MathsCBSE Class 8 · A Story of Numbers

Bases and Place Value Systems for Class 8

Updated 17 October 2026 · Fact-checked

A base-n number system has landmark numbers 1, n, n squared, n cubed, and so on. A place value system uses the position of each symbol to show its landmark number, so only a few symbols are needed, and a placeholder zero removes ambiguity. Exams ask for conversions between decimal and other bases, regrouping in addition, and the reasons for place value and zero.

Each landmark is 5 times the one before; the position of a digit tells its landmark, and the 0 holds the empty 25 place, so 143 is 1033 in base 5.

Illustration drawn for QuestionBuddy.

Understand Bases and Place Value Systems

**Base- $n$ systems.** In the Egyptian system each landmark number is 10 times the one before. Replace 10 by any positive integer $n$ and you get a **base- $n$ number system**: the first landmark number is 1 and each next landmark is $n$ times the current one. So the landmark numbers are $n^{0} = 1, n^{1}, n^{2}, n^{3}, \dots$ . For base 5 these are $1, 5, 25, 125, 625, 3125, \dots$ , and for base 7 they are $1, 7, 49, 343, \dots$ . Base 10 is the decimal system.

**Writing a number in base $n$ .** Group the number into landmarks, largest first. For instance in base 5, $143 = 125 + 5 + 5 + 5 + 1 + 1 + 1$ , that is 1 of 125, 0 of 25, 3 of 5 and 3 of 1. In any base- $n$ system no landmark needs to be used $n$ or more times, since $n$ copies of one landmark make the next landmark.

Why a base helps. Sums regroup by exactly $n$ every time, and the product of any two landmarks is another landmark ( $n^{a} \times n^{b} = n^{a + b}$ ), so multiplication reduces to the distributive property. Multiplying by $n$ simply moves each count to the next landmark.

The shortcoming of the Egyptian system. For bigger numbers you need an endless sequence of new symbols for $1 0^{k}$ .

Place value. The cure is to let the position of a symbol say which landmark it counts. The Mesopotamians, in the later period, used a base-60 (sexagesimal) system: landmarks $1, 60, 6 0^{2} = 3600, 6 0^{3} = 216000, \dots$ . A number such as 640 is $(10) \times 60 + 40$ and 7530 is $(2) \times 3600 + (5) \times 60 + 30$ . No power of 60 can occur 60 or more times; if it does, 60 of them are grouped into the next power. Our units of time, 1 hour = 60 minutes and 1 minute = 60 seconds, still show this influence.

Their written numerals had a problem: with blank spaces to show a missing power it was hard to tell how many blanks there were, so the same set of symbols could be read as different numbers. Later a placeholder symbol was introduced, working like our 0, mainly in the middle of numbers. Zero as a placeholder is essential for unambiguous place value writing.

The Mayan system (Central America, 3rd to 10th centuries CE) used a dot for 1 and a bar for 5 for the numbers 1 to 19, a shell-like symbol for zero, and positions worth 1, 20, 360, ... (the third landmark is 360, not 400). Since it is not a true base-20 system, it lacks the computing advantages of a true base.

The Chinese rod numerals (by at least the 3rd century AD, used till the 17th century) are a base-10 place value system with symbols for 1 to 9, using blank space for a skipped place, so closer to the Hindu system.

The Hindu (Indian) system is base 10 with the ten symbols 0 to 9, uses place value and has a symbol for zero, so it is a fully developed place value system. The first known use of the ten digits including 0 (as a dot) is in the Bakhshali manuscript (c. 3rd century CE).

Roman and Egyptian numerals are covered on the previous topic page.

Bases and Place Value Systems: formulas to remember

Landmark numbers of base n: $n^{0} = 1, n, n^{2}, n^{3}, \dots$; Powers of n starting from 1.
Value of a base-n numeral: $(d_{k} \dots d_{1} d_{0})_{n} = d_{k} n^{k} + \dots + d_{1} n + d_{0}$; Each digit d is less than n.
Product of landmarks: $n^{a} \times n^{b} = n^{a + b}$; Another landmark number.
Largest number with k places: $n^{k} - 1$; All k digits equal n - 1.
Mesopotamian landmarks: $1, 60, 6 0^{2} = 3600, 6 0^{3} = 216000$; Base 60.
Mayan positions: $1, 20, 360$; The third position is 360, not 400.

How to solve Bases and Place Value Systems questions

Follow these steps to move between decimal and other bases.

1List the landmark numbers (powers of the base) up to the largest one that does not exceed the number.
2Starting with the largest landmark, find how many whole copies fit and subtract.
3Repeat for each smaller landmark. Write 0 where a landmark is not used.
4To add in base n, add position by position from the right; when a position reaches n or more, carry one to the next landmark.
5To convert back to decimal, multiply each digit by its landmark and add.
6Check that no digit is n or more and that your conversion back gives the original number.

Bases and Place Value Systems in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Base 60 in everyday time

Suppose a film runs for 7530 seconds and you want to display this as hours, minutes and seconds.

Write 7530 seconds in the form (hours)(minutes)(seconds).

Try it first, then show the solutionSolution

Base-60 landmarks are $3600$ , $60$ and $1$ .
$7530 = 2 \times 3600 + 330$ , and $330 = 5 \times 60 + 30$ .
So 7530 s $=$ 2 h 5 min 30 s, which is the base-60 digit string (2)(5)(30).

Answer: 2 hours 5 minutes 30 seconds

Memory hook: Clock time is a base-60 place value numeral.

Base 5 landmarks in packing

Imagine sweets are packed in packs of 5 pieces, boxes of 25 pieces and cartons of 125 pieces, with any left over sold loose.

Pack 290 sweets using as few containers as possible. How many cartons, boxes, packs and loose pieces?

Try it first, then show the solutionSolution

$290 = 2 \times 125 + 40$ .
$40 = 1 \times 25 + 15$ and $15 = 3 \times 5 + 0$ .
So 2 cartons, 1 box, 3 packs and 0 loose pieces: base-5 digits 2, 1, 3, 0.

Answer: 2 cartons, 1 box, 3 packs, 0 loose pieces

Memory hook: Largest container first, then fill the remainder with the next size.

Tricks to remember Bases and Place Value Systems

Short memory aids. Say each one aloud once, then use it on a question the same day.

Digits stay below the base
In base n no digit can be n or more. In base 5 the digits are 0 to 4.
Use it when: Spotting an invalid numeral instantly.
Carry when you reach the base
In base 10 you carry at 10; in base 5 you carry at 5; in base 60 you carry at 60.
Use it when: Addition and clock-time arithmetic.
Zero holds the place
Zero is not nothing: it is a placeholder that keeps the other digits in their correct positions.
Use it when: Why 3607 and 67 are different numbers in base 60.
Mayan 360, not 400
Mayan places go 1, 20, 360: the third is 18 x 20.
Use it when: Avoiding the false base-20 assumption.

Bases and Place Value Systems: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Landmarks of base 7 · 1 mark · MCQBoard level

The fourth landmark number of a base-7 system, counting 1 as the first, is: (a) 28 (b) 49 (c) 343 (d) 2401

Try it first, then show the solutionSolution

Landmarks: $7^{0} = 1$ , $7^{1} = 7$ , $7^{2} = 49$ , $7^{3} = 343$ .
The fourth is $7^{3} = 343$ .

Answer: (c) 343

Example 2: Decimal to base 5 · 2 marks · Short answerBoard level

Write 289 in the base-5 system (landmarks 1, 5, 25, 125, 625) and give the digit string.

Try it first, then show the solutionSolution

Largest landmark not above 289 is 125: $289 = 2 \times 125 + 39$ . (1 mark)
$39 = 1 \times 25 + 14$ , $14 = 2 \times 5 + 4$ . Digits: 2, 1, 2, 4. (1 mark)

Answer: 2124 in base 5

Example 3: A number in base 60 · 2 marks · Short answerBoard level

Write 4000 as a Mesopotamian base-60 numeral by giving the number of 3600s, 60s and 1s.

Try it first, then show the solutionSolution

$4000 = 1 \times 3600 + 400$ . (1 mark)
$400 = 6 \times 60 + 40$ . So the groups are (1)(6)(40); check $3600 + 360 + 40 = 4000$ . (1 mark)

Answer: (1)(6)(40)

Example 4: A number in the Mayan system · 3 marks · Short answerHarder board level

The Mayan positions are worth 1, 20 and 360. Write 1000 in the Mayan system, using 0 for an empty place, and say what each position holds.

Try it first, then show the solutionSolution

$1000 \div 360 = 2$ remainder $280$ , so two 360s. (1 mark)
$280 = 14 \times 20 + 0$ , so fourteen 20s and no 1s. (1 mark)
Positions (top to bottom): 2, 14, 0. Check $720 + 280 + 0 = 1000$ . (1 mark)

Answer: (2)(14)(0)

Example 5: Addition in base 5 · 4 marks · Long answerHarder board level

Add the base-5 numerals 3412 and 124 directly in base 5 (carrying at 5), and check by converting to decimal.

Try it first, then show the solutionSolution

Ones: $2 + 4 = 6 = 1 \times 5 + 1$ , write 1, carry 1. (1 mark)
Fives: $1 + 2 + 1 = 4$ , write 4. Twenty-fives: $4 + 1 = 5 = 1 \times 5 + 0$ , write 0, carry 1. (1 mark)
125s: $3 + 1 = 4$ . The answer is 4041 in base 5. (1 mark)
Check: $341 2_{5} = 375 + 100 + 5 + 2 = 482$ , $12 4_{5} = 25 + 10 + 4 = 39$ , $482 + 39 = 521$ , and $404 1_{5} = 500 + 0 + 20 + 1 = 521$ . (1 mark)

Answer: 4041 in base 5, which is 521

Where students lose marks in Bases and Place Value Systems

The slips examiners see most often, and what to write instead.

Mistake: Writing a digit equal to or above the base
In base n the digits go only up to n - 1.
Wrong working
"7 in base 5 is written 7."
Right working
$7 = 1 \times 5 + 2$ , so it is 12 in base 5.
Do this: Check each digit against the base before finishing.
Usually costs: The answer mark.
Mistake: Forgetting the zero in an empty place
A missing power changes the value completely.
Wrong working
"3607 = (1)(7) in base 60."
Right working
$3607 = (1) (0) (7)$ with a placeholder zero for the empty 60s place.
Do this: List all landmarks and write a digit (possibly 0) for each.
Usually costs: 1 mark.
Mistake: Carrying at 10 in another base
You carry when a position reaches the base, not 10.
Wrong working
"In base 5, 4 + 3 = 7, so write 7."
Right working
$4 + 3 = 7 = 1 \times 5 + 2$ , so write 2 and carry 1.
Do this: Say the base before each addition step.
Usually costs: The part.
Mistake: Assuming the Mayan system is base 20 throughout
Its third position is 360, not 400.
Wrong working
"Third Mayan position is 20 x 20 = 400."
Right working
It is 18 x 20 = 360, so the system is not a true base-20 system.
Do this: Quote the positions 1, 20, 360.
Usually costs: 1 mark.

Exam tips

Always list the landmark numbers of the base first. A one-line list makes the rest quick and checkable.
Write the digit string and also show the sum of digit x landmark as a check.
In "why" questions, mention both ideas: a base (powers of one number) and place value (position gives the landmark).
Use 0 for every unused landmark between the first and last used.
Cite time units (60 minutes, 60 seconds) as the everyday example of base 60.

Practice questions on Bases and Place Value Systems

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · Assertion-reasonBoard level
Assertion (A): In a base-n number system, multiplying any landmark number by n gives the next landmark number.
Reason (R): The landmark numbers of a base-n system are 1, n, n squared, n cubed, and so on.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
What is n times n^k?
Answer
(a). $n \times n^{k} = n^{k + 1}$ is the next power, so A follows from R.
Q2 · 3 marks · Short answerHarder board level
In a base-5 system with landmark numbers 1, 5, 25 and 125 only, and each landmark used at most 4 times, what is the largest number that can be written? Show that it is one less than the next landmark.
Hint
Use 4 of each landmark and add.
Answer
Largest $= 4 \times 125 + 4 \times 25 + 4 \times 5 + 4 \times 1 = 500 + 100 + 20 + 4 = 624$ . The next landmark is $625 = 5^{4}$ and $625 - 1 = 624$ . This also shows that every number below 625 has a representation with at most 4 copies of each landmark.
Q3 · 4 marks · Long answerHOTS / competency
Design a base-4 system with landmarks 1, 4, 16, 64 and 256. (i) Write 100 in it. (ii) Write 400 in it. (iii) What do you notice about the digit strings for 100 and 400, and why does it happen?
Hint
Multiplying by 4 moves every count to the next landmark.
Answer
(i) $100 = 1 \times 64 + 2 \times 16 + 1 \times 4 + 0 \times 1$ , so the digit string is 1210. (ii) $400 = 1 \times 256 + 2 \times 64 + 1 \times 16 + 0 \times 4 + 0 \times 1$ , so the digit string is 12100. (iii) 12100 is 1210 followed by a 0. Since $400 = 100 \times 4$ , multiplying by the base shifts every count to the next landmark, which is exactly what appending a 0 does.
Q4 · 4 marks · Case-basedHOTS / competency
Case study: a Mesopotamian scribe writes base-60 numerals with a placeholder symbol for an empty place. Write the value of each numeral (digits listed from the highest place to the units place): (i) (1)(7) (ii) (1)(0)(7) (iii) (1)(0)(0)(7) (iv) Explain why the scribe needs the placeholder, and why leaving a gap without a symbol is unreliable.
Hint
Use landmarks 1, 60, 3600, 216000.
Answer
(i) $1 \times 60 + 7 = 67$ . (ii) $1 \times 3600 + 0 \times 60 + 7 = 3607$ . (iii) $1 \times 216000 + 0 + 0 + 7 = 216007$ . (iv) Without a placeholder the same symbols could be read as 67, 3607 or 216007 depending on how wide the gap is; spacing was inconsistent between scribes, so the placeholder (like our 0) is needed to fix each symbol to its position.
Q5 · 3 marks · Short answerHarder board level
A Mayan numeral has three positions worth 1, 20 and 360 (from the bottom). The digits are 3 (360s), 5 (20s) and 12 (1s). Find its value. Then find the value if the same digits were read in a true base-20 system with positions 1, 20 and 400, and the difference.
Hint
Multiply each digit by its position value.
Answer
Mayan value $= 3 \times 360 + 5 \times 20 + 12 = 1080 + 100 + 12 = 1192$ . In a true base-20 reading: $3 \times 400 + 5 \times 20 + 12 = 1200 + 100 + 12 = 1312$ . Difference $= 1312 - 1192 = 120$ , which is $3 \times (400 - 360)$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What are the landmark numbers of a base-n system?Show answer
1, n, n squared, n cubed, and so on.
2.What is the base of the Mesopotamian system?Show answer
60 (sexagesimal).
3.What are the first three Mayan positions?Show answer
1, 20 and 360.
4.What was the role of the placeholder symbol?Show answer
To mark an empty place so numerals are read unambiguously, like zero.
5.Which digits are used in base 5?Show answer
0, 1, 2, 3 and 4.
6.Where is the first known use of ten digits including 0 (as a dot)?Show answer
The Bakhshali manuscript (c. 3rd century CE).

Bases and Place Value Systems: frequently asked questions

Is the Roman system a base system?

No. Its landmark numbers 1, 5, 10, 50, 100, 500, 1000 do not form powers of one number, so it is not a base system.

What is the difference between a base system and a place value system?

A base system has landmark numbers that are powers of a number. A place value system also uses the position of a symbol to show which landmark it counts, so it needs only finitely many symbols.

Why did the Mesopotamians use base 60?

The chapter says it is not certain; several theories are given, such as links to calendar periods and the ease of showing fractions, so no single reason is established.

Is the Hindu system the only decimal place value system?

No. The Chinese rod numerals were also decimal place value, but the Hindu system has a symbol for zero, so it is fully developed.