Learn how counting works in base 2, 5, 7, 8, 10 and 16: convert in both directions, add, subtract and multiply in other bases, find an unknown base, and read the careful history of our own number system and zero.
Every number you write rests on a hidden choice: how many different digits you allow before you must carry into a new column. That choice is the base. We count in base 10, computers store everything in base 2, and some file codes use base 8. The quantities never change, only the way we write them. This guide teaches you to read and write numbers in any base, to convert between bases in both directions, to add, subtract and multiply in other bases, and to find an unknown base. It also tells the history of our own number system carefully, separating what is well established from what historians still debate.
The base (also called the radix) of a number system is the number of different digits it uses, and also the amount by which each column is worth more than the column on its right. In base 10 there are ten digits, 0 to 9, and the columns are worth 1, 10, 100, 1000 and so on. These are the powers of 10: 10⁰, 10¹, 10², 10³.
In any base b, the columns from the right are worth b⁰ = 1, b¹, b², b³ and so on. The digits allowed are 0 up to b − 1. When a column would reach the value b, you write 0 in it and carry 1 to the next column on the left, exactly as you do when 9 + 1 becomes 10 in decimal.
| Base | Name | Digits allowed | Column values (right to left) |
|---|---|---|---|
| 2 | Binary | 0, 1 | 1, 2, 4, 8, 16, 32 … |
| 5 | Quinary | 0 to 4 | 1, 5, 25, 125 … |
| 7 | Septenary | 0 to 6 | 1, 7, 49, 343 … |
| 8 | Octal | 0 to 7 | 1, 8, 64, 512 … |
| 10 | Decimal | 0 to 9 | 1, 10, 100, 1000 … |
| 16 | Hexadecimal | 0 to 9, then A to F | 1, 16, 256, 4096 … |
To show the base, we write it as a small number after the answer, called a subscript. So 101₂ is a binary number (it equals 5) while 101₁₀ is one hundred and one. If the base is not written, assume base 10. Hexadecimal needs six extra symbols for the values ten to fifteen: A = 10, B = 11, C = 12, D = 13, E = 14 and F = 15.
The best way to feel a base is to count in it. Notice what happens at the base itself: in base 5, the number after 4 is 10 (read “one zero”, meaning one five and no ones), and in binary the number after 1 is 10 (one two and no ones).
| Decimal | Binary (2) | Base 5 | Base 7 | Octal (8) |
|---|---|---|---|---|
| 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 | 10 | 7 |
| 8 | 1000 | 13 | 11 | 10 |
| 9 | 1001 | 14 | 12 | 11 |
| 10 | 1010 | 20 | 13 | 12 |
| 11 | 1011 | 21 | 14 | 13 |
| 12 | 1100 | 22 | 15 | 14 |
| 13 | 1101 | 23 | 16 | 15 |
| 14 | 1110 | 24 | 20 | 16 |
| 15 | 1111 | 30 | 21 | 17 |
| 16 | 10000 | 31 | 22 | 20 |
| 17 | 10001 | 32 | 23 | 21 |
| 18 | 10010 | 33 | 24 | 22 |
| 19 | 10011 | 34 | 25 | 23 |
| 20 | 10100 | 40 | 26 | 24 |
Look at the row for 8. In octal it is 10, the first number that needs two digits, and in base 7 the number 7 is also written 10. Every base writes its own base as 10.
Check each digit against the base.
The fix: octal uses only the digits 0 to 7. The digit 8 does not exist in base 8, just as base 10 has no single digit for “ten”. So 381 cannot be an octal number. In any base b, every digit must be smaller than b.
Multiply each digit by the value of its column, then add. This is called the expansion method, and it works for every base.
(a) 1101₂ = (1×8) + (1×4) + (0×2) + (1×1) = 8 + 4 + 0 + 1 = 13.
(b) 132₅ = (1×25) + (3×5) + (2×1) = 25 + 15 + 2 = 42.
(c) 245₇ = (2×49) + (4×7) + (5×1) = 98 + 28 + 5 = 131.
(d) 157₈ = (1×64) + (5×8) + (7×1) = 64 + 40 + 7 = 111.
2F₁₆ = (2×16) + (15×1) = 32 + 15 = 47. And FF₁₆ = (15×16) + 15 = 240 + 15 = 255, the biggest value that fits in 8 binary digits.
Binary digits look like decimal digits, but the columns are different.
The fix: 101₂ = (1×4) + (0×2) + (1×1) = 5. Always expand with the powers of the base you are given. Say “one zero one”, not “one hundred and one”.
Divide the number by the new base. Write down the quotient and the remainder. Divide the quotient again, and keep going until the quotient is 0. Then read the remainders from the bottom to the top.
(a) Write 100 in base 5. 100 ÷ 5 = 20 remainder 0; 20 ÷ 5 = 4 remainder 0; 4 ÷ 5 = 0 remainder 4. Reading upwards: 400₅. Check: 4 × 25 = 100.
(b) Write 200 in base 7. 200 ÷ 7 = 28 remainder 4; 28 ÷ 7 = 4 remainder 0; 4 ÷ 7 = 0 remainder 4. Reading upwards: 404₇. Check: 4×49 + 0×7 + 4 = 196 + 4 = 200.
(c) Write 255 in octal. 255 ÷ 8 = 31 remainder 7; 31 ÷ 8 = 3 remainder 7; 3 ÷ 8 = 0 remainder 3. Reading upwards: 377₈.
This method is quicker when the base is 2. List the column values, then take away the largest one that fits, put a 1 in that column, and continue with what is left.
Write 100 in binary. The columns are 64, 32, 16, 8, 4, 2, 1. The largest that fits is 64, leaving 36. Then 32 fits, leaving 4. Then 16 and 8 do not fit, 4 fits, leaving 0. So the digits are 64:1, 32:1, 16:0, 8:0, 4:1, 2:0, 1:0, giving 1100100₂.
The remainders come out in the wrong order.
The fix: the first remainder is the units digit, so it belongs on the right, and the last remainder belongs on the left. For 25, the remainders from top to bottom are 1, 0, 0, 1, 1. Reading them upwards gives 11001. Reading downwards gives 10011, which is 19, not 25. Check: 16 + 8 + 1 = 25.
The safe route goes through decimal: change to base 10 first, then to the new base.
Convert 213₅ to base 7. First, 213₅ = (2×25) + (1×5) + 3 = 50 + 5 + 3 = 58. Now 58 ÷ 7 = 8 remainder 2; 8 ÷ 7 = 1 remainder 1; 1 ÷ 7 = 0 remainder 1. Reading upwards: 112₇. Check: 49 + 7 + 2 = 58.
Because 8 = 2³, three binary digits make exactly one octal digit. Split a binary number into groups of three, starting from the right, adding zeros on the left if needed. Then convert each group separately.
(a) 1101011₂ becomes 001 101 011, so the octal number is 153₈. Check: 64 + 40 + 3 = 107 and 64 + 32 + 8 + 2 + 1 = 107.
(b) Write 45 in binary using octal. 45 ÷ 8 = 5 remainder 5, so 45 = 55₈. Replace each 5 by its binary group 101, giving 101101₂. Check: 32 + 8 + 4 + 1 = 45.
In the same way, groups of four binary digits make one hexadecimal digit, since 16 = 2⁴. This is why programmers like hexadecimal for long binary numbers.
The methods of column addition, subtraction and multiplication are exactly the same. The only change is where you carry: in base b you carry when a column total reaches b.
(a) Binary: 1011₂ + 110₂. Units: 1 + 0 = 1. Next: 1 + 1 = 2, which is 10 in binary, so write 0 and carry 1. Next: 0 + 1 + 1 (carry) = 2, so write 0 and carry 1. Next: 1 + 1 (carry) = 2, so write 0 and carry 1, giving a final 1. The answer is 10001₂. Check: 11 + 6 = 17 = 16 + 1.
(b) Base 5: 243₅ + 124₅. Units: 3 + 4 = 7 = 1×5 + 2, so write 2 and carry 1. Next: 4 + 2 + 1 = 7, so write 2 and carry 1. Next: 2 + 1 + 1 = 4. The answer is 422₅. Check: 73 + 39 = 112 = 4×25 + 2×5 + 2.
(c) Base 7: 56₇ + 34₇. Units: 6 + 4 = 10 = 1×7 + 3, so write 3 and carry 1. Next: 5 + 3 + 1 = 9 = 1×7 + 2, so write 2 and carry 1. Final digit 1. The answer is 123₇. Check: 41 + 25 = 66 = 49 + 14 + 3.
(d) Subtraction: 1101₂ − 101₂ = 1000₂, because 13 − 5 = 8.
101₂ × 11₂. Multiply by 1 in the units place: 101. Multiply by 1 in the twos place, which shifts one place left: 1010. Add: 101 + 1010 = 1111₂. Check: 5 × 3 = 15.
Multiplying in binary only ever copies the number or shifts it left. That is why binary is so simple for electronic circuits.
When the base is the unknown, treat it like a letter b and write the expansion.
(a) The number 23 in base b equals 17 in decimal. Then 2b + 3 = 17, so 2b = 14 and b = 7. (The digit 3 is smaller than 7, so the base is allowed.)
(b) 31 in base b equals 16. Then 3b + 1 = 16, so 3b = 15 and b = 5.
(c) 100 in base b equals 49. Then b² = 49, so b = 7. We reject b = −7, because a base must be positive.
(d) 42 in base b equals 26. Then 4b + 2 = 26, so b = 6. Always check that every digit is smaller than your answer: 4 and 2 are both smaller than 6.
Historians agree on the broad story, but some details are uncertain, so this section says so where that is the case.
Where base 10 comes from. The ten digits of our system are called the Hindu-Arabic numerals. The idea of a place-value system with a symbol for zero was developed in India, over several centuries. Brahmi numerals, used in India from about the 3rd century BCE, were an early ancestor of our digits, although they were not yet a place-value system. The usual explanation for counting in tens is that people have ten fingers, which is likely but cannot be proved.
Zero as a placeholder. Babylonian scribes, working in base 60, began to use a mark to show an empty column in the last few centuries BCE, so that numbers were not misread. The Maya, independently, had a zero symbol in their calendar counts. In these cases zero marked an empty place, but it was not yet a number used in calculation.
Zero as a number. The Indian mathematician Brahmagupta, in his book Brahmasphutasiddhanta of 628 CE, wrote rules for adding, subtracting and multiplying with zero and with negative numbers. His rules for dividing by zero were not correct by today’s standards, but he is rightly famous for treating zero as a number.
The Bakhshali manuscript is an ancient Indian text written on birch bark that uses a dot as a placeholder. In 2017, radiocarbon dating at the University of Oxford suggested that its oldest pages were much older than scholars had thought, possibly from about the 3rd or 4th century CE. Experts still debate what this means for the history of zero.
Travel to the Islamic world and Europe. The Persian scholar Muhammad ibn Musa al-Khwarizmi worked in Baghdad in the early 9th century CE and wrote about calculating with the Indian numerals. His name is the source of the English word algorithm. The Italian merchant-mathematician Leonardo of Pisa, called Fibonacci, learned the system while living in North Africa and explained it to European readers in his book Liber Abaci in 1202. It took several more centuries for it to replace Roman numerals in everyday use. The word zero comes from the Arabic sifr (“empty”) through Italian, and the same Arabic word is the source of the English word cipher.
Binary. The German scholar Gottfried Leibniz published a clear account of binary arithmetic in 1703. Binary became the basis of electronic computers in the 20th century.
The ancient Babylonians used base 60. Many historians link this to our 60 minutes in an hour and 360 degrees in a circle, which reached us through later Greek astronomers. The Maya counted in twenties. Some peoples have used base 5 or base 20, perhaps because we have five fingers on a hand and twenty fingers and toes in all.
Type a decimal number, or tap a number below. You see it in five bases at once, and the repeated-division steps for the base you choose.
Type a number and its base. The checker tells you whether every digit is allowed, then shows the expansion and the decimal value. In bases above 10 use the letters A to F.
The digits 6 and 7 do not exist in base 5.
The fix: that is a decimal sum. In base 5, carry whenever a column reaches 5: 3 + 4 = 7 = 1×5 + 2, so write 2 and carry 1, and so on, giving 422₅. Any answer written in base 5 must contain only the digits 0 to 4.
Tap the correct answer for each question. Your best score is saved on this device.
1. 1010₂ = 8 + 0 + 2 + 0 = 10. 34₅ = 3×5 + 4 = 19, and 19 = 16 + 2 + 1, so 10011₂.
2. 45 ÷ 8 = 5 remainder 5, so 55₈. In binary, 5 = 101, so 101101₂ (check: 32 + 8 + 4 + 1 = 45).
3. b² = 36, so b = 6.
1. (2×27) + (1×9) + (0×3) + 1 = 54 + 9 + 0 + 1 = 64.
2. 100 ÷ 6 = 16 remainder 4; 16 ÷ 6 = 2 remainder 4; 2 ÷ 6 = 0 remainder 2. Reading upwards: 244₆. Check: 2×36 + 4×6 + 4 = 72 + 24 + 4 = 100.
3. 1011₂ = 11 and 11₂ = 3. 11 + 3 = 14 = 8 + 4 + 2, so 1110₂.
4. (a) 45 ÷ 8 = 5 remainder 5, so 55₈. (b) 45 = 32 + 8 + 4 + 1, so 101101₂.
5. 1001₂ × 11₂: 1001 + 10010 = 11011. So the answer is 11011₂. Check: 9 × 3 = 27 = 16 + 8 + 2 + 1.
6. b² + 1 = 26, so b² = 25 and b = 5 (a base cannot be negative). The digits 1 and 0 are smaller than 5, so the base is valid.
7. (a) 255 ÷ 8 = 31 remainder 7; 31 ÷ 8 = 3 remainder 7; 3 ÷ 8 = 0 remainder 3, so 377₈. (b) 2×64 + 0×8 + 5 = 133. (c) Three binary digits give 2³ = 8 different patterns, 000 to 111, and octal has exactly eight digits, 0 to 7, so each pattern matches one octal digit.
Learners often feel that decimal is the only “real” way to write numbers. A strong way to break this is to start with counting: ask them to count to twenty in base 5 on one hand, writing each number as they go. They then discover carrying for themselves, and place value becomes an idea they have used, not a rule they were given. Use concrete objects such as counters grouped in fives or twos. Ask learners to write the column values above each digit every time; this habit removes most errors. The Base Converter on this page shows the repeated-division steps, so learners can compare their working with a correct one. When teaching the history, stress that mathematics was built by many cultures over many centuries, and that historians sometimes disagree about dates. This is a good example of how to treat evidence honestly.
For further reading, look at the number-base topic in your own examination syllabus and past papers, as not every board includes it. Reputable sources on the history of numerals include the MacTutor History of Mathematics archive of the University of St Andrews. This article is original EDUSAMBAM writing for students up to O-Level and similar courses.
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