Find the value of $2xy$ . 
If  $13x+17y=643$ ,$\{x,y\}\in \mathbb{N}$, then what is the value of two times the product of 
  $x$ and $y$ ?

Options
$a.)\ 744\quad \quad \quad \quad \quad 
b.)\ 844\\
\color{green}{c.)\ 924}\quad \quad \quad \quad \quad 
d.)\ 884\\$
I tried, 
$13x+17y \pmod{13}\equiv 0\\
\implies 2y  \pmod{13}\equiv 3 \\
\implies y=8
\implies y=8, x=39$
$2xy=624$
I look for a short and simple way .
I have studied maths up to $12$th grade.
 A: 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 $2xy=340$.

A: $13x+17y=643=13\cdot50-7=13\cdot50-7(13\cdot4-17\cdot3)$
$\iff13(x-22)=17(21-y)\implies \dfrac{17(21-y)}{13}=x-22$ 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?
