Question

Let $x = 111.$........11 (20 digits)
$y = 333$........33 (10 digits)
and $z = 222$.........22 (10 digits),

Answer 1

$ x =\frac{1}{9}[999 \ldots \ldots . .99]=\frac{1}{9}\left[10^{20}-1\right] \ldots . .(1) $
$y =\frac{1}{3}[999 \ldots \ldots . .99]=\frac{1}{3}\left[10^{10}-1\right] $
$z =\frac{2}{9}\left[10^{10}-1\right] $
$\therefore \frac{ x - y ^2}{ z }=\frac{\left(10^{20}-1\right)-\left(10^{10}-1\right)^2}{2\left(10^{10}-1\right)}=\frac{\left(10^{10}+1\right)-\left(10^{10}-1\right)}{2}=1$