Thinking Mathematically (6th Edition)

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

Chapter 4 - Number Representation and Calculation - 4.2 Number Bases in Positional Systems - Exercise Set 4.2: 32

Answer

$41_{\text{six}}$

Work Step by Step

The place values of base six are: $,,,, 6^3, 6^2, 6^1, 1$. The place value that are less than $25$ are $6^1$, and $1$ Divide $25$ by $6^1$ or $6$ to obtain: $25 \div 6 = 4$ remainder $1$ Divide the remainder $1$ by $1$ to obtain: $1 \div 1 = 1$ remainder $0$ Thus, $25=(4 \times 6^1) + (1 \times 1) \\25=41_{\text{six}}$
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.