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Lesson 3, Topic 3
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# Converting Numbers in Other Bases to Base Ten

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A given number can be converted from any base to base 10 using the Expansion Method.

Once you are able to work out the place values, you can easily convert other base numbers to binary.

Let’s take a look at the following examples;

Write the following numbers in expanded form.

1. 356eight
2. 10321four

Solution

1st Step: To use the expansion method for conversion we first find the place numbers.

1. 356eight

Students often find it difficult to locate the place values. The easiest way is to start from the last digit to the first digit. The last Digit always starts with a 0, the digit that follows a 1, and so on. (0, 1 2, 3, etc)

2nd Step: Multiply each digit by its base number raised to the power of its place value

356eight  = 3 x 82 + 5 x 81 + 6 x 80

Base eight is also called octal system

2. 10321four

Multiply each digit by its base number raised to the power of its place value

10321four = 1 x 44 + 0 x 43 + 3 x 42 + 2 x 41 + 1 x 40

To convert any number base to base ten, simply express it as the power of the given base. As discussed earlier the place value is important here

Example 1

Express the following binary numbers as numbers in base ten.

a. 110100two

b. 534six

c. 4307eight

Solution

a. 110100two

Using the expanding method

110100ten = 1 x 25 + 1 x 24 + 0 x 23 + 1 x 22 + 0 x 21 + 0 x 20

= (1 x 25) + (1 x 24) + (0 x 23) + (1 x 22) + (0 x 21) + (0 x 20)

= (1 x 32) + (1 x 16) + 0 + (1 x 4) + 0 + 0

= 32 + 16 + 0 + 4 + 0 + 0

= 32 + 16 + 4

= 52ten

110100two = 52ten

b. 534six

Using the expansion method:

534six = 5 x 62 + 3 x 61 + 4 x 60

Note

62 = 6 x 6 = 36

61 = 6

60 = 1

= (5 x 36) + (3 x 6) + (4 x 1)

= 180 + 18 + 4

= 202ten

534six  = 202ten

c. 4307eight

Using the expanding method: (same approach as Examples above)

4307eight = 4 x 83 + 3 x 82 + 0 x 81 + 7 x 80

= (4 x 512) + (3 x 64) + (0 x 8) + (7 x 1)

= 2048 + 192 + 0 + 7

= 2247ten

:- 4307eight = 2247ten

Example 2

Find the value of (111two)2 in base two

Using the expansion method;

111two = (1 x 22) + (1 x 21) + (1 x 20

= 1 x 4 + 1 x 2 + 1 x 1

= 4 + 2 + 1

= 7ten

Recall that (111two)2 = (7ten)2 = 7 x 7 = 49

= 49ten

Then convert 49ten to base two (binary) using the method of repeated division

49ten = 110001two error: