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: 17

Answer

$44,261$

Work Step by Step

RECALL: in the hexadecimal system. $A= 10$, $C=12$, and $E=14$. Thus, the expanded form of $ACE5_{\text{sixteen}}$ is: $=(10 \times 16^3) + (12 \times 16^2)+(14 \times 16^1) + (5 \times 1) \\=(10 \times 4096) + (12 \times 256) + (14 \times 16) + 5 \\=40,960+3,072 + 224+5 \\=44,261$
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