TITLE: If $5x+7y=2011$ show that $285<x+y<403$
QUESTION [5 upvotes]: Let $x,y\in\mathbb{N}$ two natural numbers that satisfy the equality
\begin{equation}5x+7y=2011\end{equation}
Show that 
\begin{equation}285<x+y<403\end{equation}
The only thing I obtained is $(x,y)=1$, since $2011$ is a prime number.

REPLY [2 votes]: When you are doing such questions, the best way to approach is by simply trying to manipulate what you are given and try to change it to what you want to proof. So start by the given then change it arithmetically( it should make sense) then finally find a relation and make a concluding statement. This might help you! So let's get to it,
Since $(x,y)$ is the element of natural numbers we can simply see that,
$$5x+5y<5x+7y$$
$$5x+5y<5x+7y=2011$$
$$\Rightarrow 5x+5y<2011$$
$$\Rightarrow x+y<402.2<403$$
Simillarly,
$$7x+7y>2011$$
$$\Rightarrow 287.2<285<x+y$$
As a conclusion we have $285<x+y<403$ thus proving what we wanted