Number Bases

Lessons

Senior 3 · Mathematics

Number Bases

5 lessons

Lesson 1 of 5

Numerals, digits and bases

22 min

A clay tablet that counts in sixties

The photograph below shows a small clay tablet from Mesopotamia (in modern Iraq), written in the Old Babylonian period, more than 3,500 years ago. Across the diagonal of the square the writer pressed four groups of wedge marks, which modern scholars read as 1, 24, 51 and 10. They are the digits of a number written in base sixty:

1 + \frac{24}{60} + \frac{51}{60^2} + \frac{10}{60^3} \approx 1.414213,

which is $\sqrt{2}$ correct to five decimal places. We still use sixties when we tell the time: 60 seconds make a minute and 60 minutes make an hour. Grouping in tens is only one choice. This unit studies numerals written with other groupings, above all base 2, the language of every computer and phone.

Definition

Number, numeral and digit. A number is an amount or quantity. A numeral is a written name for a number. A digit is one of the single symbols used to write numerals.

Notation: the same number can have many numerals: $7$ , $111_{2}$ , $12_{5}$ and the Roman VII all name the number of days in a week.

Example: the numeral 305 is made of the digits 3, 0 and 5.

Definition

Base (or radix) of a place-value system: the number of units of one place that are exchanged for one unit of the next place to the left. A system in base $b$ uses exactly $b$ digits: $0, 1, 2, \ldots, b - 1$ .

Notation: the base is written as a small subscript after the numeral. $243_{5}$ is read "two four three, base five". A numeral with no subscript is in base ten.

Example: base five uses only the digits 0, 1, 2, 3 and 4; base two (binary) uses only 0 and 1.

Grouping and place value

To write 23 counters in base five, put them in groups of five. There are 4 full groups and 3 counters left over, so 23 is written $43_{5}$ : 4 fives and 3 ones.

$23 = 43_{5}$ : the left digit counts groups of five, the right digit counts ones.

With more counters, five groups of five are bundled into one "twenty-five", five twenty-fives into one "hundred and twenty-five", and so on. So in base $b$ the places, from the right, are worth $b^{0} = 1$ , $b^{1}$ , $b^{2}$ , $b^{3}$ , …

Base	$b^{4}$	$b^{3}$	$b^{2}$	$b^{1}$	$b^{0}$	Digits used
2	16	8	4	2	1	0, 1
5	625	125	25	5	1	0 to 4
8	4096	512	64	8	1	0 to 7
10	10000	1000	100	10	1	0 to 9
16	65536	4096	256	16	1	0 to 9, A to F

Important

Rule (value of a numeral in base $b$ ). The numeral $d_n d_{n-1} \ldots d_1 d_0$ in base $b$ stands for

d_n \times b^n + d_{n-1} \times b^{n-1} + \cdots + d_1 \times b + d_0.

Why: $b$ units of any place are exchanged for 1 unit of the next place, so each place is worth $b$ times the place on its right. Starting from 1 on the right, the places are worth $1, b, b^2, b^3, \ldots$ , and each digit says how many of that place there are.

Definition

Hexadecimal digits: base sixteen needs sixteen single digits, so after 0 to 9 it uses letters: $\mathrm{A} = 10$ , $\mathrm{B} = 11$ , $\mathrm{C} = 12$ , $\mathrm{D} = 13$ , $\mathrm{E} = 14$ , $\mathrm{F} = 15$ .

Notation: $2\mathrm{F}_{16}$ means 2 sixteens and 15 ones.

Example: $2\mathrm{F}_{16} = 2 \times 16 + 15 = 47$ .

Worked example 1: converting to base ten

Write each numeral in base ten: (a) $243_{5}$ (b) $101101_{2}$ (c) $376_{8}$ (d) $1\mathrm{A}7_{16}$ .

Solution Multiply each digit by its place value and add.

\begin{aligned} \text{(a)}\;\; 243_5 &= 2 \times 25 + 4 \times 5 + 3 = 50 + 20 + 3 = 73 \\ \text{(b)}\;\; 101101_2 &= 32 + 8 + 4 + 1 = 45 \\ \text{(c)}\;\; 376_8 &= 3 \times 64 + 7 \times 8 + 6 = 192 + 56 + 6 = 254 \\ \text{(d)}\;\; 1\mathrm{A}7_{16} &= 1 \times 256 + 10 \times 16 + 7 = 256 + 160 + 7 = 423 \end{aligned}

In (b) only the places holding a 1 count: $2^{5} = 32$ , $2^{3} = 8$ , $2^{2} = 4$ and $2^{0} = 1$ .

Worked example 2: a faster way, digit by digit

Convert $3142_{5}$ to base ten without working out powers of 5.

Solution Start with the first digit. Repeatedly multiply by the base and add the next digit:

3 \;\xrightarrow{\times 5 + 1}\; 16 \;\xrightarrow{\times 5 + 4}\; 84 \;\xrightarrow{\times 5 + 2}\; 422.

So $3142_{5} = 422$ . Check with place values: $3 \times 125 + 1 \times 25 + 4 \times 5 + 2 = 375 + 25 + 20 + 2 = 422$ . ✓ This is the same "multiply and add" pattern as synthetic division in the unit on quadratic equations, with $b$ in place of $r$ .

Worked example 3: which numerals are possible?

(a) Which of $352_{5}$ , $1021_{2}$ , $706_{8}$ and $9G_{16}$ are valid numerals? (b) What is the smallest base in which $4063$ could be a numeral?

Solution

(a) A digit must be smaller than the base. $352_{5}$ contains 5, and $1021_{2}$ contains 2, so both are invalid. $706_{8}$ is valid (all digits are below 8). $9G_{16}$ is invalid, because the hexadecimal digits stop at F.

(b) The largest digit is 6, so the base must be at least $6 + 1 = 7$ .

Warning

"10" is the base itself. In every base $b$ , the numeral $10_{b}$ means one $b$ and no ones, that is $b$ : $10_{2} = 2$ , $10_{5} = 5$ , $10_{16} = 16$ . So never read $101_{2}$ as "one hundred and one"; it is five.

Tip

Read the digits one at a time. Say "one zero one, base two", not "one hundred and one". Words like "hundred" and "thousand" belong to base ten only, and using them in another base leads to wrong values.

Khan Academy: Introduction to number systems and binary (10 min). Watch how the place values of base ten are rebuilt as powers of 2 for binary, and how a binary numeral is turned into its base-ten value.

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Practice

The last three questions are harder.

Write the place values of the first four places, from the right, in base 7.
Convert $314_{5}$ to base ten.
Convert $11011_{2}$ to base ten.
Convert $157_{8}$ to base ten.
Convert $\mathrm{B}4_{16}$ to base ten.
Which of these is not a valid numeral: $2012_{3}$ , $777_{8}$ , $5\mathrm{A}_{11}$ , $406_{5}$ ? Explain.
Which is larger, $1000_{2}$ or $12_{5}$ ?
What is the smallest base in which 3504 could be a numeral? Find its base-ten value in that base.

Show answersHide answers

1, 7, 49, 343
$3 \times 25 + 1 \times 5 + 4 = 84$
$16 + 8 + 2 + 1 = 27$
$64 + 5 \times 8 + 7 = 111$
$11 \times 16 + 4 = 180$
$406_{5}$ , because the digit 6 is not smaller than the base 5. (In $5\mathrm{A}_{11}$ , $\mathrm{A} = 10 \lt 11$ , so it is valid.)
$1000_{2} = 8$ and $12_{5} = 7$ , so $1000_{2}$ is larger.
Base 6 (the largest digit is 5). $3 \times 216 + 5 \times 36 + 0 \times 6 + 4 = 648 + 180 + 4 = 832$ .

Sources and credits

4.0: Hindu-Arabic Positional System, 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.
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.
Photo: YBC-7289-OBV-labeled.jpg, A. Urcia, Yale Peabody Museum of Natural History, labels by Theodor Langhorne Franklin, Wikimedia Commons, released under CC0. Used unchanged.
The grouping diagram, the digit-by-digit method and the table of place values. Written by Ganzaa.
Video: Introduction to number systems and binary, Khan Academy, embedded from YouTube.

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

1A clay tablet that counts in sixties

2Grouping and place value

A clay tablet that counts in sixties

Grouping and place value