If $x,y$ are integers such that $3x+7y$ is divisible by $11$, then which of the following is divisible by $11$? I am currently studying for the GRE, and this question came up.
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) $x+y-1$
I tried looking at the equation $3x+7y\equiv 0 \pmod{11}$ and tried to manipulate it by adding terms of the form $11nx$ and $11my$ for some integers $n,m$ but got stuck. 
I would appreciate any help and would also like to hear some strategies for tackling these types of problems, since this isnt the first time I have seen it.
 A: One idea is to take for example $x=5$, $y=1$. Or $x=7$, $y=-3$. These make $3x+7y$ divisible by $11$. 
Test the "answers" with one of these. I prefer $5,1$ because the arithmetic is simpler, but let's try the much more natural $7,-3$. Only d) works. 
As a formal verification, or another tool, multiply $3x+7y$ by something that will give us a $4x +?y$ modulo $11$. Note that $5$ will do, since $(5)(3)=15\equiv 4$. So what is $(5)(7)$? Notice it is congruent to $-9$. 
By the way, the entries with non-zero constant terms cannot be right, since they break down at $x=0$, $y=0$. 
A: $3x+7y\equiv -8x-4y=-2(4x+2y)\equiv-2(4x-9y)\pmod{11}$. This shows that the original form is divisible by 11 iff form (d) is divisible by 11. An example such as $x=1$, $y=9$ shows the other options do not hold. 
A: Using the extended Euclidean algorithm and Bézout's identity you can parametrise all solutions to the equation 3x + 7y = 11n, hence you can find counterexamples and reason about all solutions using a single calculation.
In this case it is $x = -2*11n + 7k$ and $y = 1*11n - 3k$, so $4x + 6y \equiv_{11} 10 k$.
A: For the GRE a quick and dirty solution is perfectly acceptable. By substituting $x=5$ and $y=1$ you can rule out (a), (c), and (e). By substituting $x=1$ and $y=9$ you can rule out (b). This leaves only (d) as a possibility.
I chose these two solutions to $11\mid 3x+7y$ by setting $y=1$ and quickly checking to find the smallest positive $x$ that gave me a multiple of $11$, and then setting $x=1$ and doing the corresponding thing.
Given sufficient time, you can verify that $3(4x-9y)=4(3x+7y)-55y$; the righthand side of this equation is a multiple of $11$ if $3x+7y$ is, and $3$ is relatively prime to $11$, so $4x-9y$ is a multiple of $11$ if $3x+7y$ is.
A: Note that $3(4)+7(3)=12+21=33$ which is divisible by 11. Therefore, we can put $x=4$ and $y=3$ to test each of the answers. And you see that the answer is D.
