problem: Evaluate $\lceil\sqrt{10}\rceil+ \lfloor\sqrt{10}\rfloor$.
solution: Since $\sqrt{9}<\sqrt{10}<\sqrt{16}$, we know that $\sqrt{10}$ is a number between $3$ and $4$. Therefore, the smallest integer that is greater than or equal to $\sqrt{10}$ is $4$ and the greatest integer that is less than or equal to $\sqrt{10}$ is $3$. So, $\lceil\sqrt{10}\rceil+ \lfloor\sqrt{10}\rfloor=4+3=\boxed{7}$.