+0  
 
+1
418
2
avatar+3836 

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+7220 
+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+96055 
+1

Very nice, Max  !!!

 

 

cool cool cool

 Dec 29, 2017

34 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.