Question

If $x=111\dots \dots 1\left(20\text{digits}\right),$ $y=333\dots \dots 3\left(10\text{digits}\right)$ and $z=222\dots \dots 2\left(10\text{digits}\right)$, then $x-z$ is equal to

• A. $y^2$
• B. $2y^2$
• C. $y^3$
• D. $3y^3$

Answer 1

The correct option is A $y^2$

We know that,
$x=111\dots \dots 1\left(20\text{digits}\right)\Rightarrow x=\frac{999\dots \dots 9\left(20\text{digits}\right)}{9}\Rightarrow x=\frac{1}{9}({10}^{20}-1)\Rightarrow y=\frac{1}{3}({10}^{10}-1)\Rightarrow z=\frac{2}{9}({10}^{10}-1)$

Now,
$x-z=\frac{{10}^{20}}{9}-2\frac{{10}^{10}}{9}+\frac{1}{9}\Rightarrow x-z={(\frac{{10}^{10}}{3}-\frac{1}{3})}^2\Rightarrow x-z={\left[\frac{1}{3}\right({10}^{10}-1\left)\right]}^2\therefore x-z=y^2$