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)?
The number you have listed, 11011_b can be represented in ANY base from 2 and up. When you say "is multiplied by b-1", I'm assuming you mean: (11011_b) - 1. If I'm wrong in my assumptions, let us know.
11011*11010+1001 =1011000111.00 - this is in base 2
=122002111.00 - this is in base 3
=121232111.00 - this is in base 4, 5, 6, 7, 8, 9,.........etc.
Note: You may choose the base you want, or is intended.
I don't think the previous answerer got the question right. Let's take a particular example, say b=4. Then the question becomes "When the base 4 number \(11011_4\) is multiplied by 3, and then \(1001_4\) is added, what is the result (written in base 4)?"
Now\(11011_4\times 3 = 33033_4\) and adding \(1001_4\) we get \(100100_4\).
Similarly, in for example base 7, we have
\(11011_7\times 6 + 1001_7 = 66066_7+1001_7=100100_7\)
In fact for any base>1 the answer is always 100100.
+118609 Yes the second answer is definitely the correct one.
I am glad you gave it a go guest one, that is the best way to learn. I hope you see this continuation.
Good work guest 2 
I shall also try to explain.
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)?
\(11011_b=b^4+b^3+0+b+1\\ (b-1)*(b^4+b^3+0+b+1)\\ =b^5+b^4+0+b^2+b\\ \quad\;\;-(b^4+b^3+0+b+1)\\ =b^5+0-b^3+b^2+0-1\\ Now\\ 1001_b=b^3+0+0+1\\~\\ \;\;\;(b^5+0-b^3+b^2+0-1)\\ \;\;\;+\qquad\;\;(b^3+0+0+1)\\ =b^5+0+0+b^2+0+0\\ =100100_b \)
