If x=111⋯1(20 digits) , y=333⋯3(10 digits) and z=222⋯2(10 digits) , then x−y2z=

x=19(1020−1), y=39(1010−1)=13(1010−1), z=29(1010−1) x−y2=19(1010+1)(1010−1)−[13(1010−1)]2 =19(1010−1)[1010+1−1010+1] =29(1010−1)=z

If x=111⋯1(20 digits) , y=333⋯3(10 digits) and z=222⋯2(10 digits) , then x−y2z=

x=19(1020−1), y=39(1010−1)=13(1010−1), z=29(1010−1) x−y2=19(1010+1)(1010−1)−[13(1010−1)]2 =19(1010−1)[1010+1−1010+1] =29(1010−1)=z

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If $x = 111 \dots 1 (20 digits)$ , $y = 333 \dots 3 (10 digits)$ and $z = 222 \dots 2 (10 digits)$ , then $\frac{x - y^{2}}{z} =$

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Detailed Solution

x = \frac{1}{9} (10^{20} - 1), y = \frac{3}{9} (10^{10} - 1) = \frac{1}{3} (10^{10} - 1), z = \frac{2}{9} (10^{10} - 1)

$\begin{matrix} x - y^{2} = \frac{1}{9} (10^{10} + 1) (10^{10} - 1) - {[\frac{1}{3} (10^{10} - 1)]}^{2} \\ = \frac{1}{9} (10^{10} - 1) [10^{10} + 1 - 10^{10} + 1] \end{matrix}$

$= \frac{2}{9} (10^{10} - 1) = z$

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If x=111⋯1(20 digits) , y=333⋯3(10 digits) and z=222⋯2(10 digits) , then x−y2z=

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If $x = 111 \dots 1 (20 digits)$ , $y = 333 \dots 3 (10 digits)$ and $z = 222 \dots 2 (10 digits)$ , then $\frac{x - y^{2}}{z} =$