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Question Number 55316 by Kunal12588 last updated on 21/Feb/19

Commented by Kunal12588 last updated on 21/Feb/19

$p l s h e l p m e i f o u n d 2519 b u t {d o n}^{'} t k n o w i s t h i s$ $s m a l l e s t . a l s o i t t a k e s 2 h r s f o r m e$ $e v e r y a n s w e r s a r e w e l c o m e$
	Answered by $@ty@m last updated on 21/Feb/19

	$G e n e r a l f o r m u l a f o r s u c h q u e s t i o n s :$ $R e q u i r e d n u m b e r =$ $L C M o f d i v i s o r s - c o m m o n d i f f e r e n c e o f$ $d i v i s o r & r e m a i n d e r$ $= 2520 - 1$ $= 2519$
	Commented by Kunal12588 last updated on 21/Feb/19

	$t h a n k y o u s i r$
	Commented by Kunal12588 last updated on 21/Feb/19

	$i t h i n k y o u a r e d o i n g t h i s$ $n = 2 x_{1} + 1 = 2 (x_{1} + 1) - 1$ $n = 3 x_{2} + 2 = 3 (x_{2} + 1) - 1$ $n = 4 x_{3} + 3 = 4 (x_{3} + 1) - 1$ $n = 5 x_{4} + 4 = 5 (x_{4} + 1) - 1$ $n = 6 x_{5} + 5 = 6 (x_{5} + 1) - 1$ $n = 7 x_{6} + 6 = 7 (x_{6} + 1) - 1$ $n = 8 x_{7} + 7 = 8 (x_{7} + 1) - 1$ $n = 9 x_{8} + 8 = 9 (x_{8} + 1) - 1$ $n = 10 x_{9} + 9 = 10 (x_{9} + 1) - 1$ $s o, n = m u l t i p l e o f (1, 2, 3, 4, 5, 6, 7, 8, 9, 10) - 1$ $s m a l l e s t n o .$ $n = L C M (1, 2, 3 . . ., 9, 10) - 1$ $n = 2520 - 1 = 2519$
	Commented by Kunal12588 last updated on 21/Feb/19

	$i a c t u a l l y d o i t l i k e t h i s q u i t e l o n g$ $n = 10 x_{9} + 9 = 9 x_{8} + 8$ $9 x_{8} = 10 x_{9} + 1$ $a f t e r s o m e a l g e b r a$ $2 x_{1} = 10 x_{9} + 8 o r x_{1} = 5 x_{9} + 4$ $3 x_{2} = 10 x_{9} + 7$ $4 x_{3} = 10 x_{9} + 6 o r 2 x_{3} = 5 x_{9} + 3$ $5 x_{4} = 10 x_{9} + 5 o r x_{4} = 2 x_{9} + 1$ $6 x_{5} = 10 x_{9} + 4 o r 3 x_{5} = 5 x_{9} + 2$ $7 x_{6} = 10 x_{9} + 3$ $8 x_{7} = 10 x_{9} + 2 o r 4 x_{7} = 5 x_{9} + 1$ $9 x_{8} = 10 x_{9} + 1$ $a f t e r r e a r r a n g i n g a n d r e a s o n i n g$ $l i k e t h i s o n e : -$ $7 x_{6} = 10 x_{9} + 3 = 7 x_{9} + 3 (x_{9} + 1)$ $x_{6} = x_{9} + 3 \times \frac{x_{9} + 1}{7}$ $∵ n i s n a t u r a l n o .$ $∴ x_{6} \in Z$ $∴ x_{9} + 1 i s a m u l t i p l e o f 7$ $s o, i g e t$ $x_{9} + 1 i s a m u l t i p l e o f 9, 7 a n d 4$ $o r x_{9} + 1 i s a m u l t i p l e o f 252$ $l o w e s t v a l u e o f x_{9} + 1 = 252$ $x_{9} = 251$ $n o w, a t l a s t p u t t i n g t h e v a l u e o f x_{9} i n n = 10 x_{9} + 9$ $n = 10 \times 251 + 9 = 2519$
	Commented by $@ty@m last updated on 21/Feb/19

	$I {d i d n}^{'} t d o a n y o f t h e a b o v e$ $c a l c u l a t i o n s .$ $I s i m p l y r e m e m b e r t h e f o r m u l a$ $g i v e n i n A r i t h m e t i c b o o k b y$ $R . S . A g a r w a l$ $m e a n t f o r p r e p a r a t i o n o f c o m p e t e t i v e$ $e x a m i n a t i o n .$ ${Y o u}^{'} d f i n d m a n y s u c h u s e f u l$ $f o r m u l a e (o r t r i c k s) t h e r e .$
	Commented by Kunal12588 last updated on 22/Feb/19

	$g r e a t s i r . t h a n k s o n c e m o r e$
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