Menu Close

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

Tinku Tara 0 Comments
Pin




Question Number 43540 by abdo.msup.com last updated on 11/Sep/18
prove that 111 divide 10^(6n+2) +10^(3n+1) +1
$${prove}\:{that}\:\mathrm{111}\:{divide}\:\mathrm{10}^{\mathrm{6}{n}+\mathrm{2}} \:+\mathrm{10}^{\mathrm{3}{n}+\mathrm{1}} \:+\mathrm{1} \\ $$
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
$$\mathrm{10}^{\mathrm{6}{n}} ×\mathrm{100}+\mathrm{10}^{\mathrm{3}{n}} ×\mathrm{10}+\mathrm{1} \\ $$$$\left(\mathrm{10}^{\mathrm{3}} \right)^{\mathrm{2}{n}} ×\mathrm{100}+\left(\mathrm{10}^{\mathrm{3}} \right)^{{n}} ×\mathrm{10}+\mathrm{1} \\ $$$$\left(\mathrm{999}+\mathrm{1}\right)^{\mathrm{2}{n}} ×\mathrm{100}+\left(\mathrm{999}+\mathrm{1}\right)^{{n}} ×\mathrm{10}+\mathrm{1} \\ $$$$\mathrm{100}×\left\{{f}\left(\mathrm{999}\right)+\mathrm{1}\right\}+\mathrm{10}×\left\{{g}\left(\mathrm{999}\right)+\mathrm{1}\right\}+\mathrm{1} \\ $$$$=\mathrm{100}×{f}\left(\mathrm{999}\right)+\mathrm{10}×{g}\left(\mathrm{999}\right)+\mathrm{100}+\mathrm{10}+\mathrm{1} \\ $$$$\mathrm{999}\:=\mathrm{111}×\mathrm{9} \\ $$$${terms}\:{containing}\:\mathrm{999}\:{are}\:{divisble}\:{by}\:\mathrm{111}\:{and} \\ $$$$\mathrm{100}+\mathrm{10}+\mathrm{1}=\mathrm{111}\:{is}\:{divisble}\:{by}\:\mathrm{111} \\ $$$$ \\ $$
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
$$\mathrm{10}^{\mathrm{6}{n}} ×\mathrm{100}+\mathrm{10}^{\mathrm{3}{n}} ×\mathrm{10}+\mathrm{1} \\ $$$$\left(\mathrm{10}^{\mathrm{3}} \right)^{\mathrm{2}{n}} ×\mathrm{100}+\left(\mathrm{10}^{\mathrm{3}} \right)^{{n}} ×\mathrm{10}+\mathrm{1} \\ $$$$\left(\mathrm{999}+\mathrm{1}\right)^{\mathrm{2}{n}} ×\mathrm{100}+\left(\mathrm{999}+\mathrm{1}\right)^{{n}} ×\mathrm{10}+\mathrm{1} \\ $$$$\mathrm{100}×\left\{{f}\left(\mathrm{999}\right)+\mathrm{1}\right\}+\mathrm{10}×\left\{{g}\left(\mathrm{999}\right)+\mathrm{1}\right\}+\mathrm{1} \\ $$$$=\mathrm{100}×{f}\left(\mathrm{999}\right)+\mathrm{10}×{g}\left(\mathrm{999}\right)+\mathrm{100}+\mathrm{10}+\mathrm{1} \\ $$$$\mathrm{999}\:=\mathrm{111}×\mathrm{9} \\ $$$${terms}\:{containing}\:\mathrm{999}\:{are}\:{divisble}\:{by}\:\mathrm{111}\:{and} \\ $$$$\mathrm{100}+\mathrm{10}+\mathrm{1}=\mathrm{111}\:{is}\:{divisble}\:{by}\:\mathrm{111} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *