2^k can be cool

I think you need to add that they cannot be expressed as the sum of two or more consecutive POSITIVE integers.

N cannot be written as sum of 2 or more consecutive integer if and only if N is a power of 2 proof:- <i> consider all odd number >= 3

as all odd can be expressed in form of 2 L -1 = (L-1) +(L) , therefore the only odd number satisfying the equation is '1'

<ii>consider even number

we can write N= (2^k)(p), where p is an odd number

let us consider the values for which N= $\sum_{i=1}^i$ G+i = iG +i(i+1)/2 = i{(2 G+1) +i}/2

therefore

2^(k+1) *p= i{(2G+1)+i}

(as, one of the factors of RHS is odd an both the factors are greater >1) there fore for p=1 , there exist no solution

if 2^(k+1)> p>1, we can select i= p , G= {2^(k+1) - (i+1)}/2

if p>2^(k+1) , we can select 2^(k+1)= i , p= 2G+1+i

proved

therefore answer = 2^0 +..............2^10 = 2047