TITLE: question about division
QUESTION [9 upvotes]: 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?

REPLY [4 votes]: Hint $\  \mathbb F = \mathbb Z/11\:$ & $\:\mathbb Z/13\,$ are fields, so $\rm\,\mathbb F^{\:\!n}$ is a vector space, hence lines through the origin are uniquely determined by any non-origin point on the line. To find such a point, pick an easily invertible coefficient $c$ of the equation and scale the equation by $c^{-1}$ so the coefficient becomes $1$.
E.g. $\rm\ \color{#c00}4x -5y + 2z = 0\:$ over $\rm\:\mathbb Z/13.\:$ We look for an  easy invertible coef, e.g. one dividing $13\pm1,\,$ e.g.  $\,\color{#c00}4\cdot 3 = 13- 1 = -1,\,$ so scaling by $\,4^{-1} = - 3$ yields $\rm\: x +2y - 6z = 0,\,$ so $\rm\,x = -2y+6z.\,$  Picking "simple" values for $\rm\:y,z\:,\:$ say $\rm\:y=0,\:z=2\:$ yields $\rm\:x=-1,\:$ which lies only on the line  $\rm\:(D)\ -7x +12y+3z = 0.$ 
This method is both general and quick - esp. quick when there is an easily invertible coefficient dividing modulus $\pm 1\:$ (which occurs frequently for small moduli by the law of small numbers).