## 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: 43

#### Answer

$12010_{\text{three}}$

#### Work Step by Step

The place values of base three are: $...,3^5, 3^4, 3^3, 3^2, 3^1, 1$. The place values that are less than or equal to $138$ are $3^4, 3^3$, $3^2$, $3^1$, and $1$ Divide $138$ by $3^4$ or $81$ to obtain: $138 \div 81 = 1$ remainder $57$ Divide the remainder $57$ by $3^3$ or $27$ to obtain: $57 \div 27 = 2$ remainder $3$ Divide the remainder $3$ by $3^2$ or $9$ to obtain: $3 \div 9 = 0$ remainder $3$ Divide the remainder $3$ by $3^1$ or $3$ to obtain: $3 \div 3 = 1$ remainder $0$ Divide the remainder $0$ by $1$ to obtain: $0\div 1 = 0$ remainder $0$ Thus, $138=(1 \times 3^4) + (2 \times 3^3) + (0 \times 3^2) + (1 \times 3^1) + (0 \times 1) \\138=12010_{\text{three}}$

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