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

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- Mar 5th 2009, 07:27 PMBagoroCongruency
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 - Mar 5th 2009, 07:43 PMo_O
Add the modulus to $\displaystyle 7$ :

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

Since $\displaystyle \gcd (2,17) = 1$, we can cancel both sides by 2 without worrying about the modulus. - Mar 7th 2009, 11:29 AMsiegfried
- Mar 7th 2009, 11:51 AMo_O
Sorry that was a typo. Fixed it now.

- Mar 7th 2009, 01:14 PMjohn_n82
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).