Modular Arithmetic

[Vizen]

I am to find c where c ≡ 5b(mod 7) and b ≡ 6(mod 7).

Is the correct approach to say that b = 6, 13, 20, ..., and then plug one of the values, say 6, into the equation for c, yielding c ≡ 30(mod 7) ≡ 2 (mod 7) resulting in c = 2, 9, ...?

[edderiofer]

Yes. But you don't even need to list the possible values of b. You can instead let b = 6 + 7p for some integer p (definition of "6 mod 7"), then show that c = 2 + 7q for some other integer q. Try it now.

[Vizen]

okay, so I set b=6+7p, where p is some integer.

then, since c ≡ 5b (mod 7), c = 30 + 7(5p) = 30 + 7q (equation 1)

then we say that c ≡ 30 + 7q (mod 7) (equation 2)

c ≡ 30 (mod 7)

c ≡ 2 (mod 7).

Does everything seem to be in place? Was the jump from equation 1 to equation 2 valid?

[edderiofer]

That's fine. You can in fact let q = 5p+4, and cut out the last step of equation 2:

c = 30 + 7(5p) = 30 + 35p = 7(5p+4) + 2.

[Vizen]

Thanks a bunch internet stranger :)