0
$\begingroup$

Let's say I have this linear congruency: $2x + 1234 = 7 \mod 17$. Without "$+1234$" I would've used the following formulas: $x = x_0 + k(\dfrac{m}{gcd(a, m)})$, whereas $ax_0 + my_0 = b$. But I don't know what to do with $1234$? Thank you in advance.

$\endgroup$

2 Answers 2

1
$\begingroup$

No worries, just solve it like you would solve a normal equation.

$$2x+1234=7$$ $$2x=7-1234$$

and then apply your usual method.

$\endgroup$
0
$\begingroup$

I would calculate everything mod. $17$: $\;1234\equiv 10\mod 17$, so the equation becomes $$2x+10\equiv 7\mod17\iff 2x\equiv-3\equiv14\mod 17. $$ Now as $2$ is a unit mod. $17$, we may apply the cancellation law: $$2x\equiv 14=2\cdot 7\mod17\iff x\equiv 7\mod 17.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .