## 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: 38

#### Answer

$10111_{\text{two}}$

#### Work Step by Step

The place values of base two are: $...,2^4, 2^3, 2^2, 2^1, 1$. The place values that are less than or equal to $23$ are $2^4, 2^3$, $2^2$, $2^1$, and $1$ Divide $23$ by $2^4$ or $16$ to obtain: $23 \div 16 = 1$ remainder $7$ Divide the remainder $7$ by $2^3$ or $8$ to obtain: $7 \div 8 = 0$ remainder $7$ Divide the remainder $7$ by $2^2$ or $4$ to obtain: $7\div 4 = 1$ remainder $3$ Divide the remainder $3$ by $2^1$ or $2$ to obtain: $3 \div 2 = 1$ remainder $1$ Divide the remainder $1$ by $1$ to obtain: $1 \div 1 = 1$ remainder $0$ Thus, $23=(1 \times 2^4) + (0 \times 2^3) + (1 \times 2^2) + (1 \times 2^1) + (1 \times 1) \\23=10111_{\text{two}}$

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