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: 46

Answer

$525_{\text{nine}}$

Work Step by Step

The place values of base nine are: $...,9^3, 9^2, 9^1, 1$. The place values that are less than or equal to $428$ are $9^2$, $9^1$, and $1$ Divide $428$ by $9^2$ or $81$ to obtain: $428 \div 81 = 5$ remainder $23$ Divide the remainder $23$ by $9^1$ or $9$ to obtain: $23 \div 9 = 2$ remainder $5$ Divide the remainder $5$ by $1$ to obtain: $5\div 1 = 5$ remainder $0$ Thus, $428=(5 \times 9^2) + (2 \times 9^1) + (5 \times 1) \\428=525_{\text{nine}}$
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