When the base-b number 11011b is multiplied by b−1, then 1001b is added, what is the result (written in base b )?
11011b=(b4+b3+b+1)1011011b×(b−1)=((b4+b3+b+1)(b−1))10=(b5−b3+b2−1)10(b5−b3+b2−1)+(b3+1)=b5+b2(b5+b2)10=100100b
Done! Yay!
Very nice, Max !!!
+4624 
