## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 4 - Number Representation and Calculation - 4.2 Number Bases in Positional Systems - Exercise Set 4.2 - Page 226: 39

#### Answer

$111001_{\text{two}}$

#### Work Step by Step

The place values of base two are: $...,2^5, 2^4, 2^3, 2^2, 2^1, 1$. The place values that are less than or equal to $57$ are $2^5, 2^4, 2^3$, $2^2$, $2^1$, and $1$ Divide $57$ by $2^5$ or $32$ to obtain: $57 \div 32 = 1$ remainder $25$ Divide the remainder $25$ by $2^4$ or $16$ to obtain: $25 \div 16 = 1$ remainder $9$ Divide the remainder $9$ by $2^3$ or $8$ to obtain: $9 \div 8 = 1$ remainder $1$ Divide the remainder $1$ by $2^2$ or $4$ to obtain: $1\div 4 = 0$ remainder $1$ Divide the remainder $1$ by $2^1$ or $2$ to obtain: $1 \div 2 = 0$ remainder $1$ Divide the remainder $1$ by $1$ to obtain: $1 \div 1 = 1$ remainder $0$ Thus, $57=(1\times 2^5) + (1 \times 2^4) + (1 \times 2^3) + (0 \times 2^2) + 01 \times 2^1) + (1 \times 1) \\57=111001_{\text{two}}$

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