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: 47

Answer

$4443_{\text{seven}}$

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 $1599$ are $7^3$, $7^2$, $7^1$, and $1$ Divide $1599$ by $7^3$ or $343$ to obtain: $1599 \div 343 = 4$ remainder $227$ Divide the remainder $227$ by $7^2$ or $49$ to obtain: $227 \div 49 = 4$ remainder $31$ Divide the remainder $31$ by $7^1$ or $7$ to obtain: $31 \div 7 = 4$ remainder $3$ Divide the remainder $r$ by $1$ to obtain: $3 \div 1 = 4$ remainder $0$ Thus, $1599=(4 \times 7^3) + (4 \times 7^2) + (4 \times 7^1) + (3 \times 1) \\1599=4443_{\text{seven}}$
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