Find all integers that leave a remainder of 3 when divided by 5 , a remainder of 5 when divided by 7 , and a remainder of 7 when divided by 11.
Let x be such an integer. Since x leaves a remainder of 3 when divided by 5, x= 5n+ 3 for some integer 3. That is the same as the Diophantine equation x- 5n= 3. It is obvious that 8- 5= 3 so x= 8, n= 1 is a solution. In fact, x= 8+ 5k, n= 1+ k is the general solution since then x- 5n= (8+ 5k)- 5(1+ k)= 8- 5+ 5k- 5k= 3.
Since x leaves a remainder of 5 when divided by 7, x= 7m+ 5. Since x= 8+ 5k, that is 8+ 5k= 7m+ 5 which is the same as the Diophantine equation 7m- 5k= 3, 5 divides into 7 once with remainder 2 and 2 divides into 5 twice with remainder 1: 5- 2(2)= 1; 5- 2(7- 5)= 3(5)- 2(7)= (-2)(7)- (-3)(5)= 1. Multiplying by 3, (-5)(7)- (-9)(5)= 3. m= -5 and k= -9 is a solution and the general solution is m= -5+ 5j, k= -9+ 7j. Since x= 8+ 5k, x= 8+ 5(-9+ 7j)= -37+ 35j. Notice that if we take j= 2, x= -37+ 70= 33 which has remainder 3 when divided by 5 and remainder 5 when divided by 7. Since 35= 5(7) has remainder 0 when divided by either, x= -37+ 35j has remainder 3 when divided by 5 and 5 when divided by 7 for all j.
Since x leaves a remainder of 7 when divided by 11, x= 11p+ 7. Since x= -37+ 35j, we have -37+ 35j= 11p+ 7 which is the same as the Diophantine equation 35j- 11p= 44. 11 divides into 35 three times with remainder 2 and 2 divides into 11 5 times with remainder 1: 11- 5(2)= 1; 11- 5(35- 3(11))= (-5)(35)- (-16)(11)= 1. Multiplying by 44, (-220)(35)- (-704)(11)= 44. The general solution to the Diophantine equation is j= -220+ 11i, p= -704+35i.
Since x= 11p+ 7, we have x= 11(-704+ 35i)+ 7= -7737+ 385i. If we take i= 21, we get x= -7737+ 8085= 348. 5 divides 348 69 times with a remainder or 3, 7 divides 348 49 times with a remainder of 5, and 11 divides 348 31 times with a remainder of 7. since 385= 5(7)(11) has remainder 0 when divided by 5, 7, or 11, x= -7737+ 385i will have those remainders for all i.