+816 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 $256$?
+36 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 $255$
