Solving for a variable in a modular arithmetic equation

Question

$\fbox{$13x + 1 \equiv 0 \pmod {100}$}$

I solved the equation above by trying different multiples to isolate $x$ until I found something that worked. I have two questions:

$\fbox{$1.$}\ $ What if there was no solution for $x$? How would I be able to prove it?

$\fbox{$2.$}\ $ Are there a set of steps that I could program a computer to follow and get an answer if other similar modular equations are inputted?

My solution is below:

$13x +1 \equiv 0 \pmod {100}$

$13x \equiv 99 \pmod {100}$ (added $99$ to both of equation and applied the $\mod 100$ to the left side)

$104x \equiv 792 \pmod {100}$ (multiplied both sides by $8$)

$4x \equiv 792 \pmod {100}$ (removed a $100$ from the left side)

$x \equiv 198 \pmod {100}$ (divided both side by $4$)

Like I said, I believe I got the right solution but only through trial and error. I was wondering if there is a more systematic way of solving these problems.

Thank you for any help.

For different modular equations, we can check if it holds certain modular properties or not to confirm the existence of solutions. Here, for example, the equation is $13x + 1 \equiv 0 \pmod{100}$ $\implies 13x \equiv -1 \pmod{100}$. Since $13$ is co-prime to $100$, it obliviously has a solution for $13x \equiv 1 \pmod{100}$ (existence of inverse). Replace $x$ with $-x$ and you're done. — thesagniksaha, Dec 15 '18 at 09:22
https://math.stackexchange.com/questions/3040108/what-are-the-integral-solutions-of-239x-111y-1#comment6268260_3040108 — lab bhattacharjee, Dec 15 '18 at 09:27
The current answers failed to point out that your attempt was completely incorrect. In particular, your last step (division by $4$) is invalid. Counter-example: $4·31 \equiv 24$ mod $100$ but $31 \not\equiv 6$ mod $100$. At one point you multiplied by $8$, which has a common factor with $100$, so it became irreversible (and the resulting equations have more solutions). — user21820, Dec 15 '18 at 17:49

score 5 · Answer 1 · answered Dec 15 '18 at 09:17

5

Writing $$x\equiv -\frac{1}{13}\mod 100$$ adding the module to the numerator we get $$x\equiv \frac {99}{13}\equiv \frac{199}{13}\equiv \frac{299}{13}\equiv 23\mod 100$$ so $$x\equiv 23\mod 100$$

answered Dec 15 '18 at 09:17

Dr. Sonnhard Graubner

97,058

Thank you. I understand that it better now. – Dan Dec 15 '18 at 22:19

score 3 · Answer 2 · answered Dec 15 '18 at 09:55

This is equivalent to the congruence equation $$13x\equiv -1\mod 100,$$ so you only have to find an inverse of $13\bmod 100$. As $13$ and $100$ are coprime, this inverse exists by Bézout's identity (this answers negatively your first question). You'll find it with the extended Euclidean algorithm: \begin{array}{rrrl} r_i & u_i & v_i & q_i \\\hline 100 & 0 & 1 \\ 13 & 1 & 0 & 7 \\\hline 9 & -7 & 1 & 1 \\ 4 & 8 & -1 & 2 \\ 1 & \color{red}{-23} & 3 \\ \hline \end{array}

Thus the inverse of $13$ is $-23$, and $$13x\equiv -1\mod 100\iff x\equiv(-23)(-1)= 23\mod 100.$$

Solving for a variable in a modular arithmetic equation

2 Answers2