# linear congruent modulo with only one solution

$$n \cdot x ≡ k \mod{20} \\n,k \in \mathbb{Z}$$

Choose $$n,k$$ so there is only $$(a)$$ one, $$(b)$$ no solution in $$\mathbb{Z_{20}}$$.

If $$n = 0$$, then two solution exist, and that is $$k=20$$ and $$k=-20$$, as we are in $$\mathbb{Z_{20}}$$

If $$n,k = 0$$, then there is one solution, because $$0 ≡\mod{20} = 0$$

Stuck on the question how to find no solution. Basically you could just put $$n \geq 21$$ and $$\mod{20}$$ will never be able to generate a number $$\geq 20$$

• $20 \equiv -20\pmod{20}$ so they are regarded as the same. Dec 10, 2012 at 6:56

For one solution, say $n=1$, $k=1$. Clearly, $x\equiv 1\pmod{20}$ is the only solution.

For no solution, $n=2$, $k=1$. If $n=2$, then $nx$ is always even, so cannot be congruent to any odd number modulo $20$.

For more than $1$ solution, $n=4$, $k=0$. Note that $x\equiv 0\pmod{20}$ is a solution, but so is $x\equiv 5\pmod{20}$. Also, $10$ and $15$ work.

For exactly two solutions, $n=2$, $k=0$ works, the only solutions are $x\equiv 0\pmod{20}$ and $x\equiv 10\pmod{20}$.

Surely you could have found these, or ones like them. For all the questions, there are several answers that work.

Your calculation was not correct, $20$ and $-20$ and $0$ are all the same modulo $20$. In fact, if we take $n=0$, $k=0$, there are $20$ solutions (any $x$ works).

• no solution, because even on the left and odd on the right? i'm such a dumbass. Dec 10, 2012 at 7:01
• Well, it's a bit more subtle, because for example $18\equiv 3\pmod{15}$. The key is that $20$ is even. Dec 10, 2012 at 7:07
• Thank you for your detailed answer. much appreciated :) Dec 10, 2012 at 7:15

Hint: $ax \equiv b \pmod{m}$ has a unique solution if and only if $\gcd(a,m) = 1$.

More generally, using Linear congruence theorem, this equation is solvable $\iff$ $(20,n)\mid k$
In that case,we have exactly $(20,n)$ solutions.