<s>[INST] question about division The question is from the following two problems:

Let $x$ and $y$ be positive integers such that $3x+7y$ is divisible by $11$. Which of the following must also be divisible by $11$?
  A. $4x+6y$  B. $x+y+5$  C. $9x+4y$  D. $4x-9y$  E. $x+y-1$
If $x,y$ and $z$ are positive integers such that $4x-5y+2z$ is divisible by $13$, then which one of the following must also be divisible by $13$?
  A. $x+13y-z$ B. $6x-10y-z$ C.$x-y-2z$ D.$-7x+12y+3z$ E. $-5x+3y-4z$

One way I know is that one can substitute some specific number to rule out the wrong answers. Or alternatively, one can try the properties of congruence(addition and subtraction), which I think is not easy to figure out the steps. Just as for the second problem, how on earth can one figure out one needs to add 
$$-39x\equiv 0\pmod{13}$$
$$65y\equiv 0\pmod{13}$$
$$13z\equiv 0\pmod{13}$$
to
$$4x-5y+2z\equiv 0\pmod{13}$$
and then divides by $5$
in order to get the answer
$$-7x+12y+3z\equiv 0\pmod{13}?$$
So, here is my question:

Is there a general method (which might be also quick) to solve this kind of problems?

 [/INST] This is what I would do:
If $3x+7y$ is divisible by $11$, then $3x + 7y\equiv 0\pmod{11}$, so $3x\equiv -7y = 4y\pmod{11}$. Multiplying through by $4$, we get $12x \equiv 16y\pmod{11}$, which simplifies to $x\equiv 5y\pmod{11}$.
Then one can plug into the congruences: $4x+6y \equiv 26y\equiv 3y\pmod{11}$; $x+y+5\equiv 6y+5\pmod{11}$; $9x+4y\equiv 49y\equiv 5y\pmod{11}$; $4x-9y\equiv 11y\equiv 0\pmod{11}$; and $x+y-1\equiv 6y-1\pmod{11}$. Clearly, the answer is D.
For the second one, from $4x-5y+2z\equiv 0\mod{13}$ we have $4x\equiv 5y-2z\pmod{13}$; multiplying through by $-3$ we get
$$-12x \equiv -15y+6z\pmod{13}$$
which is equivalent to $x \equiv -2y+6z\pmod{13}$.
Plugging into the different possibilities, we have:
$$\begin{align*}
\text{A. }&& x+13y - z &\equiv -2y+5z\pmod{13}\\
\text{B. }&& 6x-10y-z &\equiv 4y +9z\pmod{13}\\
\text{C. }&& x-y-2z &\equiv -3y+4z\pmod{13}\\
\text{D. }&&-7x+12y+3z &\equiv 26y -39z \equiv 0\pmod{13}\\
\text{E. }&& -5x+3y-4z &\equiv 13y-34z \equiv 5z\pmod{13},
\end{align*}$$
so again, the answer is D.

Alternatively.
Once you have $x-5y\equiv 0\pmod{11}$ in the first problem, multiply by a suitable $k$ to try to match the coefficient of $x$ in the alternatives, until you get a match:


*

*For $4x+6y$, multiply $x-5y$ by $4$ to get $4x-20y \equiv 4x+9y$, not equal.

*For $x+y+5$, not equal.

*For $9x+4y$, multiply $x-5y$ by $9$ to get $9x-45y \equiv 9x-y$; not equal.

*For $4x-9y$, multiply $x-5y$ by $4$ to get $4x - 20y \equiv 4x -9y$; equal. This is the answer.


For the second one, again, from $x+2y-6z\equiv 0$, we try to match the coefficient of $x$ in each:


*

*Multiply by $1$ to try to get $x+13y-z$; no luck.

*Multiply by $6$ to try to get $6x-10y-z$; we get $6x+12y-36z\equiv 6x+12y-3z$; no luck.

*Multipy by $1$ to try to get $x-y-2z$; no luck.

*Multiply by $-7$ to try to get $-7x+12y+3z$; we get $-7x-14y+42z \equiv -7x -y+3z \equiv -7x+12y+3z$. Equal, so this is the answer.

</s>