### 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: I notice that the problem involves both multiplication and addition in base $b$, so I will need to use the rules for those operations in base $b$.
To start, I will rewrite $11011_b$ and $1001_b$ using the place value of each digit.
In base $b$, each digit represents a power of $b$, so I have $11011_b = b^4 + b^3 + b + 1$ and $1001_b = b^3 + 1$.
Now, to multiply $11011_b$ by $b-1$, I can use the distributive property and multiply each term by $b-1$.
This gives me $(b^4 + b^3 + b + 1)(b-1) = b^5 - b^4 + b^4 - b^3 + b^2 - b + b - 1$.
I can simplify this by canceling out the $-b^4$ and $+b^4$ terms, and the $-b$ and $+b$ terms.
This leaves me with $b^5 - b^3 + b^2 - 1$.
Now, to add $1001_b$ to this, I can align the digits according to their place value and add them column by column, starting from the rightmost column.
In the rightmost column, I have $-1 + 1 = 0$.
In the next column, I have $0 + 0 = 0$.
In the next column, I have $b^2 + 0 = b^2$.
In the next column, I have $-b^3 + b^3 = 0$.
In the leftmost column, I have $b^5 + 0 = b^5$.
So the final result is $b^5 + b^2$.