With x---3(mod11) and x---4(mod13), how can I conclude that x---69(mod143)?

Question

I know 143 is [11,13] but don't know how to get 69. (I don't know how to type the triple symbol).

Answer 1

x≡3(mod11) ---> x-3 divides by 11 or x-3=11 m, 

x≡4(mod13) ---> x-4 divides by 13 or x-4=13 n,  m,n are Intgers

(x-3)=11m (multiply by 13)

(x-4)=13n  (multiply by 11)

13(x-3)=11*13m ---> 13x-3*13= 11*13 m --> 13x ≡39(mod 143) (A)

11(x-4)=11*13n  ---> 11x-4*11= 11*13 n ---> 11x ≡44(mod 143) (B)

If a1≡b1 (mod n) and  a2≡b2( mod n), then a1+a2= b1+b2 (mod n)

From (A) and (B):

24x≡83 mod(143)

 

Next, if a≡b (mod m) and b≡c (mod m), then a≡c (mod m)

83≡-60 (mod 143) -->24x≡-60 ( mod 143) 

[Here we use this property: if a≡b ( mod m) and k is an arbitrary integer, then ka≡kb (mod m) ]

2x≡-5 (mod143) and again -5≡138 (mod 143),

Therefore, 2x≡138 (mod 143) -->x≡69 ( mod 143).

 

Of course,  there is a shorter way to show this, but I thought this is more clear.

 

A short solution is this:

 

x≡3(mod11) <---- (1)
x≡4(mod13) <----(2)

From (2) x=4+13t <--- (3)  (t integer) and replace x in RHS (1), that is,

4+13t=3mod(11)

2t=-1(mod 11)≡10 (mod 11)  [-1≡10(mod11)]

t=5 (mod 11) --> t=5+11y, (y another integer) and substitute t in (3) 

x=4+13t=4+13(5+11y)=69+143y

So we have x=69+143y. This means that x≡69( mod 143).