The combinatorial Gem

Corresponding to $2^n$ , $n$ numbers can be obtained by subtracting the numbers : $1 , 2, 2^2 ,...,2^{n-1}$ from it.

From $2^{10}$ onwards each difference is greater than $1000$ . So, considering the cases till $2^{10}$ .

By using numbers ranging from $2$ to $2^{10}$ we obtain : $\left(\displaystyle\sum_{k=1}^{10} k \right)$ numbers $=55$ .

Out of these $55$ numbers, $5$ are greater than $1000$ : $(2^{10}-1 , 2^{10}-2^1, ..., 2^{10}-2^4)$

Hence, the answer is : $55-5=\boxed{50}$