Topic Content:
- Converting Numbers in Other Bases to Base Ten
- Worked Examples
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:
Worked Example 3.3.1:
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 × 82 + 5 × 81 + 6 × 80
Base eight is also called the octal system
2. 10321four


Multiply each digit by its base number raised to the power of its place value
10321four = 1 × 44 + 0 × 43 + 3 × 42 + 2 × 41 + 1 × 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.
Worked Example 3.3.2:
Express the following binary numbers as numbers in base ten.
a. 110100two
b. 534six
c. 4307eight
a. 110100two
Solution
Using the expanding method
Place Value | 5 | 4 | 3 | 2 | 1 | 0 |
Binary Number | 1 | 1 | 0 | 1 | 0 | 0 |
110100ten = 1 × 25 + 1 × 24 + 0 × 23 + 1 × 22 + 0 × 21 + 0 × 20
= (1 × 25) + (1 × 24) + (0 × 23) + (1 × 22) + (0 × 21) + (0 × 20)
= (1 × 32) + (1 × 16) + 0 + (1 × 4) + 0 + 0
= 32 + 16 + 0 + 4 + 0 + 0
= 32 + 16 + 4
= 52ten
⇒ 110100two = 52ten
b. 534six
Solution
Using the expansion method:
Place Value | 2 | 1 | 0 |
Binary Number | 5 | 3 | 4 |
534six = 5 × 62 + 3 × 61 + 4 × 60
= (5 × 36) + (3 × 6) + (4 × 1)
= 180 + 18 + 4
= 202ten
⇒ 534six = 202ten
Note
62 = 6 × 6 = 36
61 = 6
60 = 1
c. 4307eight
Solution
Place Value | 3 | 2 | 1 | 0 |
Binary Number | 4 | 3 | 0 | 7 |
Using the expanding method: (same approach as Examples above)
4307eight = 4 × 83 + 3 × 82 + 0 × 81 + 7 × 80
= (4 × 512) + (3 × 64) + (0 × 8) + (7 × 1)
= 2048 + 192 + 0 + 7
= 2247ten
⇒ 4307eight = 2247ten
Worked Example 3.3.3:
Find the value of (111two)2 in base two
Solution
1st Step: The first thing to do is to convert the number to base ten.
2nd Step: After converting to base ten we will then square the value.
3rd Step: Finally we will convert the squared value back to base two.
1st Step:
Place Value | 2 | 1 | 0 |
Binary Number | 1 | 1 | 1 |
Using the expansion method;
111two = (1 × 22) + (1 × 21) + (1 × 20)
= 1 × 4 + 1 × 2 + 1 × 1
= 4 + 2 + 1
= 7ten
2nd Step:
Recall that (111two)2 = (7ten)2 = 7 × 7 = 49
= 49ten
3rd Step:
Then convert 49ten to base two (binary) using the method of repeated division

⇒ 49ten = 110001two