## 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: 60

#### Answer

$561_{\text{eight}}$

#### Work Step by Step

To convert a numeral in base two to a numeral in base eight, the following steps must be performed: (1) Convert the numeral in base two to base ten. $101110001_{\text{two}} \\=(1\times 2^8) + (0 \times 2^7)+(1\times 2^6) + (1 \times 2^5)+(1\times 2^4)+(0\times 2^3) + (0\times 2^2) + (0\times 2^1)+(1\times 1) \\=(1 \times 256) +0 + (1 \times 64)+(1 \times 32)+(1\times 16) +0+0+0+1 \\=256+64+32+16+1 \\=369$ (2) Convert the result in Step (1) to a numeral in base eight using division. The place values in base eight are: $8^3, 8^2, 8^1, 1$. The place values less than or equal to 369 are: $8^2, 8^1,$ and $1$. Divide $369$ by $8^2$ or $64$: $369 \div 64 = 5$ remainder $49$ Divide $49$ by $8^1$ or $8$: $49 \div 8 = 6$ remainder $1$ Divide $1$ by $1$: $1 \div 1 =1$ Thus, $369 = (5 \times 8^2) + (6 \times 8^1) + (1 \times 1) \\369= 561_{\text{eight}}$

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