Summing Lesser Divisors

Proper divisor of a number doesn't include itself.

So the sum of proper divisors of $2^{10}$ are

$1+ 2+4+8+16+ 32+ 64+128+256+512 = 1023 = 2^{10} -1$

$\text{All positive divisors of } 2^{10} \text { (less than itself), }$

are the powers of 2 (since 2 is its only prime factor), where the power n is between 0 and 9, inclusive. )

Hence, our sum (of a geometric progression) is:

$2^0 + 2^1 + 2^2 + ... + 2^8 + 2^9 = \frac { 2^{10} - 1 }{2 -1} = \boxed { 2^{10} - 1 }$

Look at this problem in binary form. For example,

$32-1=31=11111$ in binary

Each digit is a divisor of $32$