$6x + 13 = 7 \pmod{24}$

Question

What method can I use to solve this problem?

I tried with the method I used here but it won't work because I can't use Euclidean algorithm on this problem.

There are $4$ values to check $x=0,1,2,3$ after that it repeats. — kingW3, Apr 25 '19 at 09:35
This is same as $6x\equiv-6\pmod{24}$. It might help to recall some general theory of solving $ax\equiv b\pmod{m}$. Let $d=\gcd(a,m)$. Recall that $ax\equiv b\pmod{m}$ has a solution iff $d\mid b$. If $d\mid b$, there are $d$ distinct solutions mod $m$, and a unique solution mod $m/d$. First solve $\frac{a}{d}x\equiv \frac{b}{d}\pmod{\frac{m}{d}}$ for $x$ (you should at least be able to do this as $\gcd\left(\frac{a}{d}, \frac{m}{d}\right)=1$). Then if $x_0$ is a solution to this, the $d$ distinct solutions to the original congruence are ($x_0 + k\frac{m}{d}$) mod $m$, for $k=0,1,\ldots, d-1$. — Minus One-Twelfth, Apr 25 '19 at 09:36
To write $6x + 13 \equiv 7 \pmod{24}$, type $6x + 13 \equiv 7 \pmod{24}$ . Please [edit] your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site. — N. F. Taussig, Apr 26 '19 at 13:28

score 0 · Answer 1 · answered Apr 25 '19 at 09:12

0

Hint: $6x+13\equiv7\pmod{24}\iff6x\equiv-6\pmod{6\times4}$.

answered Apr 25 '19 at 09:12

score 0 · Answer 2 · answered Apr 25 '19 at 18:47

0

$x+2=4y+1$ by direct division by 6 when turned to linear polynomial form. Therefore, $x\equiv 3\bmod 4$

answered Apr 25 '19 at 18:47

score 0 · Answer 3 · answered Jul 01 '19 at 21:09

0

$$6x+13\equiv7\pmod{24}\iff 6x+6\equiv0\pmod{24}\iff 24|6x+6\iff 4|x+1$$

answered Jul 01 '19 at 21:09

J. W. Tanner

3 Answers3