prove that 111 divide 10^(6n+2) +10^(3n+1) +1


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Question Number 43540 by abdo.msup.com last updated on 11/Sep/18

$p r o v e t h a t 111 d i v i d e 10^{6 n + 2} + 10^{3 n + 1} + 1$
	Answered by tanmay.chaudhury50@gmail.com last updated on 12/Sep/18
	$10^(6n) ×100+10^(3n) ×10+1 (10^3 )^(2n) ×100+(10^3 )^n ×10+1 (999+1)^(2n) ×100+(999+1)^n ×10+1 100×{f(999)+1}+10×{g(999)+1}+1 =100×f(999)+10×g(999)+100+10+1 999 =111×9 terms containing 999 are divisble by 111 and 100+10+1=111 is divisble by 111$
	$10^{6 n} \times 100 + 10^{3 n} \times 10 + 1$ ${(10^{3})}^{2 n} \times 100 + {(10^{3})}^{n} \times 10 + 1$ ${(999 + 1)}^{2 n} \times 100 + {(999 + 1)}^{n} \times 10 + 1$ $100 \times {f (999) + 1} + 10 \times {g (999) + 1} + 1$ $= 100 \times f (999) + 10 \times g (999) + 100 + 10 + 1$ $999 = 111 \times 9$ $t e r m s c o n t a i n i n g 999 a r e d i v i s b l e b y 111 a n d$ $100 + 10 + 1 = 111 i s d i v i s b l e b y 111$
	Answered by tanmay.chaudhury50@gmail.com last updated on 12/Sep/18
	$10^(6n) ×100+10^(3n) ×10+1 (10^3 )^(2n) ×100+(10^3 )^n ×10+1 (999+1)^(2n) ×100+(999+1)^n ×10+1 100×{f(999)+1}+10×{g(999)+1}+1 =100×f(999)+10×g(999)+100+10+1 999 =111×9 terms containing 999 are divisble by 111 and 100+10+1=111 is divisble by 111$
	$10^{6 n} \times 100 + 10^{3 n} \times 10 + 1$ ${(10^{3})}^{2 n} \times 100 + {(10^{3})}^{n} \times 10 + 1$ ${(999 + 1)}^{2 n} \times 100 + {(999 + 1)}^{n} \times 10 + 1$ $100 \times {f (999) + 1} + 10 \times {g (999) + 1} + 1$ $= 100 \times f (999) + 10 \times g (999) + 100 + 10 + 1$ $999 = 111 \times 9$ $t e r m s c o n t a i n i n g 999 a r e d i v i s b l e b y 111 a n d$ $100 + 10 + 1 = 111 i s d i v i s b l e b y 111$
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