Solve for $x, \;23+35 x ≡ 4$ (mod $ 13)$

Question

I can simplify the equation, but I get stuck at the last step: $\;x≡ 9^{−1} (4-10)$ (mod $13$).

My issue is that I am unsure how to compute this by hand.

Answer 1

You are looking for $9^{-1} \pmod {13}$. You can find that using the Euclidean algorithm:

$$13 - \color{red}{9} = \color{violet}4$$ and $$\color{red}9 - 2\times \color{violet}4=1.$$ Therefore,

$$\color{red}9 - 2 \times {(13 - 9)} = 1$$

i.e., $$3\times 9 - 2 \times 13 = 1.$$

Therefore $3 \times 9 \equiv 1 \pmod {13}.$

Addendum to answer comment:

$$\color{green}{9^{-1}}\times\color{orange}{(4-10)}\equiv\color{green}3\times\color{orange}{-6}\equiv\color{green}3\times\color{orange}7=21\equiv8 \pmod {13}.$$

Answer 2

You want $1/9$ mod $13.$ There are several ways to find this, but one way is nto use that you can add $13$ to a number without changing it mod $13.$ Thus $$\frac{1}{9}=\frac{14}{9}=\frac{27}{9}=3.$$

Answer 3

$9^{-1} = 3 \pmod{13}$ as $27 = 1 \pmod{13}$ (e.g. found via the extended Euclidean algorithm).

Next, $4-10 = -6 = 7 \pmod{13}$, so your computation comes down to $7 \cdot 3 = 21 = 8 \pmod{13}$ indeed.

Just compute like you normally would going back to a representative between $0$ and $12$ whenever you need to, adding or substracting multiples of $13$.

Answer 4

$\begin{align}\bmod 13\!:\,\ &\dfrac{-6}9\equiv \dfrac{-2}3\equiv \dfrac{-15}3\equiv -5\ \ \\[.4em] {\bf Or\!:}\qquad\ &\!\!\dfrac{-6}{-4}\equiv \dfrac{3}{2}\equiv\dfrac{16}{2}\equiv 8\\[.4em] {\bf Or\!:}\qquad\ &\!\!\dfrac{-6}{-4}\equiv \dfrac{-18}{-12}\equiv\dfrac{-5}{1} \end{align}\ $

The prior used Gauss's algorithm. Generally the extended Euclidean Algorithm, or its fractional form is efficient whether for man or machine. See here for more examples.

Beware $\ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.

Answer 5

We have 23 +35x =4(mod13) It gives 35x =-19 =7(mod13) It further gives 9x =7(mod13) Implies 90x=70 (mod13) gives -x=5(mod13) implies x=8(mod13)