Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 5 - Number Representation and Calculation - Chapter Summary, Review, and Test - Review Exercises: 23

Answer

$9348 = 111140_{\text{six}}$

Work Step by Step

The place values for base six are $...,6^6, 6^5, 6^4, 6^3, 6^2, 6^1, 1$. The place values that are less than or equal to $9348$ are $6^5, 6^4, 6^3, 6^2, 6^1, 1$. Divide $9348$ by $6^5$ or $7776$: $9348 \div 7776 = 1$ remainder $1572$ Divide the remainder $1572$ by $6^4$ or $1296$: $1572 \div 1296 =1$ remainder $276$ Divide the remainder $276$ by $6^3$ or $216$: $276 \div 216 =1$ remainder $60$ Divide the remainder $60$ by $6^2$ or $36$: $60 \div 36 = 1$ remainder $24$ Divide the remainder $24$ by $6^1$ or $6$: $24 \div 6 = 4$ remainder $0$ Divide the remainder $0$ by $1$: $0 \div 1 = 0$ Thus, $9348 = (1 \times 6^5) + (1 \times 6^4) + (1 \times 6^3) + (1 \times 6^2) +(4 \times 6^1)+ (0 \times 1) \\9348 = 111140_{\text{six}}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.