## 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: 59

#### Answer

$623_{\text{eight}}$

#### 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. $110010011_{\text{two}} \\=(1\times 2^8) + (1 \times 2^7)+(0\times 2^6) + (0 \times 2^5)+(1\times 2^4)+(0\times 2^3) + (0\times 2^2) + (1\times 2^1)+(1\times 1) \\=(1 \times 256) +(1\times 128) + 0+0+(1\times 16) +0+0+(1\times 2)+1 \\=256+128+16+2+1 \\=403$ (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 403 are: $8^2, 8^1,$ and $1$. Divide $403$ by $8^2$ or $64$: $403 \div 64 = 6$ remainder $19$ Divide $19$ by $8^1$ or $8$: $19 \div 8 = 2$ remainder $3$ Divide $3$ by $1$: $3 \div 1 =3$ Thus, $403 = (6 \times 8^2) + (2 \times 8^1) + (3 \times 1) \\403= 623_{\text{eight}}$

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