Thinking Mathematically (6th Edition)

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

Chapter 4 - Number Representation and Calculation - 4.2 Number Bases in Positional Systems - Exercise Set 4.2: 34

Answer

$151_{\text{seven}}$

Work Step by Step

The place values of base seven are: $,,,, 7^3, 7^2, 7^1, 1$. The place values that are less than or equal to $85$ are $7^2$, $7^1$, and $1$ Divide $85$ by $7^2$ or $49$ to obtain: $85 \div 49 = 1$ remainder $36$ Divide the remainder $36$ by $7^1$ or $7$ to obtain: $36 \div 7 = 5$ remainder $1$ Divide the remainder $1$ by $1$ to obtain: $1 \div 1 = 1$ remainder $0$ Thus, $85=(1 \times 7^2) + (5 \times 7^1) + (1 \times 1) \\85=151_{\text{seven}}$
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