# Thread:

1. ## Congruency

I'm having trouble solving this problem.

Solve the congruence 2x is equivalent to 7 (mod 17)

Can someone work it out for me in step by step form with some explinations

2. Add the modulus to $7$ :

$\begin{array}{rcll} 2x & \equiv & 7 & \left(\text{mod } 17\right) \\ 2x & \equiv & 7 {\color{red}\ + \ 17} & \left(\text{mod } 17\right) \\ 2x & \equiv & 24 & \left(\text{mod } 17\right) \\ 2x & \equiv & 2 \cdot 12 & \left(\text{mod } 17\right) \end{array}$

Since $\gcd (2,17) = 1$, we can cancel both sides by 2 without worrying about the modulus.

3. Originally Posted by o_O
$\begin{array}{rcll}2x & \equiv & 24 & \left(\text{mod } 17\right) \\ 2x & \equiv & 2 \cdot 17 & \left(\text{mod } 17\right) \end{array}$
how does this follow

4. Sorry that was a typo. Fixed it now.

5. Congruence ax = b (mod c) for some a, b, c in Z can be written as a linear diophantine equation ax - cy = b. And that is easy to solve using the box method.

For this problem, 2x = 7 (mod 17) can be written as 2x - 17y = 7.
Using the box method, we found the Bezout relation: -56.2 + 7.17 = 7
Therefore, x = -56 = 12 (mod 17).
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