Question Number 133897 by bemath last updated on 25/Feb/21
![Find the least positive integer that leaves a remainder 3 when divided by 7 , 4 when divided by 9 , and 8 when divided by 11.](https://www.tinkutara.com/question/Q133897.png)
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{least}\:\mathrm{positive}\:\mathrm{integer} \\ $$$$\mathrm{that}\:\mathrm{leaves}\:\mathrm{a}\:\mathrm{remainder}\:\mathrm{3}\:\mathrm{when}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{7} \\ $$$$,\:\mathrm{4}\:\mathrm{when}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{9}\:,\:\mathrm{and}\:\mathrm{8}\:\mathrm{when} \\ $$$$\mathrm{divided}\:\mathrm{by}\:\mathrm{11}. \\ $$
Answered by john_santu last updated on 25/Feb/21
![by Use Chinese Remainder theorem determinant ((i,n_i ,N_i ,(y_i N_i ≡1 mod n_1 ),c_i ,(N_i y_i c_i )),(1,7,(9.11),(y_1 .9.11≡1 mod 7→y_1 =1),3,(297)),(2,9,(7.11),(y_2 .7.11≡1 mod 9→y_2 =2),4,(616)),(3,(11),(7.9),(y_3 .7.9≡1 mod 11→y_3 =7),8,(3528))) X = Σ_i N_i y_i c_i = 4441 we have x≡X mod N ⇒x≡ 4441 (mod 693) x≡ 283 (mod 693) or x = 693k + 283 ; k∈Z then the least positive integer is x = 283](https://www.tinkutara.com/question/Q133899.png)
$${by}\:{Use}\:{Chinese}\:{Remainder}\:{theorem}\: \\ $$$$\begin{array}{|c|c|c|c|}{{i}}&\hline{{n}_{{i}} }&\hline{{N}_{{i}} }&\hline{{y}_{{i}} {N}_{{i}} \equiv\mathrm{1}\:{mod}\:{n}_{\mathrm{1}} }&\hline{{c}_{{i}} }&\hline{{N}_{{i}} {y}_{{i}} {c}_{{i}} }\\{\mathrm{1}}&\hline{\mathrm{7}}&\hline{\mathrm{9}.\mathrm{11}}&\hline{{y}_{\mathrm{1}} .\mathrm{9}.\mathrm{11}\equiv\mathrm{1}\:{mod}\:\mathrm{7}\rightarrow{y}_{\mathrm{1}} =\mathrm{1}}&\hline{\mathrm{3}}&\hline{\mathrm{297}}\\{\mathrm{2}}&\hline{\mathrm{9}}&\hline{\mathrm{7}.\mathrm{11}}&\hline{{y}_{\mathrm{2}} .\mathrm{7}.\mathrm{11}\equiv\mathrm{1}\:{mod}\:\mathrm{9}\rightarrow{y}_{\mathrm{2}} =\mathrm{2}}&\hline{\mathrm{4}}&\hline{\mathrm{616}}\\{\mathrm{3}}&\hline{\mathrm{11}}&\hline{\mathrm{7}.\mathrm{9}}&\hline{{y}_{\mathrm{3}} .\mathrm{7}.\mathrm{9}\equiv\mathrm{1}\:{mod}\:\mathrm{11}\rightarrow{y}_{\mathrm{3}} =\mathrm{7}}&\hline{\mathrm{8}}&\hline{\mathrm{3528}}\\\hline\end{array} \\ $$$$\:{X}\:=\:\sum_{{i}} \:{N}_{{i}} {y}_{{i}} {c}_{{i}} \:=\:\mathrm{4441} \\ $$$${we}\:{have}\:{x}\equiv{X}\:{mod}\:{N}\:\Rightarrow{x}\equiv\:\mathrm{4441}\:\left({mod}\:\mathrm{693}\right) \\ $$$${x}\equiv\:\mathrm{283}\:\left({mod}\:\mathrm{693}\right)\:{or}\:{x}\:=\:\mathrm{693}{k}\:+\:\mathrm{283}\:;\:{k}\in\mathbb{Z} \\ $$$${then}\:{the}\:{least}\:{positive}\:{integer}\:{is}\:{x}\:=\:\mathrm{283}\: \\ $$
Commented by talminator2856791 last updated on 25/Feb/21
![why is it called chinese remainder theorem?](https://www.tinkutara.com/question/Q133922.png)
$$\:\mathrm{why}\:\mathrm{is}\:\mathrm{it}\:\mathrm{called}\:\mathrm{chinese}\:\mathrm{remainder}\:\mathrm{theorem}? \\ $$
Answered by talminator2856791 last updated on 25/Feb/21
![7a+3 = 9b+4 = 11c+8 7a = 9b+1 b = 7k−4 9b+4 = 11c+8 9b = 11c+4 c = 9j−2 9(7k−4) = 11(9j−2)+4 63k−36 = 99j−18 63k = 99j+18 7k = 11j+2 j = 7d−4 9(7k−4) = 11(9(3)−2)+4 63k−36 = 279 63k = 315 k = 5 b = 7k−4 b = 31 9b+4 = 9(31)+4 = 283](https://www.tinkutara.com/question/Q133908.png)
$$\:\mathrm{7}{a}+\mathrm{3}\:=\:\mathrm{9}{b}+\mathrm{4}\:=\:\mathrm{11}{c}+\mathrm{8} \\ $$$$\:\mathrm{7}{a}\:=\:\mathrm{9}{b}+\mathrm{1} \\ $$$$\:{b}\:=\:\mathrm{7}{k}−\mathrm{4} \\ $$$$\: \\ $$$$\:\mathrm{9}{b}+\mathrm{4}\:=\:\mathrm{11}{c}+\mathrm{8} \\ $$$$\:\mathrm{9}{b}\:=\:\mathrm{11}{c}+\mathrm{4} \\ $$$$\:{c}\:=\:\mathrm{9}{j}−\mathrm{2} \\ $$$$\: \\ $$$$\:\mathrm{9}\left(\mathrm{7}{k}−\mathrm{4}\right)\:=\:\mathrm{11}\left(\mathrm{9}{j}−\mathrm{2}\right)+\mathrm{4} \\ $$$$\:\mathrm{63}{k}−\mathrm{36}\:=\:\mathrm{99}{j}−\mathrm{18} \\ $$$$\:\mathrm{63}{k}\:=\:\mathrm{99}{j}+\mathrm{18} \\ $$$$\:\mathrm{7}{k}\:=\:\mathrm{11}{j}+\mathrm{2} \\ $$$$\:{j}\:=\:\mathrm{7}{d}−\mathrm{4} \\ $$$$\: \\ $$$$\:\mathrm{9}\left(\mathrm{7}{k}−\mathrm{4}\right)\:=\:\mathrm{11}\left(\mathrm{9}\left(\mathrm{3}\right)−\mathrm{2}\right)+\mathrm{4} \\ $$$$\:\mathrm{63}{k}−\mathrm{36}\:=\:\mathrm{279} \\ $$$$\:\mathrm{63}{k}\:=\:\mathrm{315} \\ $$$$\:{k}\:=\:\mathrm{5} \\ $$$$\: \\ $$$$\:{b}\:=\:\mathrm{7}{k}−\mathrm{4} \\ $$$$\:{b}\:=\:\mathrm{31} \\ $$$$\: \\ $$$$\:\mathrm{9}{b}+\mathrm{4}\:=\:\mathrm{9}\left(\mathrm{31}\right)+\mathrm{4}\: \\ $$$$\:=\:\mathrm{283} \\ $$$$\: \\ $$