Two positive integers $x$ and $y$ are such that:

$$
\frac{2010}{2011}<\frac{x}{y}<\frac{2011}{2012}
$$

Find the smallest possible value for the sum $x+y$.

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Solution

As $\frac{2011}{2012}4021 d \\
& \geq 12063
\end{aligned}
$$

Consequently, the minimum value of the sum is obtained with $d=2$ and in this case $x+y=8044$.

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