11. Let $a$ and $b$ be integers for which $\frac{a}{2}+\frac{b}{1009}=\frac{1}{2018}$. Find the smallest possible value of $|a b|$.
Answer: 504
Solution: Clear denominators to write this as $1009 a+2 b=1$. Clearly, $a=1, b=-504$ is a solution, and so our solutions are of the form $a=1+2 k, b=-504-1009 k$. Now, clearly $|a| \geq 1$, and $|b| \geq 504$, so $|a b| \geq 504$, and equality is attained when $a=1$ and $b=-504$.