Question

Prove the equality:
${\left(\mathrm{666...6}\right)}^2+\mathrm{888....8}=\mathrm{444...4}$
$n$ digits $n$ digits $2n$ digits

Answer 1

Represent each number as a $G.P$ e.g., we write $6666$ as
$6+6.10+{6.10}^2+++{6.10}^{n-1}$
$=\frac{6({10}^n-1)}{10-1}=\frac{2}{3}({10}^n-1)$ etc.
Hence we have to prove that
${\left[\frac{2}{3}\right({10}^n-1\left)\right]}^2+\frac{8}{9}({10}^n-1)=\frac{4}{9}({10}^{2n}-1)$
Now cancel $\therefore \frac{1}{9}({10}^n-1)$
$\therefore 4({10}^n-1)+8=4\therefore \frac{1}{9}({10}^n+1)$
Above is clearly true.