### Instruction: When the base-$b$ number $11011_b$ is multiplied by $b-1$, then $1001_b$ is added, what is the result (written in base $b$)?### Response: The first step is to expand the base-$b$ numbers as powers of $b$.
I have

\begin{align*}
11011_b &= b^4 + b^3 + b + 1\\
1001_b &= b^3 + b^0
\end{align*}
Next, I multiply the first number by $b-1$.
Using the distributive property, I get

\begin{align*}
(b^4 + b^3 + b + 1)(b-1) &= b^5 - b^4 + b^4 - b^3 + b^2 - b + b - 1\\
&= b^5 - b^3 + b^2 - 1
\end{align*}
Then, I add $1001_b$ to this result.
Using the base-$b$ addition rules, I get

\begin{align*}
(b^5 - b^3 + b^2 - 1) + (b^3 + b^0) &= b^5 + (b^3 - b^3) + b^2 + (b^0 - 1)\\
&= b^5 + b^2 + (1 - 1)\\
&= b^5 + b^2
\end{align*}
Finally, I write this answer in base $b$.
I notice that there are only two nonzero digits, one in the $b^5$ place and one in the $b^2$ place.
So, the answer is