If $13x+17y=643$ ,$\{x,y\}\in \mathbb{N}$, then what is the value of two times the product of $x$ and $y$ ?


$a.)\ 744\quad \quad \quad \quad \quad b.)\ 844\\ \color{green}{c.)\ 924}\quad \quad \quad \quad \quad d.)\ 884\\$

I tried,

$13x+17y \pmod{13}\equiv 0\\ \implies 2y \pmod{13}\equiv 3 \\ \implies y=8 \implies y=8, x=39$


I look for a short and simple way .

I have studied maths up to $12$th grade.

  • 2
    $\begingroup$ have you solved the equation? i have found $(x,y)=(5,34),(22,21),(39,8))$ $\endgroup$ – Dr. Sonnhard Graubner Jul 18 '15 at 14:31
  • $\begingroup$ @Dr. Sonnhard:Ok i see my fault , but what could I have done in case the solution pairs would have been more than $6$ , then it would be cumbersome to try and see each case for me. $\endgroup$ – R K Jul 18 '15 at 14:43

When applying mod $13$, the equation $13x+17y=643$ becomes $$4y\equiv 6\pmod{13}$$ or $$40y\equiv 60\pmod {13}$$ that is, $y\equiv 8\pmod {13}$.

Now, to find $x$, apply mod $17$:

$$13x\equiv 14\pmod{17}$$ or $$4\cdot 13x\equiv 56\pmod {17}$$ thus, $x\equiv 5\pmod{17}$.

Now we are to find the concrete values of $x$ and $y$: $$13(17u+5)+17(13v+8)=643$$ which yields $$221(u+v)+201=643$$ therefore, $u+v=2$. Since $u$ and $v$ must not be negative, we have three possibilities:

  • $u=2$, $v=0$. Then $x=39$, $y=8$, so $2xy=624$.
  • $u=v=1$. Then $x=22$, $y=21$. Then $2xy=924$.
  • $u=0$, $v=2$. Then $x=5$, $y=34$. Then $2xy=340$.


$\iff13(x-22)=17(21-y)\implies \dfrac{17(21-y)}{13}=x-22$ which is an integer

$\implies13|17(21-y),\implies13|(21-y)$ as $(13,17)=1$

$\implies21-y=13m\iff y=21-13m$

and consequently, $x=22+17m$ where $m$ is an integer

We need $21-13m=y>0\iff13m <21\implies m\le0$ as $m$ is an integer

and $22+17m=x>0\iff17m>-22\implies m\ge-1$ as $m$ is an integer

$\implies-1\le m\le0$

Can you take it home from here?


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