[Algebra]
Problem: Given that $p=6.6 \times 10^{-27}$, then $\sqrt{p}=a \times 10^{b}$, where $1 \leq a<10$ and $b$ is an integer. Compute $10 a+b$ rounded to the nearest integer.

Solution: Note that $p=6.6 \times 10^{-27}=66 \times 10^{-28}$, so $a=\sqrt{66}$ and $b=-14$. Note that $\sqrt{66}>\sqrt{64}=8$. Because $8.1^{2}=65.61$ and $8.15^{2}=66.4225>66$, conclude that $81<10 \sqrt{66}<81.5$, hence $10 a$ rounded to the nearest integer is 81 , and the answer is $81-14=\mathbf{6 7}$.

Final answer: 67