Thinking Mathematically (6th Edition)

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

Chapter 5 - Number Representation and Calculation - Chapter Summary, Review, and Test - Review Exercises - Page 244: 21

Answer

$473 = 122112_{\text{three}}$

Work Step by Step

The place values for base three are $...,3^5, 3^4, 3^3, 3^2, 3^1, 1$. The place values that are less than or equal to $473$ are $3^5, 3^4, 3^3, 3^2, 3^1, 1$. Divide $473$ by $3^5$ or $243$: $473 \div 243 = 1$ remainder $230$ Divide the remainder $230$ by $3^4$ or $81$: $230 \div 81 = 2$ remainder $68$ Divide the remainder $28$ by $3^3$ or $27$: $68 \div 27 = 2$ remainder $14$ Divide the remainder $14$ by $3^2$ or $9$: $14 \div 9 = 1$ remainder $5$ Divide the remainder $5$ by $3^1$ or $3$: $5 \div 3 = 1$ remainder $2$ Divide the remainder $2$ by $1$: $2 \div 1 = 2$ Thus, $473 = (1 \times 3^5) + (2 \times 3^4) + (2 \times 3^3) + (1 \times 3^2) +(1 \times 3^1)+ (2 \times 1) \\473 = 122112_{\text{three}}$
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