Digits of a fraction

Note that $\begin{aligned} \frac {25}{99} & = \frac {25}{100} \left(\frac 1{1-\frac 1{100}}\right) = 0.25 \left(1 + 0.01 + 0.0001 + 0.000001 + \cdots \right) = 0.25252525 \cdots = 0.\dot 2 \dot 5 \end{aligned}$ Then the sum of its first $2^{12} - 50$ decimal digits is:

$\begin{aligned} S & = (2+5) \times \frac {2^{12} - 50}2 \\ \sqrt S & = \sqrt{7(2^{11}-25)} \\ & = \sqrt{7(4(2^9) - 3(2^3) - 1)} & \small \blue{\text{Let } x = 2^3} \\ & = \sqrt{7(4x^3-3x-1)} \\ & = \sqrt{7(x-1)(4x^2 + 4x+1)} \\ & = \sqrt{7(x-1)(2x+1)^2} & \small \blue{\text{Put back }x = 2^3 = 8} \\ & = \sqrt{7^2 \times 17^2} \\ & = 7 \times 17 = \boxed{119} \end{aligned}$

Since $\frac{25}{99}=0.252525 \cdots$ , and $2^{12}-50$ is an even number, the sum of first $2^{12}-50$ digits is equal to $(2+5)\left(\dfrac{2^{12}-50}{2}\right)=7(2^{11}-25)=14161$ and therefore $\sqrt{14161}=\boxed{119}$ .