Question Number 139737 by mathsuji last updated on 30/Apr/21
$${Find}\:{the}\:{value}\:{of}\:{X}\:{such}\:{that}: \\ $$$$\mathrm{1}\centerdot\mathrm{5}\centerdot\mathrm{9}\centerdot\mathrm{13}\centerdot\mathrm{17}\centerdot...\centerdot\mathrm{2021}\equiv{X}\left({mod}\mathrm{1000}\right) \\ $$
Answered by mindispower last updated on 01/May/21
![A=Π_(k=0) ^(505) (1+4k)≡X[1000] 1000=8.5^3 k=2n⇒1+4k≡1[8] k=2n+1⇒1+4k≡5[8] Π_(k=0) ^(505) (1+4k)≡Π_(k=0) ^(252) (8k+1).Π_(k=0) ^(252) (5+8k)≡Π_(k=0) ^(252) 5[8] =5^(252) [8] 5^2 =1[8]⇒5^(252) ≡1[8] ⇒A≡1[8] 1+4k=5s⇒5s−4k=1 (s,k)=(1,1) solution ⇒5(s−1)−4(k−1)=0⇒k=5l+1 ⇒1+4(5l+1)=5(4l+1)=1+4k l=0,l=1,l=2⇒5,45,25∈{1,5,9,13,17^ .....} ⇒(1.5.9......2021)≡0[125] A≡0[125] A≡1[8] A=8k+1=125d 125d−8k=1 (d,k)=(5,78) solution k=125m+78 A≡8(125m+78)+1=1000m+625[1000]≡X[100] X=625](Q139813.png)
$${A}=\underset{{k}=\mathrm{0}} {\overset{\mathrm{505}} {\prod}}\left(\mathrm{1}+\mathrm{4}{k}\right)\equiv{X}\left[\mathrm{1000}\right] \\ $$$$\mathrm{1000}=\mathrm{8}.\mathrm{5}^{\mathrm{3}} \\ $$$${k}=\mathrm{2}{n}\Rightarrow\mathrm{1}+\mathrm{4}{k}\equiv\mathrm{1}\left[\mathrm{8}\right] \\ $$$${k}=\mathrm{2}{n}+\mathrm{1}\Rightarrow\mathrm{1}+\mathrm{4}{k}\equiv\mathrm{5}\left[\mathrm{8}\right] \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\mathrm{505}} {\prod}}\left(\mathrm{1}+\mathrm{4}{k}\right)\equiv\underset{{k}=\mathrm{0}} {\overset{\mathrm{252}} {\prod}}\left(\mathrm{8}{k}+\mathrm{1}\right).\underset{{k}=\mathrm{0}} {\overset{\mathrm{252}} {\prod}}\left(\mathrm{5}+\mathrm{8}{k}\right)\equiv\underset{{k}=\mathrm{0}} {\overset{\mathrm{252}} {\prod}}\mathrm{5}\left[\mathrm{8}\right] \\ $$$$=\mathrm{5}^{\mathrm{252}} \left[\mathrm{8}\right] \\ $$$$\mathrm{5}^{\mathrm{2}} =\mathrm{1}\left[\mathrm{8}\right]\Rightarrow\mathrm{5}^{\mathrm{252}} \equiv\mathrm{1}\left[\mathrm{8}\right] \\ $$$$\Rightarrow{A}\equiv\mathrm{1}\left[\mathrm{8}\right] \\ $$$$\mathrm{1}+\mathrm{4}{k}=\mathrm{5}{s}\Rightarrow\mathrm{5}{s}−\mathrm{4}{k}=\mathrm{1} \\ $$$$\left({s},{k}\right)=\left(\mathrm{1},\mathrm{1}\right)\:{solution} \\ $$$$\Rightarrow\mathrm{5}\left({s}−\mathrm{1}\right)−\mathrm{4}\left({k}−\mathrm{1}\right)=\mathrm{0}\Rightarrow{k}=\mathrm{5}{l}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{1}+\mathrm{4}\left(\mathrm{5}{l}+\mathrm{1}\right)=\mathrm{5}\left(\mathrm{4}{l}+\mathrm{1}\right)=\mathrm{1}+\mathrm{4}{k} \\ $$$${l}=\mathrm{0},{l}=\mathrm{1},{l}=\mathrm{2}\Rightarrow\mathrm{5},\mathrm{45},\mathrm{25}\in\left\{\mathrm{1},\mathrm{5},\mathrm{9},\mathrm{13},\mathrm{1}\bar {\mathrm{7}}.....\right\} \\ $$$$\Rightarrow\left(\mathrm{1}.\mathrm{5}.\mathrm{9}......\mathrm{2021}\right)\equiv\mathrm{0}\left[\mathrm{125}\right] \\ $$$${A}\equiv\mathrm{0}\left[\mathrm{125}\right] \\ $$$${A}\equiv\mathrm{1}\left[\mathrm{8}\right] \\ $$$${A}=\mathrm{8}{k}+\mathrm{1}=\mathrm{125}{d} \\ $$$$\mathrm{125}{d}−\mathrm{8}{k}=\mathrm{1} \\ $$$$\left({d},{k}\right)=\left(\mathrm{5},\mathrm{78}\right)\:{solution} \\ $$$${k}=\mathrm{125}{m}+\mathrm{78} \\ $$$${A}\equiv\mathrm{8}\left(\mathrm{125}{m}+\mathrm{78}\right)+\mathrm{1}=\mathrm{1000}{m}+\mathrm{625}\left[\mathrm{1000}\right]\equiv{X}\left[\mathrm{100}\right] \\ $$$${X}=\mathrm{625} \\ $$$$ \\ $$$$ \\ $$
Commented by mathsuji last updated on 02/May/21
$${thank}\:{you}\:{very}\:{much}\:{sir} \\ $$