prove-that-111-divide-10-6n-2-10-3n-1-1-

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^{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^{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