Prove if $56x = 65y$ then $x + y$ is divisible by $11$ If $x$ and $y$ are natural numbers, and $56x = 65y$, prove that $x + y$ is divisible by $11$.
I tried  taking the $\gcd(56x,65y)$ using the Euclidean algorithm, but I got nowhere with it and do not know where to head. 
 A: Why not just $x+y=(56x-55x)+(66y-65y)=11(6y-5x)+(56x-65y)=11(6y-5x)$ (since $56x-65y=0$).   
A: Since $11$ does not divide $56$ and $11$ is prime, $11$ divides $x+y$ if and only if it divides $56(x+y)$.  But $56(x+y)=56x+56y=65y+56y=121y$, which in fact is divisible by $11^2$.
A: We have $56x\equiv x\pmod{11}$ and $65y\equiv -y \pmod{11}$. If $56x=65y$, it follows that $x\equiv -y\pmod{11}$, which is what we needed to show.
A: Another variation (short, strong result, explicitly invoking Gauss lemma) :
Adding $56y$ to both members gives $56(x+y) = 121y$. Since $56$ and $121$ are relatively prime, by Gauss, $121$ divides $(x+y)$.
A: $56$ and $65$ are relatively prime, so if $56x=65y$, then $65\mid x$ and $56\mid y$; say $x=65m$ and $y=56n$. Then 
$$56\cdot65m=56x=65y=65\cdot56n\;,$$
so $m=n$. Thus, the solutions are of the form $x=65k,y=56k$ for integers $k$, and $$x+y=(65+56)k=121k=11(11k)\;.$$
Thus, $x+y$ is even divisible by $11^2$.
A: Let's work in $\mathbb R^2$. We have a linear map given by the matrix $$A=\begin{pmatrix}65 & -56\\1&1\end{pmatrix}$$ And we are interested in the solutions to the equation $A\mathbf x=\begin{pmatrix}0\\ b\end{pmatrix}$ where $b$ is an integer. Since the determinant of $A$ is $121$, Cramer's rule implies that the first coordinate of the solution $\mathbf x$ is $\frac {56b}{121}$. Thus, if $\mathbf x$ has integer coordinates, then $b$ is a multiple of $121$.
From here the idea is pretty straight forward to generalize. If $r,s$ are integers such that $d=65r+56s$ is relatively prime to both $56$ and $65$, then any pair of integers $x,y$ for which $65x=56y$ satisfies $rx+sy\equiv 0\pmod d$.
A: $56x = 65y \implies x + y = 11(6y - 5x)$
A: This is basically the same as the answer by Brian M. Scott:
$$65(x+y)=65x+65y=65x+56x=121x$$
Thus $11^2 | 65(x+y)$ and since it is relatively prime to $65$....
A: $$56x=65y\tag1$$
$$x+y=\frac{56}{56}(x+y)=\frac{56x+56y}{56}$$
from (1) we have
$$\frac{65y+56y}{56}=\frac{121y}{56}=\frac{11^2y}{56}\tag2$$
As $121$ is not divisible by 56, so $56|y$ or $y = 56 n$ which helps us to deduce
$$x+y = 11^2n$$
or $11|x+y$
A: 56x = 65y
x = (65/56) y
x + y = (65/56)y + (56/56)y = (121/56)y
It is obvious that (121/56) is divisible by 11.
