Tricky Powers of Two

On the LHS, let us factor out $2^{20}$ to obtain:

$2^{20} (1+2^4+2^5+2^{x-20}) = 2^{20}(1 + 16 + 32 + 2^{x-20}) = 2^{20}(49 + 2^{x-20})$

At $x=25$ , we get:

$2^{20}(49 + 2^{25-20}) = 2^{20}(49 + 32) = 2^{20}[7^2 + 2(7)(2) + 2^2] = 2^{20}(7+2)^2 = 2^{20}9^{2} = y^2 \Rightarrow y = 2^{10} \cdot 9 = 9216.$

Hence, $x+y=25+9216 = 9241 \Rightarrow \boxed{241}$ for the last three digits.

$$ There are 2 solutions for the equation. $x_1 = 24 \text{ ; } y_1 = 2^{13}$ and $x_2 = 25 \text{ ; } y_2 = 2^{10}\cdot 3^2$ . The solution for this problem is $x_2$ and $y_2$ .