mod multiplication

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A multiplication has been performed incorrectly, but the answer is correct (mod 9), (mod 10), and (mod 11). What is the closest that the incorrect result can possibly be to the correct result?

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If $a$ is the correct and $b$ is the incorrect then $(a-b)\equiv 0\text{ mod }9,10,11$

By using Chinese Reminder Theorem, one gets $(a-b)\equiv 0\text{ mod }9\cdot 10\cdot 11=990$ (because $9$, $10$, and $11$ are relatively prime)

The closest the incorrect answer can possibly be to the correct result is $990$