How to show steps for 11x - 7 $\equiv$ 0 (mod 13)?

Question

I know the answer is $x = 3$, but I'm having trouble showing the steps for it. All I have is $11x \equiv 7 (\mod 13)$.

Answer 1

There are general methods here based on Euclid's division algorithm, and these always work. You can look that up.

Sometimes there are short cuts you can spot, as here. Don't forget that there are negative numbers in congruence classes. So here, modulo $13$ $$11x\equiv -2x$$ and you then only need to divide by $2$. So $7\equiv -6$ or $7\equiv 20$ get you to an even residue.

Whence the equation can be rewritten as $-2x+6\equiv 0$ with $x\equiv 3$ by simple division ($2$ and $13$ are coprime), or you could write $-2x-20\equiv 0$, from which you obtain $x\equiv -10\equiv 3$.

Answer 2

Your question $$11x\equiv 7\mod{13}$$ is a special case of the congruent equation $$ax\equiv b \mod{p},$$ where $p$ is a prime. Here is a standard way to find the solution:

1. If $p$ divides $a$, then $b$ must be a multiple of $p$, since otherwise the equation would have no solution.
2. Suppose $$\gcd(a,p)=1,$$ and we consider the congruent equation $$ax\equiv 1 \mod{p},$$ i.e., $x$ is the modular multiplicative inverse of $a$ wrt modulus $p$, which can be computed by the extended Euclidean algorithm. The existence is provided by the Bezout's identity. We denote it by $M$.
3. Now we have the equation $$aM\equiv 1 \mod{p},$$ times $b$ on the both sides we get $$a(bM)\equiv b \mod{p},$$ one can see that $bM$ is a solution to the require congruent equation (modulus $p$)

So in the case $$11x\equiv 7\mod{13},$$ the multiplicative inverse of $11$ wrt $13$ is $6$: $$1=11-2\times 5=11-(13-11)\times 5=11\times 6-13\times 5,$$ and hence $$6\times 7\equiv 3 \mod{13}$$ is a solution to the requirement.

Answer 3

To solve the congruence $11x \equiv 13 \pmod{13}$, we will find the inverse of $11$ modulo $13$ using the extended Euclidean algorithm, then multiply both sides of the congruence by the inverse of $11$ modulo $13$ to find $x$.

First, we use the Euclidean algorithm to find $\gcd(11, 13)$. \begin{align*} 13 & = 1 \cdot 11 + 2\\ 11 & = 5 \cdot 2 + 1\\ 2 & = 2 \cdot 1 \end{align*} Hence, $\gcd(11, 13) = 1$. Next, we wish to express $1$ as a linear combination of $11$ and $13$. Working backwards, we obtain \begin{align*} 1 & = 11 - 5 \cdot 2\\ & = 11 - 5(13 - 11)\\ & = 6 \cdot 11 - 5 \cdot 13 \end{align*} Since $6 \cdot 11 = 1 + 5 \cdot 13$, $6 \cdot 11 \equiv 1 \pmod{13}$. Therefore, $6$ is the multiplicative inverse of $11$ modulo $13$.
\begin{align*} 11x & \equiv 7 \pmod{13}\\ 6 \cdot 11x & \equiv 6 \cdot 7 \pmod{13}\\ x & \equiv 42 \pmod{13}\\ x & \equiv 3 \pmod{13} \end{align*} Check. If $x \equiv 3 \pmod{13}$, then $11x \equiv 11 \cdot 3 \equiv 33 \equiv 7 \pmod{13}$.

Answer 4

The extended Euclidean algorithm is an efficient algorithmic way to compute modular fractions and inverses, but often there are much quicker methods for small numbers. Here it is extremely simple, just reduce the numerator and denominator to reps having smallest magnitude, namely

$$\bmod 13\!:\ 11x\equiv 7\iff x \equiv \dfrac{7}{11} \equiv \dfrac{-6}{-2} \equiv 3\qquad$$

That generally simplifies matters because the smaller the denominator the higher the probability that it will (exactly) divide the numerator. If you're not familiar with modular fractions then you can simply replace $\,x\equiv a/b\,$ by the equation $\,bx\equiv a$ and work instead with the equations (similar to Mark's answer). You can find many other worked examples in my prior answers, e.g. here and here.

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

Answer 5

The problem 11x − 7 ≡ 0 (mod 13) is equivalent to 11 x0 − 7 = 13 x1 where x0, x1 ∈ . One way to solve the linear equation would be to express the original variables (x0, x1) in terms of parameter variables (t0, t1) where each of the parameter variables take on any value from . To find such as expression you can judiciously introduce the parameter variables and corresponding equation one at a time.

After generating the additional equations (the ones with parameter variables) you can take them plus the original equation as a system on linear equations. The system of equations can be solved using back substitution. After each back substitution step if the resulting equation has a greatest common factor then the greatest common factor should be removed before continuing.

The PDF document "Showing the Steps to Solve 11x − 7 ≡ 0 (mod 13)" applies these ideas in a step-by-step fashion. The following diagram is a snap shop of the PDF page: