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: 20

Answer

$21 = 10101_{\text{two}}$

Work Step by Step

The place values for base two are $...,2^5, 2^4, 2^3, 2^2, 2^1, 1$. The place values that are less than $21$ are $2^4, 2^3, 2^2, 2^1, 1$. Divide $21$ by $2^4$ or $16$: $21\div 16 = 1$ remainder $5$ Divide the remainder $5$ by $2^3$ or $8$: $5 \div 8 = 0$ remainder $5$ Divide the remainder $5$ by $2^2$ or $4$: $5 \div 4 = 1$ remainder 1 Divide the remainder $2^1$ by $2$: $1 \div 2 = 0$ remainder $1$ Divide the remainder $1$ by $1$: $1 \div 1 = 1$ Thus, $21 = (1 \times 2^4) + (0 \times 2^3) + (1 \times 2^2) +(0 \times 2^1)+ (1 \times 1) \\21 = 10101_{\text{two}}$
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