Find the required digit What is the digit at the $50^{th}$ place from left of $(\sqrt50 +7)^{50}$  

I thought of binomial expansion but it was way too lengthy. Can anyone suggest any other  way?


Note that $a=7+\sqrt{50}$ and $b=7-\sqrt{50}$ satisfy $a+b=14$ and $ab=-1$. Note that $a\gt 14$ so that $a^n \gt  10^n$ (by some margin) so that $a^{50}\gt 10^{50}$ and $b^{50}\lt 10^{-50}$
Note that if $Y_n=a^n+b^n$ we have $Y_n=14Y_{n-1}+Y_{n-2}$ so the $Y_r$ are integers. ($a$ and $b$ satisfy $x^2-14x-1=0$, $Y_0=2, Y_1=14$)
Conclude.
