Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 4 - Number Representation and Calculation - Chapter Summary, Review, and Test - Review Exercises - Page 245: 35



Work Step by Step

To multiply two numbers those are not of base 10, follow the below steps, To multiply such numbers, multiply just as it is done in base ten. That is first multiply the right most column numbers and continue same as it is done for base ten but in base four there are only 0,1,2, 3 and 4, if multiplication exceeds 4, then change it in base five. \[\begin{align} {{123}_{five}} & \\ \times {{4}_{five}} & \\ \end{align}\] To solve this, follow the below step, \[\begin{align} & {{3}_{five}}\times {{4}_{five}}={{12}_{ten}} \\ & {{4}_{five}}\times {{2}_{five}}={{8}_{ten}} \\ & {{4}_{five}}\times {{1}_{five}}={{4}_{five}} \end{align}\] And here 12 and 8 are not in base five, so convert it. \[{{12}_{ten}}=\left( 5\times 2 \right)+\left( 2\times 1 \right)={{22}_{five}}\] As 2 is the carry part, add 2 to 8, that is \[8+2=10\] \[{{10}_{ten}}=\left( 5\times 2 \right)+\left( 0\times 1 \right)={{20}_{five}}\] As 2 is the carry part, add 2 to 4, that is \[4+2=6\] \[{{6}_{ten}}=\left( 5\times 1 \right)+\left( 1\times 1 \right)={{11}_{five}}\] \[\begin{align} & \underline{\begin{align} {{123}_{five}} & \\ \times {{4}_{five}} & \\ \end{align}} \\ & \underline{{{1102}_{five}}} \\ \end{align}\] So, here is the solution of the multiplication.
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