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The proper divisors of $12$ are $1,$ $2,$ $3,$ $4$ and $6$. A proper divisor of an integer $N$ is a positive divisor of $N$ that is less than $N$. What is the sum of the proper divisors of the sum of the proper divisors of $256$?

 Aug 9, 2023
 #1
avatar+126971 
+1

Note that an integer that can  be  expressed as 2^n   has the sum of its proper divisors  =  2^n  - 1

 

Example   8 =  2^3

 

Proper   divisors =  1,2,4    sum = 7 =   2^3 - 1  

 

So

 

256  =  2^8

 

Sum of  proper divisors =  2^8 - 1 =   255

 

cool cool cool

 Aug 9, 2023
edited by CPhill  Aug 9, 2023
 #2
avatar+36 
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We have the number $256

The proper divisors of $256 are $1$, $2$, $4, $8$, $16$, $32$, $64$, and $128$

Now 1 + 2 + 4 + ..... + 128; Here first term = 1 

                                                     common ratio = 2

       = \( {1 (2^8 - 1) \over 2 - 1}\)         [ a + ar + ar2 + ... + arn-1

                                  = \({a(r^n - 1) \over r - 1}\);(r >1)

       = \( {256 - 1 \over 1}\)

       = 255

Thus the sum of proper divisor of $256$ is $255$

 Aug 10, 2023

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