Converting between bases – Number Bases (Senior 3 Mathematics) | Ganzaa

Number Bases

Lessons

Senior 3 · Mathematics

Number Bases

5 lessons

Lesson 2 of 5

Converting between bases

24 min

Writing 2026 in base two

In base two the year 2026 is written $11111101010_{2}$ . Converting into base ten, as in the last lesson, means multiplying by powers of the base and adding. Converting out of base ten reverses this: you divide by the base again and again, and the remainders are the digits. This lesson also shows how to move between two bases that are not ten, and a shortcut that links bases 2, 8 and 16.

Definition

Octal and hexadecimal: the names of base eight and base sixteen. Binary is base two, and decimal is base ten.

Notation: $3752_{8}$ is an octal numeral; $7EA_{16}$ is a hexadecimal (often shortened to "hex") numeral.

Example: $17_{8} = 8 + 7 = 15$ , and 15 in hexadecimal is the single digit $\mathrm{F}$ .

Important

Rule (base ten to base $b$ ). Divide the number by $b$ and write down the remainder. Divide the quotient by $b$ again, and repeat until the quotient is 0. The remainders, read from the last to the first, are the digits of the number in base $b$ .

Why: each division writes a number as $\text{quotient} \times b + \text{remainder}$ . So $N = q_{0} b + r_{0}$ , then $q_{0} = q_{1} b + r_{1}$ , and so on. Substituting, $N = r_0 + r_1 b + r_2 b^2 + \cdots$ : the first remainder is the ones digit, the second is the $b$ s digit, and the last remainder is the leading digit.

Worked example 1: 2026 in base two

Division	Quotient	Remainder
$2026 \div 2$	1013	0
$1013 \div 2$	506	1
$506 \div 2$	253	0
$253 \div 2$	126	1
$126 \div 2$	63	0
$63 \div 2$	31	1
$31 \div 2$	15	1
$15 \div 2$	7	1
$7 \div 2$	3	1
$3 \div 2$	1	1
$1 \div 2$	0	1

Read the remainders from the bottom up: $2026 = 11111101010_{2}$ .

Check: $1024 + 512 + 256 + 128 + 64 + 32 + 8 + 2 = 2026$ . ✓

Worked example 2: 2026 in base five and in base sixteen

Base five

Division	Quotient	Remainder
$2026 \div 5$	405	1
$405 \div 5$	81	0
$81 \div 5$	16	1
$16 \div 5$	3	1
$3 \div 5$	0	3

So $2026 = 31101_{5}$ . Check: $3 \times 625 + 1 \times 125 + 1 \times 25 + 0 \times 5 + 1 = 1875 + 125 + 25 + 1 = 2026$ . ✓

Base sixteen (remainders of 10 or more become letters)

Division	Quotient	Remainder
$2026 \div 16$	126	10 (= A)
$126 \div 16$	7	14 (= E)
$7 \div 16$	0	7

So $2026 = 7EA_{16}$ . Check: $7 \times 256 + 14 \times 16 + 10 = 1792 + 224 + 10 = 2026$ . ✓

A conversion table

Learn to read this table in every direction. Notice where each base "rolls over" to a new place: at 2, 4, 8 and 16 in base two, at 5 in base five, at 8 in base eight and at 16 in base sixteen.

Base ten	Base two	Base five	Base eight	Base sixteen
0	0	0	0	0
1	1	1	1	1
2	10	2	2	2
3	11	3	3	3
4	100	4	4	4
5	101	10	5	5
6	110	11	6	6
7	111	12	7	7
8	1000	13	10	8
9	1001	14	11	9
10	1010	20	12	A
11	1011	21	13	B
12	1100	22	14	C
13	1101	23	15	D
14	1110	24	16	E
15	1111	30	17	F
16	10000	31	20	10

Worked example 3: between two bases that are not ten

Convert $1324_{5}$ to base eight.

Solution Go through base ten. First, $1324_5 = 1 \times 125 + 3 \times 25 + 2 \times 5 + 4 = 125 + 75 + 10 + 4 = 214$ . Then divide by 8:

Division	Quotient	Remainder
$214 \div 8$	26	6
$26 \div 8$	3	2
$3 \div 8$	0	3

So $1324_{5} = 326_{8}$ . Check: $3 \times 64 + 2 \times 8 + 6 = 192 + 16 + 6 = 214$ . ✓

The shortcut between bases 2, 8 and 16

Because $8 = 2^{3}$ , three binary places together are worth exactly one octal place; because $16 = 2^{4}$ , four binary places are worth one hexadecimal place. So each octal digit is a block of 3 bits, and each hexadecimal digit is a block of 4 bits.

Octal digit	3 bits	Hex digit	4 bits	Hex digit	4 bits
0	000	8	1000	0	0000
1	001	9	1001	1	0001
2	010	A	1010	2	0010
3	011	B	1011	3	0011
4	100	C	1100	4	0100
5	101	D	1101	5	0101
6	110	E	1110	6	0110
7	111	F	1111	7	0111

Worked example 4: grouping bits

(a) Write $11111101010_{2}$ in octal and in hexadecimal. (b) Write $2\mathrm{B}_{16}$ in binary.

Solution

(a) Octal: group the bits in threes from the right: $11 \;|\; 111 \;|\; 101 \;|\; 010$ , which gives $3,\ 7,\ 5,\ 2$ . So $11111101010_{2} = 3752_{8}$ .

Hexadecimal: group in fours from the right: $111 \;|\; 1110 \;|\; 1010$ , which gives $7,\ \mathrm{E},\ \mathrm{A}$ . So $11111101010_2 = 7EA_{16}$ , the same as in Worked example 2. ✓

(b) Replace each hex digit by 4 bits: $2 = 0010$ and $\mathrm{B} = 1011$ . So $2\mathrm{B}_{16} = 0010\,1011_2 = 101011_2$ (the leading zeros can be dropped). Check: $2 \times 16 + 11 = 43 = 32 + 8 + 2 + 1$ . ✓

Each octal digit is a block of 3 bits and each hexadecimal digit a block of 4 bits, so $11111101010_2 = 3752_8 = \mathrm{7EA}_{16}$ without going through base ten.

Warning

Two direction mistakes. Read the remainders from the bottom up: the first remainder is the ones digit, not the leading digit. And group bits from the right, adding zeros on the left if the last group is short: $11\,111_2$ is $37_{8}$ , not $76_{8}$ .

Tip

Always convert back. After converting out of base ten, turn your answer back into base ten with place values. If you do not get the number you started with, look for a remainder written in the wrong order.

Khan Academy: Converting from decimal to binary (4 min). Watch how a base-ten number is built from the largest power of 2 that fits, one place at a time. It is a second method you can use to check the repeated-division method in this lesson.

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Practice

Check every answer by converting back. The last three are harder.

Convert 45 to base two.
Convert 100 to base five.
Convert 500 to base eight.
Convert 700 to base sixteen.
Convert $231_{5}$ to base two.
Write $1101110_{2}$ in octal and in hexadecimal, without going through base ten.
Write $3\mathrm{F}_{16}$ in binary and in octal.
Convert $432_{5}$ to base eight.

Show answersHide answers

$101101_{2}$
$400_{5}$ ( $4 \times 25 = 100$ )
$764_{8}$ ( $7 \times 64 + 6 \times 8 + 4 = 500$ )
$2BC_{16}$ ( $2 \times 256 + 11 \times 16 + 12 = 700$ )
$231_{5} = 66 = 1000010_{2}$
$1\,|\,101\,|\,110 \to 156_8$ ; $110\,|\,1110 \to 6\mathrm{E}_{16}$ . Both equal 110.
$3 = 0011$ , $\mathrm{F} = 1111$ : $111111_{2}$ ; regroup in threes: $111\,|\,111 \to 77_8$ . All equal 63.
$432_{5} = 100 + 15 + 2 = 117$ ; $117 = 165_{8}$ .

Sources and credits

4.2: Converting with Base Systems, from Contemporary Mathematics by Donna Kirk and contributors, OpenStax, provided by LibreTexts, licensed under CC BY 4.0 (licence read on the page, 30 September 2026). Adapted by Ganzaa.
The division tables, the conversion table and the bit-grouping shortcut. Written by Ganzaa.
Video: Converting from decimal to binary, Khan Academy, embedded from YouTube.

The Rwandan worked examples, the diagrams, the practice items and the quiz questions were written by Ganzaa.

1Writing 2026 in base two

2A conversion table

3The shortcut between bases 2, 8 and 16

Writing 2026 in base two

A conversion table

The shortcut between bases 2, 8 and 16