## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 4 - Number Representation and Calculation - 4.2 Number Bases in Positional Systems - Exercise Set 4.2: 33

#### Answer

$322_{\text{five}}$

#### Work Step by Step

The place values of base five are: $,,,, 5^3, 5^2, 5^1, 1$. The place values that are less than or equal to $87$ are $5^2$, $5^1$, and $1$ Divide $87$ by $5^2$ or $25$ to obtain: $87 \div 25 = 3$ remainder $12$ Divide the remainder $12$ by $5^1$ or $5$ to obtain: $12 \div 5 = 2$ remainder $2$ Divide the remainder $2$ by $1$ to obtain: $2 \div 1 = 2$ remainder $0$ Thus, $87=(3 \times 5^2) + (2 \times 5^1) + (2 \times 1) \\87=322_{\text{five}}$

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