Solve the congruence $6x+15y \equiv 9\pmod {18}$


$(6,18)=6$, so $$15y \equiv 9\pmod 6$$ $$15y \equiv 3\pmod 6$$

So the equation will have $(15,6)$ solutions. Now we divide by 3

$$5y \equiv 1\pmod 2$$.

Solving the Diophantine equation we get $y \equiv1\pmod 2 $, so $y=1+2m$

$$6x \equiv 9-15y\pmod {18}$$ $$6x \equiv 9-15(1+2m)\pmod {18}$$ $$6x \equiv -6-30m\pmod {18}$$

Divide by 6

$$x \equiv -1-5m\pmod 3$$

The right solution is $x-m \equiv 2\pmod 3$. I have $x+5m \equiv 2\pmod 3$ $$x-m+6m \equiv 2\pmod 3$$ $$x-m \equiv 2\pmod 3$$

Wolfram alpha says $y=1+2t$ and $x=t+3d+2$


3 Answers 3


Dividing everything by $3$, including the modulus, we get the equivalent congruence $$2x+5y\equiv 3\pmod{6}.$$ It is convenient to rewrite this as $2x-y\equiv 3\pmod{6}$, or equivalently $y\equiv 2x-3\equiv 2x+3\pmod{6}$.

Now we have a parametric solution: $x\equiv t\pmod{6}$, $y\equiv 2t+3\pmod{6}$. To write it out at length, set $t=0,1,2,3,4,5$. The first two solutions are then $x\equiv 0\pmod{6}$, $y\equiv 3\pmod{6}$ and $x\equiv 1\pmod{6}$, $y\equiv 5\pmod{6}$. Four more to go.


I hope my answer is okay. $6x+15y \equiv 9(mod 18)$

$=> 2x+5y \equiv 3(mod 6)$

$=> 2x-y \equiv 3(mod 6)$

i.e $2x-y-3 \equiv 0(mod 2)$ and $\equiv 0(mod 3)$

this means $y+1 \equiv 0(mod 2)$ and $x+y \equiv 0(mod 3)$

Write $y=2p+1$ and so $x-p +1 \equiv 0(mod 3)$

Then use parametric equations.

  • $\begingroup$ why is $2x-y\equiv3(mod6)$ $\endgroup$ Jun 30, 2016 at 6:02
  • $\begingroup$ write $5y \equiv 6y-y$ $\endgroup$
    – null
    Jun 30, 2016 at 6:07
  • $\begingroup$ you get $2x-y\equiv 0(mod 3)$. How do you get $x+y\equiv 0(mod 3)$. Is it because $-x-y\equiv 0(mod 3)$ implies $x+y\equiv 0(mod 3$)? $\endgroup$ Jun 30, 2016 at 6:14
  • $\begingroup$ $2x-y-3 \equiv 0(mod3)$ implies $3x-x-y-3 \equiv 0(mod3)$ i.e $-x-y \equiv 0(mod3)$ i.e $x+y \equiv 0(mod3)$ $\endgroup$
    – null
    Jun 30, 2016 at 6:19
  • $\begingroup$ can you provide another exercise?. My book just has that one? $\endgroup$ Jun 30, 2016 at 6:21

Wolfram answer:
$y = 2c + 1$
$x = c + 3d + 2$

  • $\begingroup$ so the y is right. What's wrong with the x?. I got 3d+2 $\endgroup$ Jun 30, 2016 at 5:14

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