1
$\begingroup$

Use the Euclidean algorthim to find a solution to $x$ (in terms of $r$ and $s$) if $$x\equiv r \mod(28) \;\;\text{and}\;\;\; x\equiv s \mod(45).$$

  • 0
    What have you tried so far? What problems have you been having with working this out? – 2017-02-09
  • 0
    The question i originally posted (before someone modified it) had equality signs, not equivalence signs.. I'm not sure if those two things are the same? I believe the original question I had meant, x=r in mod(28) and x=s in mod(45) – 2017-02-09
  • 0
    they are the same in this context – 2017-02-09
  • 0
    Why are they the same in this context if you don't mind expanding on this? – 2017-02-09
  • 0
    Basically it comes from how you are treating "mod". The more correct way is to regrad "mod" as an equivalence class on $\mathbb{Z}$. In this reading "1" is equivalent to "3" in mod(2). There is also the slightly more sloppy way of considering "mod(x)" to be the act of applying the remainder operator for x to all the values and then comparing for equality this is why sometimes you see = but it is less correct. – 2017-02-09
  • 0
    oh that makes sense. I also prefer the equivalence sign or equality. – 2017-02-09

1 Answers 1

1

Since $\gcd(28,45)=1$, it follows that there exist $m,n \in \Bbb Z$ such that $28m + 45n = 1$. Consider then $x = 28mr + 45ns$. Then

$$x = 28mr + 45ns = (1-45n)r + 45ns = r + 45(-nr+ns) \equiv r \pmod{45}$$

and

$$x = 28mr + 45ns = 28mr + (1-28m)s = s + 28(mr-ms) \equiv s \pmod{28} .$$

This solution is not unique because $x + 28 \cdot 45 k$ will also be a solution for every $k \in \Bbb Z$.

To concretely find $m$ and $n$ one usually uses the extended Euclidian algorithm: it will give $m = -8$ and $n=5$, therefore the solutions are $x = -224r + 225s + 1260k$ with $k \in \Bbb Z$ arbitrary.

  • 0
    how did you come up with the equation x=28mr+45ns, and how are you manipulating it to its final form? – 2017-02-13
  • 0
    @Nicole: Your problem is a particular case of the Chinese remainder theorem, and my answer to it is a particular case of its usual [constructive proof](https://en.wikipedia.org/wiki/Chinese_remainder_theorem#Existence_.28constructive_proof.29). – 2017-02-13
  • 0
    Yes! I ended up figuring it out. Thank you very much – 2017-02-13