We need to find two consecutive integers $a$ and $b$ such that $a < \sqrt{111} < b$.

1. Problem Description

We need to find two consecutive integers $a$ and $b$ such that $a < \sqrt{111} < b$.

2. Solution Steps

We are looking for perfect squares close to

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1. We know that $10^2 = 100$ and $11^2 = 121$.

Since $100 < 111 < 121$, we have $\sqrt{100} < \sqrt{111} < \sqrt{121}$.

This means $10 < \sqrt{111} < 11$.

Thus, the two consecutive integers are 10 and

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3. Final Answer

$10 < \sqrt{111} < 11$