find the smallest number which when we divide by 12,15,18, and 27 leaves a remainder of 8,11,14 and 23 respectively?


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Question Number 136040 by zakirullah last updated on 18/Mar/21

$f i n d t h e s m a l l e s t n u m b e r w h i c h w h e n w e d i v i d e b y 12, 15, 18, and 27 leaves a remainder$ $of 8, 11, 14 and 23 respectively ?$
	Answered by mr W last updated on 18/Mar/21

	$N = 27 a + 23 = 27 a + 9 + 14$ $27 a + 9 = 18 b \Rightarrow 3 a + 1 = 2 b \Rightarrow a = 2 c + 1$ $N = 27 (2 c + 1) + 23 = 54 c + 39 + 11$ $54 c + 39 = 15 d \Rightarrow 18 c + 13 = 5 d \Rightarrow c = 5 e - 1$ $N = 54 (5 e - 1) + 50 = 270 e - 12 + 8$ $270 e - 12 = 12 f \Rightarrow 45 e - 2 = 2 f \Rightarrow e = 2 n$ $N = 270 \times 2 n - 4 = 540 n - 4$ $N_{m i n} = 536$
	Answered by Rasheed.Sindhi last updated on 19/Mar/21

	$T h i s i s a s p e c i a l c a s e i n w h i c h d i f f e r e n c e$ $b e t w e e n a d i v i s o r a n d i t s c o r r e s p o n d i n g$ $r e m a i n d e r i s s a m e .$ $12 - 8 = 15 - 11 = 18 - 14 = 27 - 23 = 4$ $I n t h i s c a s e :$ $R e q u i r e d n u m b e r = L C M (d i v i s o r s) - c o m m o n d i f f e r e n c e$ $= L C M (12, 15, 18, 27) - 4 = 540 - 4 = 536$
	Commented by mr W last updated on 19/Mar/21

	$g r e a t s i r!$
	Commented by Rasheed.Sindhi last updated on 19/Mar/21

	$TH \forall NKS s s S i r!$ $P l e a s e s e e m y a n s w e r t o$ $You can't use 'macro parameter character #' in math mode$ $& g i v e c r i t i c a l r e m a r k .$
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