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I have been tasked with solving a linear congruence:

$$-12x\equiv-3\pmod{26}$$

How do I do this? I've never done linear congruences with minus signs so I'm quite confused.

Usually I would find the inverse of the LHS and multiply the RHS by the inverse however obviously since we have a negative number, the number isn't in $\mathbb{Z}_{26}$ so we can't find an inverse.

user2850514
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  • $-12 = 14$ in $\mathbb{Z}_{26}$, so why are negative signs a problem? But $14$ is even, so it's not invertible modulo $26$. –  Apr 14 '16 at 16:15
  • Reducing modulo $2$ gives $0 \equiv 1 \pmod 2$, so the equation is inconsistent. – Travis Willse Apr 14 '16 at 16:15
  • @T.Bongers Ah so we have $$14x\equiv 23\pmod{26}$$ however again $14$ does not have an inverse so what do I do now? – user2850514 Apr 14 '16 at 16:17

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For some integer $k$, $$\begin{align*} -12x&\equiv -3 \pmod{26}\\ -12x &= -3 + 26k \end{align*}$$ The left hand side is divisible by $2$, but the right hand side is not, so there is no solution for $x$.

peterwhy
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