$18 x \equiv 43 \pmod {23 }$.

Question

Slightly similar to a question I submitted a few minutes ago, however this time my value for $x$ is $-602$, and I'm not sure how to write the general solution for $x$ .. help please?

Help

score 2 · Answer 1 · answered Feb 08 '14 at 20:58

2

There is a unique solution to this equation modulo 23, so therefore, in general,

$$x\equiv 19\pmod{23}$$

answered Feb 08 '14 at 20:58

Mayank Pandey

1,967

why 19? what happend to the 43 and the 18 if you don't mind me asking a stupid question – user122661 Feb 08 '14 at 21:04
1

You found that -602 was a solution to this equation. Therefore, all $x$ that are -602 $\pmod{23}$ satisfy $18x\equiv 43\pmod{23}$. The 19 comes from the fact that $-602\equiv 19\pmod{23}$ – Mayank Pandey Feb 08 '14 at 23:05

score 1 · Answer 2 · answered Feb 08 '14 at 22:46

First, let's simplify the equivalence:

$18x\equiv43$ mod 23

$18x\equiv20$ mod 23

Now, find the multiplicative inverse of 18 mod 23 which is 9 as $18 * 9 \equiv 162 \equiv 23*7 + 1 \equiv 1$ (This can be done by brute force if necessarily as 23 is a prime so the non-zero values do have multiplicative inverses with 1 as the identity.

$9*18x\equiv9*20$ mod 23

$x \equiv 180$ mod 23

$x \equiv 19$ mod 23 as 23*7+19 = 180

score 0 · Answer 3 · edited Apr 13 '17 at 12:20

0

To get other solutions, the situation here is even simpler than my answer to your previous question: If $$ x_1\equiv x_2\pmod{23} $$ then $$ 23\mid18(x_1-x_2) $$ and therefore $$ 18x_1\equiv18x_2\pmod{23} $$ we don't even need to worry about $\gcd(18,23)$ for the preceding, though if we do consider it, we can get all solutions.

edited Apr 13 '17 at 12:20

Community

1

answered Feb 08 '14 at 21:04

robjohn

345,667

How exactly would I go about finding the general solution. I've found the unique values for s and t in the formula 18s +23t = 43 [s = 473 and t=-602] however I'm not sure how to figure out the values for x (in general) that will yield the R.S. – user122661 Feb 08 '14 at 21:10
@user122661: If the divisor ($18$ here) and the modulus ($23$ here) are relatively prime, then the solutions will all differ by multiples of the modulus. In general, the solutions will all differ by multiples of the modulus divided by the gcd of the modulus and the divisor. – robjohn Feb 09 '14 at 00:43

score 0 · Answer 4 · answered Feb 08 '14 at 23:08

Try this chain (all mod $23$, which is prime)

$$18x\equiv 43$$ $$18x\equiv 66$$ $$3x\equiv 11$$ $(11+23=34, 34+23=57)$ $$3x\equiv 57$$ $$x\equiv 19$$

Knocking out prime factors of $2$ and $3$ in this way is easy, and systematic. I took the easy opportunity to knock out a factor of $6$ (could have started $18x \equiv 20$).

$18 x \equiv 43 \pmod {23 }$.

4 Answers4