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When the base-\(b\) number \(11011_b\)  is multiplied by \(b-1 \), then \(1001_b\) is added, what is the result (written in base \(b\) )?

 Dec 28, 2017
 #1
avatar+7677 
+2

\(11011_b = (b^4 + b^3 + b + 1)_{10}\\ \quad 11011_b \times (b-1)\\ =\left((b^4+b^3+b+1)(b-1)\right)_{10}\\ =(b^5 - b^3 + b^2 - 1)_{10}\\ \\ (b^5-b^3+b^2-1) + (b^3+1) = b^5 + b^2\\ (b^5+b^2)_{10} = 100100_b\)

Done! Yay!

 Dec 29, 2017
 #2
avatar+101746 
+1

Very nice, Max  !!!

 

 

cool cool cool

 Dec 29, 2017

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