what is the remainder when (111..)+(222..)+(333..)+....+(77..) is divided by 37


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Question Number 34044 by NECx last updated on 29/Apr/18

$w h a t i s t h e r e m a i n d e r w h e n$ $(111 . .) + (222 . .) + (333 . .) + . . . . + (77 . .)$ $i s d i v i d e d b y 37$
Commented by Rasheed.Sindhi last updated on 30/Apr/18

$N = (111 . .) + (222 . .) + (333 . .) + . . . . + (77 . .)$ $= 1 (111 . . .) + 2 (111 . .) + 3 (111 . . .) . . . + 7 (111 . .)$ $= (1 + 2 + 3 + . . . + 7) (11 . . .)$ $= 28 \times 111 . . .$ $Let 111 . . . is of n - digits number .$ $For n = 1$ $N = 28.1 = 28,$ $If 28 is divided by 37 the remainder is 28$ $. . . . . . .$ $. . . .$ $If n = 3 k, the remainder is 0$ $If n = 3 k + 1, the remainder is 28$ $If n = 3 k + 2, the remainder is 12$ $The remainders : 0, 28, 12$
Commented by NECx last updated on 30/Apr/18

$t h a n k y o u s o m u c h b u d d y$
	Answered by tanmay.chaudhury50@gmail.com last updated on 30/Apr/18

	$(111 . . .) + (222 . . .) + (333 . . .) + . . . . + (777 . . .)$ $(111 . . .) + 2 (111 . . .) + 3 (111 . . .) + . . . + 7 (111 . . .)$ $l e t v a l u e o f (111 . . .) s a y = t$ $t (1 + 2 + 3 + . . . + 7)$ $t (28)$ $n o w l e t u s f i n d t h e v a l u e o f (111 . . . u p t o n) t h e n$ $p u t n \to \infty$ $(111 . . . u p t o n t e r m s)$ $1 + 10 + 10^{2} + 10^{3} + . . . .$ $\frac{(10^{n} - 1)}{10 - 1}$ $28 t$ $(37 - 9) t$ $(37 - 9) \frac{(10^{n} - 1)}{9}$ $\frac{37}{9} (10^{n} - 1) + (1 - 10^{n})$ $w h e n d e v i d e d b y 37 f i r s t p a r t i s d i v i s b l e b y 37$ $r e m a i n i n g i s r e m a i n d e r u n d e r t h e c o n d k t i o n n \to \infty$ $p l s r e c h e c k t h e q u e s t i o n p l s$
	Commented by NECx last updated on 30/Apr/18

	$w o w . . . . l e m m e c h e c k i t . . R e a l l y$ $I l o v e w h a t y o u j u s t d i d . T h a n k s$
	Commented by Rasheed.Sindhi last updated on 30/Apr/18

	$The remainder should be positive, whereas$ $1 - 10^{n} is negative for n > 0$
	Answered by 5a3k last updated on 30/Apr/18

	$[(111 . .) + (222 . .) + (333 . .) + . . . + (777 . .)]$ $\frac{37 [3 + 6 + 9 + . . . + 21]}{37}$ $3 + 6 + 9 + . . . . + 21$ $7 / 2 [2 \times 3 + (7 - 1) 3]$ $84$
	Commented by NECx last updated on 30/Apr/18

	$n o t c l e a r p l e a s e$
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