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Question Number 80068 by Pratah last updated on 30/Jan/20

Commented by mr W last updated on 30/Jan/20

$s i r P r a t a h :$ $a r e y o u s t u d e n t ? t e a c h e r ? p r o f e s s o r ?$ ${w h a t}^{'} s t h e s o u r c e o f y o u r q u e s t i o n s ?$
Commented by Pratah last updated on 30/Jan/20

$student . plz solution$
Commented by Pratah last updated on 30/Jan/20

$sir ? john santu$
	Answered by mr W last updated on 30/Jan/20

	$N_{1} = 2 n t i m e s 1 = 11 . . . 1$ $N_{2} = n t i m e s 2 = 22 . . . 2$ $N_{3} = n t i m e s 3 = 33 . . . 3$ $N_{1} = \frac{\overset{2 n t i m e s}{99 . . . 9}}{9} = \frac{10^{2 n} - 1}{9}$ $N_{2} = \frac{2 \times \overset{n t i m e s}{99 . . . 9}}{9} = \frac{2 (10^{n} - 1)}{9}$ $N_{3} = \frac{\overset{n t i m e s}{99 . . . 9}}{3} = \frac{10^{n} - 1}{3}$ $N_{1} - N_{2} = \frac{10^{2 n} - 1 - 2 (10^{n} - 1)}{9}$ $= \frac{{(10^{n})}^{2} - 2 \times 10^{n} + 1}{9} = {(\frac{10^{n} - 1}{3})}^{2}$ $\sqrt{N_{1} - N_{2}} = \frac{10^{n} - 1}{3} = N_{3}$ $\Rightarrow p r o v e d!$
	Commented by Pratah last updated on 30/Jan/20

	$thanks$
	Answered by naka3546 last updated on 31/Jan/20

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