MathsCBSE Class 8 · A Story of Numbers

Counting, Roman and Egyptian Numerals for Class 8

Updated 17 October 2026 · Fact-checked

Early number systems count by one-to-one mapping with a standard sequence of objects, names or symbols, then speed up by grouping. The Roman system groups by landmark numbers (1, 5, 10, 50, 100, 500, 1000), and the Egyptian system uses landmark numbers that are powers of 10, which makes addition and multiplication much simpler. Exams ask for conversions, regrouping in addition, and comparisons of efficiency.

Understand Counting, Roman and Egyptian Numerals

Counting by one-to-one mapping. To count a collection (a herd of cows, say) without number names, pair each object with one item from a standard sequence: a stick per cow, a letter of the alphabet per cow, or a written symbol per cow. Such a sequence with a fixed order is a number system. Pairing each cow with a different stick, with no stick used twice, is called a one-to-one mapping; the leftover sticks answer questions such as "Has a cow gone missing?" and "Who has fewer cows?".

A good number system should be unending and easy to count with. Sticks are unending but need as many sticks as objects. Letters are convenient but run out after 26. The numerals are the written symbols (such as 0, 1, 5, 36), and written numerals go hand in hand with number names.

Counting in groups. The Gumulgal people of Australia counted in 2s: 3 is *ukasar-urapon* (2 + 1), 4 is *ukasar-ukasar* (2 + 2), and so on, and any number above 6 was called *ras*. Counting in groups of a fixed size (2, 5, 10 or 20 in different systems) is more efficient than tally marks, but it is still clumsy for large numbers like 1345 if you group only by 5s. Humans find it hard to see 5 or more objects at a glance, which may be why groups of 5 became symbols.

The Roman system. It uses I for 1, V for 5, X for 10, L for 50, C for 100, D for 500 and M for 1000. These are landmark numbers: numbers with their own symbol. To write a number, take as many of the largest landmark as possible, then the next, and so on. For 27: $27 = 10 + 10 + 5 + 1 + 1$ , so XXVII. Following the pattern that 4 is IV (one less than 5), 40 is XL (ten less than 50), though the text notes that people were not always consistent and sometimes wrote 40 as XXXX. The system grew from the older Greek system in about the 8th century BCE in Rome.

Roman numerals are much more efficient than tallies, but arithmetic is hard because a landmark does not always group into the next one in the same size: five Is make a V but two Vs make an X, and five Cs make a D. Multiplication is especially awkward, because products such as $L \times D$ ( $50 \times 500 = 25000$ ) are not landmark numbers. Roman calculators used an abacus.

The Egyptian system (about 3000 BCE). Its landmark numbers are $1, 10, 1 0^{2}, 1 0^{3}, \dots$ , each 10 times the previous, so they are all powers of 10. A number is written by grouping into landmarks, largest first: $324 = 100 + 100 + 100 + 10 + 10 + 4$ . Because each group of 10 makes the next landmark, addition is just regrouping: ten of one symbol are replaced by one of the next. The product of two landmark numbers is another landmark number ( $1 0^{a} \times 1 0^{b} = 1 0^{a + b}$ ), so multiplication follows from the distributive property. No symbol can appear 10 or more times in a numeral, since 10 of them are replaced by the next symbol. The drawback is that bigger numbers need ever more new symbols; the Egyptian system gave efficient numerals up to a crore ( $1 0^{7}$ ).

Abacus. Around the 11th century even users of Roman numerals used an abacus based on powers of 10: each line stands for a power of 10, counters are placed according to the grouping, and a counter above a line is worth 5 of that line's value.

The idea of a base and of place value is developed on the next topic page of this chapter.

Counting, Roman and Egyptian Numerals: formulas to remember

Roman landmark numbers: $I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, M = 1000$; Group the number into these, largest first.
Egyptian landmark numbers: $1, 10, 1 0^{2}, 1 0^{3}, \dots$; Each is 10 times the previous one.
Product of two Egyptian landmarks: $1 0^{a} \times 1 0^{b} = 1 0^{a + b}$; Always another landmark number.
Regrouping rule: $10 of one landmark = 1 of the next landmark$; The basis of addition in the Egyptian system.
Roman regroupings: $5 \times I = V, 2 \times V = X, 5 \times X = L, 2 \times L = C, 5 \times C = D, 2 \times D = M$; The sizes alternate between 5 and 2, which complicates regrouping.

How to solve Counting, Roman and Egyptian Numerals questions

Use these routines for conversion and arithmetic with old numeral systems.

1To write a number in Roman numerals, split it by place value (thousands, hundreds, tens, ones) and write each part using the landmarks, remembering IV, IX and XL as used in the chapter.
2To read a Roman numeral, add the values from the left, and subtract a symbol that stands before a larger one (IV is 4).
3To add in a system with landmarks, count the number of each symbol in both numerals, then regroup from the smallest landmark upwards.
4In the Egyptian system regroup ten of any symbol into one of the next; in the Roman system regroup by 5 or 2 as the landmark pair requires.
5To multiply, split one number into landmarks, multiply each landmark by the other number using the distributive property, and regroup.
6Check any conversion by converting back to ordinary Hindu numerals.

Counting, Roman and Egyptian Numerals 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.

One-to-one mapping

Suppose a shepherd with no number names keeps a pebble in a bag for every sheep that leaves the pen in the morning.

In the evening 44 sheep return, and the bag has 47 pebbles. How does the shepherd know that some sheep are missing, and how many? A neighbour keeps 63 pebbles for her flock: who has more sheep, and how many more?

Try it first, then show the solutionSolution

Pair one returning sheep with one pebble. After 44 pairs, 3 pebbles remain unpaired, so 3 sheep are missing: $47 - 44 = 3$ .
Pair the shepherd's 47 pebbles with the neighbour's 63 pebbles. The neighbour's heap has pebbles left over, so her flock is larger, by $63 - 47 = 16$ sheep.

Answer: 3 sheep are missing; the neighbour has 16 more sheep.

Memory hook: Leftover tokens after pairing show the difference without any number names.

Grouping into landmarks

Imagine a clerk writing a ledger total of 1247 in Roman numerals so that it can be read by a trader in ancient Rome.

Write 1247 in Roman numerals (using XL for 40 as in the chapter) and count the symbols.

Try it first, then show the solutionSolution

$1247 = 1000 + 200 + 40 + 7$ .
$1000 = M$ , $200 = C C$ , $40 = X L$ , $7 = V I I$ .
So 1247 is MCCXLVII, with $1 + 2 + 2 + 3 = 8$ symbols.

Answer: MCCXLVII (8 symbols)

Memory hook: Split by place value first, then convert each part.

Tricks to remember Counting, Roman and Egyptian Numerals

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

Five and two alternate
In Roman numerals the landmarks go 1, 5, 10, 50, 100, 500, 1000: times 5, times 2, times 5, times 2, and so on.
Use it when: Regrouping during addition.
MDCLXVI
Remember the symbols in order of size as M D C L X V I (a nonsense word with 1000 first).
Use it when: Writing a Roman numeral from the largest landmark down.
Ten makes a new symbol
In the Egyptian system, collect ten of any symbol and swap them for one of the next landmark.
Use it when: Addition by regrouping.
Powers make products easy
If landmarks are powers of one number, any product of two landmarks is another landmark.
Use it when: Explaining why the Egyptian system multiplies easily.

Counting, Roman and Egyptian Numerals: solved examples

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

Example 1: Reading a Roman numeral · 1 mark · MCQBoard level

The value of CXLIV is: (a) 164 (b) 144 (c) 154 (d) 134

Try it first, then show the solutionSolution

C $= 100$ , XL $= 40$ , IV $= 4$ .
Total $= 100 + 40 + 4 = 144$ .

Answer: (b) 144

Example 2: Writing in Roman numerals · 2 marks · Short answerBoard level

Write 137 in Roman numerals by grouping it into landmark numbers, and state how many symbols are used.

Try it first, then show the solutionSolution

$137 = 100 + 10 + 10 + 10 + 5 + 1 + 1$ . (1 mark)
So 137 is CXXXVII, which has 7 symbols. (1 mark)

Answer: CXXXVII, 7 symbols

Example 3: Adding Roman numerals by regrouping · 3 marks · Short answerHarder board level

Add DCCLXV and CCCXLVIII without first writing them in Hindu numerals, then check using Hindu numerals.

Try it first, then show the solutionSolution

Ones: V and VIII give $5 + 8 = 13$ = X and III. Write III, carry X. (1 mark)
Tens: LX (60) and XL (40) give 100, and the carried X makes 110 = C and X. Write X, carry C. Hundreds: DCC (700) and CCC (300) give 1000 = M, and the carried C makes 1100 = M and C. (1 mark)
Result: M, C, X, III = MCXIII. Check: $765 + 348 = 1113$ . (1 mark)

Answer: MCXIII (1113)

Example 4: Multiplying landmark numbers · 3 marks · Short answerHarder board level

Find V x L and L x D in Roman numerals. Explain with these results why multiplication is hard in the Roman system, but easy in the Egyptian system.

Try it first, then show the solutionSolution

$V \times L = 5 \times 50 = 250 = C C + L$ , i.e. CCL. (1 mark)
$L \times D = 50 \times 500 = 25000$ , which is greater than M (1000) by far; there is no landmark for it, so it must be written as 25 Ms, or a new symbol must be invented. (1 mark)
In the Egyptian system every landmark is a power of 10, so the product of any two landmarks is a landmark ( $1 0^{a} \times 1 0^{b} = 1 0^{a + b}$ ). (1 mark)

Answer: CCL and 25000 (25 Ms); products of landmarks are again landmarks in the Egyptian system but not in the Roman system.

Example 5: Comparing efficiency of two systems · 4 marks · Long answerHOTS / competency

Write 4589 using Roman landmarks additively (use V and IIII for 9, and so on, with no subtractive forms) and also as an Egyptian-style grouping into powers of 10. Count the symbols in each and comment on which numeral is shorter and which is easier for arithmetic.

Try it first, then show the solutionSolution

$4589 = 4000 + 500 + 80 + 9$ . (1 mark)
Roman, additive: MMMM D L XXX V IIII, which is $4 + 1 + 1 + 3 + 1 + 4 = 14$ symbols. (1 mark)
Egyptian: 4 thousands, 5 hundreds, 8 tens, 9 ones: $4 + 5 + 8 + 9 = 26$ symbols. (1 mark)
The Roman numeral is shorter because it has landmarks for 5, 50 and 500, but its landmarks do not follow one rule, so regrouping and multiplication are harder; the Egyptian numeral is longer but every regrouping is by exactly 10. (1 mark)

Answer: Roman additive: 14 symbols; Egyptian: 26 symbols. Roman is shorter, Egyptian is easier for arithmetic.

Where students lose marks in Counting, Roman and Egyptian Numerals

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

Mistake: Writing four Is where a subtractive form is expected, or the reverse, without stating the rule
The chapter writes 4 as IV and 40 as XL but notes that old users were not always consistent. Choose one convention and say so.
Wrong working
"4 = IIII in one line and IV in the next line."
Right working
State the convention (IV, IX, XL as in the chapter) and use it throughout.
Do this: Write the convention in a few words at the top of the answer.
Usually costs: 1 mark.
Mistake: Ignoring regrouping by 5 and 2 in Roman addition
Five Is make a V, but two Vs make an X; the grouping size changes.
Wrong working
"VIII + VII: 8 + 7 gives fifteen single marks, so the answer is a row of fifteen Is."
Right working
VIII + VII: 15 ones = 3 Vs = 1 X + 1 V, so XV.
Do this: Count symbols then regroup upwards, alternating 5 and 2.
Usually costs: 1 to 2 marks.
Mistake: Putting 10 or more of one Egyptian symbol in a numeral
Ten of a landmark is replaced by one of the next landmark.
Wrong working
"Sum = 12 ones and 3 tens"
Right working
12 ones = 1 ten + 2 ones, so 4 tens and 2 ones.
Do this: Check every count is below 10 before you write the answer.
Usually costs: The answer mark.
Mistake: Confusing landmark numbers with place values
In a landmark system the symbol carries its own value; position matters only in place value systems.
Wrong working
"The Roman X in XI is worth 10 only because it is on the left."
Right working
X is always 10; the I is added because it follows.
Do this: Say which kind of system the question is about.
Usually costs: The explanation mark.

Exam tips

In conversion questions, split the number by place value first and show that line.
Always verify an arithmetic answer by converting the numerals to Hindu numerals and back.
For explanation questions, name the property: landmark numbers being powers of one number is what makes products easy.
Write counts of each symbol before regrouping; examiners give a mark for the regrouping step.
Use dates and facts exactly as in the chapter (Roman system from the 8th century BCE; Egyptian about 3000 BCE) or leave them out.

Practice questions on Counting, Roman and Egyptian Numerals

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 the Egyptian system the product of two landmark numbers is always another landmark number.
Reason (R): The landmark numbers of the Egyptian system are powers of 10, and the product of two powers of 10 is a power of 10.
(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
Use 10^a x 10^b = 10^(a+b).
Answer
(a). $1 0^{a} \times 1 0^{b} = 1 0^{a + b}$ , which is again a landmark number. R is the reason for A.
Q2 · 3 marks · Short answerHarder board level
In the Egyptian system an accountant has 6 hundreds, 7 tens and 8 ones in one numeral and 3 hundreds, 5 tens and 9 ones in another. Add them by regrouping symbols, writing the answer in terms of counts of each landmark and in ordinary numerals.
Hint
Regroup from the smallest landmark: 10 ones make a ten.
Answer
Ones: $8 + 9 = 17$ , so 7 ones and 1 ten carried. Tens: $7 + 5 + 1 = 13$ , so 3 tens and 1 hundred carried. Hundreds: $6 + 3 + 1 = 10$ , so 0 hundreds and 1 thousand carried. Result: 1 thousand, 0 hundreds, 3 tens, 7 ones $= 1037$ . Check: $678 + 359 = 1037$ .
Q3 · 4 marks · Long answerHOTS / competency
In the Egyptian system, a number is 2 tens and 4 ones, and another is 1 ten and 3 ones. Use the distributive property on landmark numbers to find their product as counts of each landmark, regroup, and check with ordinary multiplication.
Hint
Expand (2 x 10 + 4 x 1)(1 x 10 + 3 x 1) and use 10 x 10 = 100.
Answer
$(2 \times 10 + 4) (10 + 3) = 2 \times 10 \times 10 + 2 \times 10 \times 3 + 4 \times 10 + 4 \times 3 = 2 hundreds + 6 tens + 4 tens + 12 ones$ . That is 2 hundreds, 10 tens, 12 ones. Regroup 12 ones as 1 ten and 2 ones: 2 hundreds, 11 tens, 2 ones. Regroup 10 tens as 1 hundred: 3 hundreds, 1 ten, 2 ones $= 312$ . Check: $24 \times 13 = 312$ .
Q4 · 4 marks · Case-basedBoard level
Case study: a Stone Age herder has no number names. She keeps one pebble in a pouch for each cow that leaves the shed.
(i) In the evening 44 cows return while the pouch has 47 pebbles. How does she find out how many cows are missing? (1 mark)
(ii) Another herder has 63 cows. Without number names, how can she tell whose herd is larger, and by how many? (1 mark)
(iii) She now replaces every 5 pebbles with one large stone. Show how to record 47 cows with these stones and pebbles. (2 marks)
Hint
Pair objects one-to-one and look at what is left over.
Answer
(i) Pair each returning cow with a pebble; 3 pebbles stay unpaired, so 3 cows are missing. (ii) Pair the two sets of pebbles one-to-one; the herder whose pebbles are left over has the larger herd, and the leftover pebbles show the difference, $63 - 47 = 16$ cows. (iii) $47 = 5 \times 9 + 2$ , so 9 large stones and 2 pebbles.
Q5 · 3 marks · Short answerHarder board level
Using the Gumulgal style of counting in 2s, with urapon for 1 and ukasar for 2, write the number names for 11 and 12. Then evaluate (7 + 5) and (11 - 4) in this system and give the answers as names.
Hint
11 = 2 + 2 + 2 + 2 + 2 + 1.
Answer
11 is five ukasars and one urapon: ukasar-ukasar-ukasar-ukasar-ukasar-urapon. 12 is six ukasars. For $7 + 5$ : 7 is three ukasars and an urapon, 5 is two ukasars and an urapon; together 5 ukasars and 2 urapons, and two urapons make one ukasar, so 6 ukasars, which is 12. For $11 - 4$ : 4 is two ukasars; removing them from five ukasars and an urapon leaves three ukasars and an urapon, which is 7: ukasar-ukasar-ukasar-urapon.

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 is a one-to-one mapping?Show answer
Pairing each object of a collection with a different item of a standard sequence, with none left out or used twice.
2.What are the Roman landmark numbers?Show answer
1, 5, 10, 50, 100, 500, 1000 (I, V, X, L, C, D, M).
3.What are the Egyptian landmark numbers?Show answer
Powers of 10: 1, 10, 100, 1000, and so on.
4.Why is multiplication easy in the Egyptian system?Show answer
The product of two landmark numbers is another landmark number.
5.What was the abacus used for?Show answer
Calculating, with lines standing for powers of 10 and counters placed for each landmark.
6.How many times can one symbol appear in an Egyptian numeral?Show answer
At most 9 times, since ten of them make the next symbol.

Counting, Roman and Egyptian Numerals: frequently asked questions

Did the Roman system have a symbol for zero?

No symbol for zero appears in the chapter's Roman system. It uses landmark numbers with their own symbols, so a placeholder is not needed.

Was each system an improvement on the one before it?

No. The chapter stresses that the actual history is more complex and the systems did not form a simple chain of improvements.

Which names did Europeans and Arab scholars give to the digits 0 to 9, and why do the names differ?

European scholars learned them from the Arab world and called them Arabic numerals; Arab scholars such as Al-Khwarizmi called them Hindu numerals. Today Hindu, Indian or Hindu-Arabic are used.

Where did the digits 0 to 9 originate?

In India. The chapter notes the first known use of ten digits including a dot for 0 in the Bakhshali manuscript (c. 3rd century CE).