Find the sum of $327_8$ and $73_8$ in base $8$.
I know that base $8$ means that each digit represents a power of $8$, starting from the rightmost one. For example, $327_8 = 3 \times 8^2 + 2 \times 8^1 + 7 \times 8^0$. To add two numbers in base $8$, I need to align them by their digits and add each pair of digits from right to left, carrying over any extra values that are greater than $7$. So I write $327_8$ and $73_8$ as follows: \begin{align*} 327_8 \\ + \quad 73_8 \end{align*} Starting from the rightmost column, I add $7$ and $3$, which gives me $10$. But $10$ is not a valid digit in base $8$, so I need to carry over the $8$ and write the remainder $2$ in the sum. Moving to the next column, I add $2$, $7$, and the carried $1$, which gives me $10$ again. Moving to the leftmost column, I add $3$ and $0$, and the carried $1$, which gives me $4$. This time, $4$ is a valid digit in base $8$, so I write it below the sum and I'm done. \begin{align*} 327_8 \\ + \quad 73_8 \\ \hline \quad 422_8 \end{align*} 