Where did the six come from in the Chinese Remainder Theorem?

[chaotixmonjuish]

I need help making sense of my notes:

x congruent 4 mod 11
x congruent 3 mod 13

ai mi Mi yi aiMiyi
4 11 13 6 4*13*6
3 13 11 6 3*11*6

I'm not sure where the six came from

 

[lurflurf]

no clue
usually one solves
13a=4 (mod 11) (a=0,1,...10)
11b=3 (mod 13) (b=0,1,...12)
ie a and be are found by modular division
so that
n*gcd(11,13)+(gcd(11,13)-13a-11b)
with n=1,2,3,...
solves the original problem

 

[robert Ihnot]

Chaotixmonjuist:
ai mi Mi yi aiMiyi
4 11 13 6 4*13*6
3 13 11 6 3*11*6 I'm not sure where the six came from

Some number repeats are going on there. But we have 1/11==6 Mod 13, and 1/13==6 Mod 11.

 

[robert Ihnot]

I was hoping someone would go a little further with this problem. It may not be clear what is being discussed. (Apparently the teacher must have been writing on the blackboard.)

What I take is being said above is X==4 Mod 11 and X==3 Mod 13. Find X congruent to 11x13= 143.

Setting up the first part of the problem, we want to find 4/13 ==X(1) Mod 11, and 3/11==X(2) Mod 13.

The first case gives x(1)==24==2 Mod 11, and the second gives X(2) ==18==5 Mod 13. So we get x(1)=2, X(2)=5.

We then set up the equation X =13*x(1) + 11*x(2) +k143. (k is chosen so as to give us the smallest positive value.)

Now if we checked this product Mod 11 the last part on x(2) would go out. On the other hand on the first part, 13*x(1), we have included the inverse of 13 mod 11 in our calcuations, so we are only going to get the value of X ==4 Mod 11. Similarly for the other part of the equation.

So that working out the equation: X=13*2 + 11*5 = 81. This value satisfies the conditions.