Thinking Mathematically (6th Edition)

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

Chapter 4 - Number Representation and Calculation - 4.3 Computation in Positional Systems - Exercise Set 4.3 - Page 235: 63

Answer

Required solution is\[{{56}_{\text{seven}}}\].

Work Step by Step

Let's calculate: First, write the multiples of \[31\] in base seven. \[\begin{align} & 31\times 1=31 \\ & 31\times 2=62 \\ & 31\times 3(62+31)=123 \\ & 31\times 4(123+31)=154 \end{align}\] \[31\times 5(154+31)=215\] \[31\times 6(215+31)=246...\] ….. (1) Now, divide the given base seven numerals. \[\begin{align} & {{31}_{\text{seven}}}\overset{{{56}_{\text{seven}}}}{\overline{\left){\begin{align} & {{2426}_{\text{seven}}} \\ & \underline{215} \\ \end{align}}\right.}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,246 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{246} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\,\,\,\,\,\,0} \\ \end{align}\] Here, first divide 242 by 31, highest number less than \[{{242}_{\text{seven}}}\]is \[{{215}_{\text{seven}}}\]from the table of 31.
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