Solving congruences in Z

Question

I've been stuck on this question for quite a bit of time. I have tried using euclids algorithm, solving for x, just about everything. Hints appreciated.

$$19x\equiv 341\;(\text{mod}\;401)$$

I have used Euclid's Algorithm using gcd(19,401) and then applying the extended algorithm to find 19m and 401k but it does not give me the correct solution set according to my answer book

What have you achieved so far? You need to apply the extended Euclidean algorithm. — Wuestenfux, May 31 '19 at 07:31
@discombobulator If you show us the attempt detailed (what did you get for the inverse), we can find the error. — Peter, May 31 '19 at 07:52
@peter I found the inverse of 19 mod 401 to be 190, and I do not know how to proceed from here. — discombobulator, May 31 '19 at 07:58
All you have to do then is to multiply the equation with this inverse — Peter, May 31 '19 at 07:59
What was the solution set you got, and what does the answer say? — Mark Bennet, May 31 '19 at 08:00
@Peter how would I do that? Just multiply 190 in the original equation? — discombobulator, May 31 '19 at 08:03
@Mark Bennet I have not been able to find a solution set and the answer says [229]subscript 401. — discombobulator, May 31 '19 at 08:03
Yes, then on the left side $19$ cancels out and on the right side you get the result by calculating $190\cdot 341\mod 401$ — Peter, May 31 '19 at 08:04
@Peter I think I'm missing something here. I still dont get the correct solution. I multiplied 190 on both sides to get 190(19) and 190(341) mod 401. — discombobulator, May 31 '19 at 08:13
@Peter it did. I was inputting the modulo calculation in my calculator incorrect the whole time. — discombobulator, May 31 '19 at 08:38
@discombobulator In this case, you can delete the question now (despite of the answers you received) , if you want. — Peter, May 31 '19 at 08:40
It can be done with simple mental arithmetic if you use Gauss's algorithm - see my answer. — Bill Dubuque, May 31 '19 at 15:19

Wuestenfux · Accepted Answer · 2019-05-31T08:20:21.280

1

Well, the extended Euclidean algorithm gives $$190\cdot 19 + (-9)\cdot 401 = 1.$$ Now you see that $$190\equiv 19^{-1}\mod 401$$ and so $$x \equiv 19^{-1}\cdot 341\mod 401,$$ i.e., $$x\equiv 190\cdot 341\mod 401.$$ Thus $$x\equiv 229\mod 401.$$

edited May 31 '19 at 08:20

answered May 31 '19 at 08:07

Wuestenfux

20,964

Yes, thats exactly what I have. How do I proceed from there? – discombobulator May 31 '19 at 08:14

Bill Dubuque · Answer 2 · 2019-05-31T19:20:00.900

$\!\bmod 401\!:\,\ x\equiv \dfrac{341}{19}\equiv \dfrac{-60}{19}\equiv \dfrac{-60\cdot 21}{\ \ \ \underbrace{19\cdot 21}_{\large -2\ \ \ \ }}\equiv 30\cdot 21\equiv 229\ $ by Gauss's algorithm

Remark $\,\bmod a^2\!+\!1\!:\,\ \dfrac{1}{a-1}\equiv \dfrac{a+1}{-2}\ $ by $\,(a\!-\!1)(a\!+\!1)\equiv -2\,$ as above (where $\, n = 20).$

This is a prototypical "easy inverse" where Gauss's or Euclid's algorithm takes only $1$ or $2$ steps because we are inverting a factor of modulus$\pm k\,$ for very small $k$.

Solving congruences in Z

2 Answers2