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: 41

Answer

$1011010_{\text{two}}$

Work Step by Step

The place values of base two are: $...,2^6, 2^5, 2^4, 2^3, 2^2, 2^1, 1$. The place values that are less than or equal to $90$ are $2^6, 2^5, 2^4, 2^3$, $2^2$, $2^1$, and $1$ Divide $90$ by $2^6$ or $64$ to obtain: $90 \div 64 = 1$ remainder $26$ Divide the remainder $26$ by $2^5$ or $32$ to obtain: $26 \div 32 = 0$ remainder $26$ Divide the remainder $26$ by $2^4$ or $16$ to obtain: $26 \div 16 = 1$ remainder $10$ Divide the remainder $10$ by $2^3$ or $8$ to obtain: $10 \div 8 = 1$ remainder $2$ Divide the remainder $2$ by $2^2$ or $4$ to obtain: $2 \div 4 = 0$ remainder $2$ Divide the remainder $2$ by $2^1$ or $2$ to obtain: $2 \div 2 = 1$ remainder $0$ Divide the remainder $0$ by $1$ to obtain: $0 \div 1 = 0$ remainder $0$ Thus, $90=(1 \times 2^6)+(0\times 2^5) + (1 \times 2^4) + (1 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (0 \times 1) \\90=1011010_{\text{two}}$
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