Thread: Needing some help with congruences

1. Needing some help with congruences

Solve the following set of linear congruences:
x≡4 (mod 24)
x≡7 (mod 11)

Ok now I that we can say that x=24k+4 and then plug this into the second congruence to get 24k+4=7 (mod 11). My teacher wrote that 2k = 3 = 14 (mod 11)
k = 7 (mod 11); however, I do not see the reasoning behind this at all.

I keep wanting to do solve it like you would a general equation..in my head 24k+4=7 (mod 11) becomes 24k=3 (mod 11) and then you would divide everything by 24 and so on to solve for k. But this is obviously not how the problem is done. Can someone please clearly explain just how exactly the value for k was obtained?

2. Originally Posted by steph3824
Solve the following set of linear congruences:
x≡4 (mod 24)
x≡7 (mod 11)

Ok now I that we can say that x=24k+4 and then plug this into the second congruence to get 24k+4=7 (mod 11). My teacher wrote that 2k = 3 = 14 (mod 11)
k = 7 (mod 11); however, I do not see the reasoning behind this at all.

I keep wanting to do solve it like you would a general equation..in my head 24k+4=7 (mod 11) becomes 24k=3 (mod 11) and then you would divide everything by 24 and so on to solve for k. But this is obviously not how the problem is done. Can someone please clearly explain just how exactly the value for k was obtained?

Google "Chinese Remainder Theorem" and apply the proof here:

Chinese remainder theorem - Wikipedia, the free encyclopedia

Tonio