math Focus: Abstract algebra  

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