### 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: To make the arithmetic easier, I will start by converting the given numbers from base $b$ to base 10, using the fact that each digit is a power of $b$.
The base-$b$ number $11011_b$ has five digits, so the highest power of $b$ is $b^4$.
The value of this number in base 10 is:

\[11011_b = 1 \cdot b^4 + 1 \cdot b^3 + 0 \cdot b^2 + 1 \cdot b + 1 = b^4 + b^3 + b + 1\]
Similarly, the base-$b$ number $1001_b$ has four digits, so the highest power of $b$ is $b^3$.
The value of this number in base 10 is:

\[1001_b = 1 \cdot b^3 + 0 \cdot b^2 + 0 \cdot b + 1 = b^3 + 1\]
Now I can multiply $11011_b$ by $b-1$ and add $1001_b$, using base 10 arithmetic:

\begin{align*}
(11011_b)(b-1) + 1001_b &= (b^4 + b^3 + b + 1)(b-1) + (b^3 + 1)\\
&= b^5 - b^4 + b^4 - b^3 + b^2 - b + b^3 - 1 + b^3 + 1\\
&= b^5 + b^2
\end{align*}
The result is $b^5 + b^2$, which is a base 10 expression.
To convert it back to base $b$, I need to find the digits that correspond to the powers of $b$.
The highest power of $b$ in the result is $b^5$, which means the result has six digits.
The first digit is the coefficient of $b^5$, which is 1.
The second digit is the coefficient of $b^4$, which is 0, since there is no $b^4$ term in the result.
The third digit is the coefficient of $b^3$, which is also 0, for the same reason.
The fourth digit is the coefficient of $b^2$, which is 1, since there is a $b^2$ term in the result.
The fifth and sixth digits are the coefficients of $b$ and $b^0$, which are both 0, since there are no $b$ or constant terms in the result.
Therefore, the result in base $b$ is $100100_b$.