# If $5x+7y=2011$ show that $285<x+y<403$

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.

• I hate people who put the year in their questions. Seriously, the bastards. (Not you @OP - I mean the prick that wrote this question initially) Jul 18, 2015 at 19:11
• @alec teal that's a petty reason to hate someone
– user223391
Jul 18, 2015 at 19:15
• thank you for your opinion, I'm sure someone is thinking the same way about their younger men. Jul 18, 2015 at 19:20
• Personally I find the sums with years nice, it gives a "fresh" feeling to the problem :) Jul 18, 2015 at 19:21
• It is common for the Putnam Mathematics Exams to include years in their problems as a way to "ensure" (or try to ensure) that the precise problems have not been posed in the past and thus ensure that the test taker is not working from memory. Jul 18, 2015 at 20:22

Note that $$5x+5y\leq 5x+7y=2011<2015\implies x+y<2015/5=403.$$ Similarly, $$7x+7y\geq 5x+7y=2011>1995\implies x+y>1995/7=285.$$
$5x+5y+2y=2011 \implies 5(x+y) < 2011 \implies x+y < 402.5$. Or, $x+y <403$, similarly $7x+7y > 2011 \implies x+y > 285$.
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