If x-111 -120 digits- y-333 -310 digits and z-222 -210 digits- then x-z is equal to

We know that, x=111 ……1(20 digits) ⇒ x = 999 …… 9(20 digits)9 ⇒ x=19(10^20 1) ⇒ y=13(10^10 1) ⇒ z=29(10^10 1) Now, nbsp; x z = 10^209 210^109+1 9 ⇒ x ...

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Byju's Answer

Standard XIII

Mathematics

Multinomial Expansion

If x=111 ……12...

Question

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

y^{2}

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2 y^{2}

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y^{3}

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3 y^{3}

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Solution

The correct option is A $y^{2}$
We know that,
$x = 111 \dots \dots 1 (20 digits) \Rightarrow x = \frac{999 \dots \dots 9 (20 digits)}{9} \Rightarrow x = \frac{1}{9} (10^{20} - 1) \Rightarrow y = \frac{1}{3} (10^{10} - 1) \Rightarrow z = \frac{2}{9} (10^{10} - 1)$

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

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Similar questions

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If x=111……1(20 digits), y=333……3(10 digits) and z=222……2(10 digits), then x−z is equal to

The correct option is A y2We know that, x=111……1(20 digits)⇒x=999……9(20 digits)9⇒x=19(1020−1)⇒y=13(1010−1)⇒z=29(1010−1) Now, x−z=10209−210109+19⇒x−z=(10103−13)2⇒x−z=[13(1010−1)]2∴x−z=y2

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