MathsCBSE Class 10 · Real Numbers

HCF and LCM by Prime Factorisation for Class 10

Q: Do I have to use prime factorisation, or can I list factors?

When a question says by prime factorisation, you must show the factorisations to get the marks. Otherwise listing is allowed, but prime factorisation is faster and easier to check for larger numbers.

Q: Why does HCF times LCM not equal the product for three numbers?

For two numbers, each prime's smaller and larger powers add up to its total power in the product. With three numbers, the middle power is counted in neither the HCF nor the LCM, so the totals stop matching.

Q: Is Euclid's division algorithm for HCF in the syllabus?

No. Euclid's division lemma and algorithm are not in the 2026-27 syllabus, so use prime factorisation for HCF and LCM.

Q: How do I know whether a word problem needs HCF or LCM?

If you are splitting quantities into the largest equal parts, use HCF. If things repeat and must meet again, or you want the least amount every size divides, use LCM.

Updated 14 October 2026 · Fact-checked

HCF and LCM by prime factorisation means writing each number as a product of prime powers, then taking the common primes at their smallest powers for the HCF and all primes at their greatest powers for the LCM. The board asks MCQs, missing-number and unknown-exponent questions, the HCF times LCM relation, and word problems and case studies.

Primes in the overlap multiply to the HCF, all the primes together give the LCM, and HCF × LCM equals the product of the two numbers.

Illustration drawn for QuestionBuddy.

Understand HCF and LCM by Prime Factorisation

The HCF (highest common factor) of some numbers is the largest number that divides all of them. The LCM (lowest common multiple) is the smallest number that all of them divide. Listing factors or multiples works for small numbers, but it is slow and error-prone for numbers like 216 and 288. Prime factorisation gives a quick, exact method.

The method rests on the Fundamental Theorem of Arithmetic: every number above 1 has exactly one prime factorisation, apart from the order of the primes. So a number's primes and their powers are like its fingerprint. To compare two numbers, you compare their fingerprints.

HCF = product of the primes common to all the numbers, each taken to its smallest power.
LCM = product of every prime that appears in any number, each taken to its greatest power.

Why smallest for HCF? A divisor of a number can only use that number's primes, and no more copies of each than the number has. To divide every number, it must respect the number with the fewest copies. Why greatest for LCM? A multiple of a number must contain all of its primes with at least its powers. To be a multiple of every number, it needs the largest power any number demands, and the smallest such multiple uses exactly that power.

For two numbers a and b this gives a neat result: $HCF (a, b) \times LCM (a, b) = a \times b$ . For each prime, the HCF takes the smaller power and the LCM the larger, and smaller plus larger is the total power in $a \times b$ . This rule is not true in general for three or more numbers, so there you always work from the factorisations. Two checks follow from the method: the HCF divides every number and the LCM, and every number divides the LCM.

The same idea works when numbers are given in factorised form, even with letters such as $a^{2} b^{3}$ , or with unknown powers such as $2^{x} \times 3^{2}$ . It also powers word problems: greatest equal share means HCF, earliest common repeat or least common quantity means LCM.

HCF and LCM by Prime Factorisation: formulas to remember

HCF by prime factorisation: $HCF = \prod (common primes)^{smallest power}$; Works for any number of numbers. A prime missing from even one number is not in the HCF.
LCM by prime factorisation: $LCM = \prod (all primes present)^{greatest power}$; Works for any number of numbers.
Product relation (two numbers only): $HCF (a, b) \times LCM (a, b) = a \times b$; Use to find a missing number. Not valid for three numbers.
Exponent form: $a = p^{m} q^{r}, b = p^{n} q^{s} \Rightarrow HCF = p^{m i n (m, n)} q^{m i n (r, s)}, LCM = p^{m a x (m, n)} q^{m a x (r, s)}$; Use for letter powers and unknown exponents.
Divisibility checks: $HCF ∣ a, a ∣ LCM, HCF ∣ LCM$; Reject any answer that fails these.
Co-prime numbers: $HCF (a, b) = 1 \Rightarrow LCM (a, b) = ab$; Consecutive numbers are always co-prime.
Pairs with given HCF and LCM: $a = hm, b = hn, HCF (m, n) = 1, mn = \frac{L}{h}$; List co-prime pairs m, n whose product is L ÷ h to find every possible pair.
Remainder versions: $greatest divisor = HCF (x - r_{1}, y - r_{2}), least number greater than r = LCM + r$; The divisor must be greater than each remainder. The number r itself also leaves remainder r (quotient 0), so LCM + r is the smallest such number greater than r.

How to solve HCF and LCM by Prime Factorisation questions

Use this routine on any HCF or LCM question, from a 1-mark MCQ to a case study.

1Decide what is asked: HCF (greatest common divisor, equal sharing) or LCM (least common multiple, repeating events). Adjust for remainders first if there are any.
2Factorise each number by repeated division by 2, 3, 5, 7, and so on. Write each in power form, primes in increasing order, on its own line.
3Line up the same primes in a column so each power is visible.
4For HCF, take only primes present in every number, at the smallest power. For LCM, take every prime, at the greatest power.
5Multiply out only if a value is needed.
6Check: HCF divides each number; each number divides the LCM. For two numbers, also check HCF × LCM = product.
7In a word problem, answer the actual question (number of boxes, time, pieces) with units and a sentence.

HCF and LCM by Prime Factorisation 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.

HCF by prime factorisation

Imagine your family is packing Diwali sweet boxes for neighbours. You have 156 laddoos, 204 barfis and 90 pieces of kaju katli. Every box must contain the same number of each sweet, and nothing should be left over.

Find the greatest number of boxes you can make, and how many of each sweet go in each box. Your sister had planned 12 boxes before the kaju katli arrived. Explain why 12 boxes no longer work.

Try it first, then show the solutionSolution

The number of boxes must divide 156, 204 and 90, and we want the greatest, so find HCF(156, 204, 90).
$156 = 2^{2} \times 3 \times 13$ , $204 = 2^{2} \times 3 \times 17$ , $90 = 2 \times 3^{2} \times 5$ .
Primes common to all three, at their smallest powers: $2 \times 3 = 6$ .
Per box: $156 \div 6 = 26$ laddoos, $204 \div 6 = 34$ barfis and $90 \div 6 = 15$ kaju katli.
Without the kaju katli, HCF(156, 204) $= 2^{2} \times 3 = 12$ , which is why 12 boxes worked before.
But 12 does not divide 90 ( $90 = 12 \times 7 + 6$ ), so with 12 boxes 6 pieces of kaju katli would be left over.

Answer: 6 boxes, each with 26 laddoos, 34 barfis and 15 kaju katli. 12 boxes no longer work because 12 is not a factor of 90.

Memory hook: Sharing out sweets equally: the box count must divide every pile, so take the HCF of all of them.

LCM by prime factorisation

Suppose your mother wants to buy the same total weight of two tea brands for a family function. One brand comes only in 250 g packets and the other only in 400 g packets.

Find the least equal weight of each brand she can buy, and the number of packets of each brand needed. If she instead needs at least 5 kg of each brand, what is the least equal weight of each she must buy?

Try it first, then show the solutionSolution

The weight must be a multiple of both 250 and 400, so find LCM(250, 400).
$250 = 2 \times 5^{3}$
$400 = 2^{4} \times 5^{2}$
LCM $= 2^{4} \times 5^{3} = 16 \times 125 = 2000$ g
Packets: $2000 \div 250 = 8$ and $2000 \div 400 = 5$ .
Common weights are multiples of 2000 g: 2000, 4000, 6000, ... The first that is at least 5000 g is 6000 g.

Answer: Least equal weight 2 kg (8 packets of 250 g and 5 packets of 400 g); for at least 5 kg, she must buy 6 kg of each.

Memory hook: Matching totals of different pack sizes: the total must be a multiple of both, so LCM.

Tricks to remember HCF and LCM by Prime Factorisation

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

H for Humble, L for Large
HCF is humble: it takes only shared primes and the smallest power. LCM is large: it takes every prime and the biggest power.
Use it when: Whenever you choose powers from a column of factorisations.
Missing in one, missing in HCF
If a prime is absent from even one number, its power there is 0, so the smallest power is 0 and it drops out of the HCF.
Use it when: Questions with three numbers or with primes like 7 or 13 in only one number.
Two only for the product rule
Say it aloud: HCF times LCM equals product, for TWO numbers only.
Use it when: Stops you using the rule on three numbers in MCQs and assertion-reason.
Divide or share means HCF, meet or match means LCM
Cutting, sharing, packing into the largest equal groups: HCF. Bells meeting, buying equal totals, least number divisible: LCM.
Use it when: Deciding the method in word problems and case studies.

HCF and LCM by Prime Factorisation: solved examples

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

Example 1: LCM of three numbers in factorised form · 1 mark · MCQBoard level

The LCM of $2^{3} \times 3^{2}$ , $2^{2} \times 3^{3} \times 5$ and $2 \times 3 \times 5^{2}$ is
(a) 1080
(b) 2700
(c) 5400
(d) 10800

Try it first, then show the solutionSolution

Primes present: 2, 3, 5.
Greatest power of 2 is $2^{3}$ , of 3 is $3^{3}$ , of 5 is $5^{2}$ .
LCM $= 2^{3} \times 3^{3} \times 5^{2} = 8 \times 27 \times 25 = 5400$

Answer: (c) 5400

Example 2: Finding unknown exponents from HCF and LCM · 3 marks · Short answerHarder board level

Let $m = 2^{a} \times 3^{2} \times 5$ and $n = 2^{3} \times 3^{b} \times 7$ , where a and b are natural numbers. The HCF of m and n is 12. Find a and b, and hence find the LCM of m and n.

Try it first, then show the solutionSolution

$12 = 2^{2} \times 3$ . Power of 2 in HCF is the smaller of a and 3, so it equals 2, giving $a = 2$ . (1 mark)
Power of 3 in HCF is the smaller of 2 and b, so it equals 1, giving $b = 1$ . (1 mark)
So $m = 2^{2} \times 3^{2} \times 5$ and $n = 2^{3} \times 3 \times 7$ . LCM $= 2^{3} \times 3^{2} \times 5 \times 7 = 2520$ . (1 mark)

Answer: $a = 2$ , $b = 1$ , LCM = 2520.

Example 3: Least number leaving the same remainder · 3 marks · Short answerHarder board level

Find the smallest natural number greater than 5 which leaves remainder 5 when divided by each of 28, 42 and 63. Also find the smallest four-digit number with this property.

Try it first, then show the solutionSolution

$28 = 2^{2} \times 7$ , $42 = 2 \times 3 \times 7$ , $63 = 3^{2} \times 7$ . (1 mark)
LCM $= 2^{2} \times 3^{2} \times 7 = 252$ . Numbers leaving remainder 5 are $252 k + 5$ ; $k = 0$ gives 5, which is excluded, so the smallest is $252 + 5 = 257$ . (1 mark)
$252 \times 3 + 5 = 761 < 1000$ and $252 \times 4 + 5 = 1013$ . (½ mark)
Check: $1013 - 5 = 1008 = 28 \times 36 = 42 \times 24 = 63 \times 16$ . (½ mark)

Answer: 257; smallest four-digit number 1013.

Example 4: Missing number and all possible pairs · 5 marks · Long answerHarder board level

The HCF of two numbers is 23 and their LCM is 1449.
(i) If one number is 161, find the other number and verify your answer by prime factorisation.
(ii) Find all pairs of natural numbers with HCF 23 and LCM 1449.
(iii) Which of these pairs has the smaller sum?

Try it first, then show the solutionSolution

(i) Using HCF × LCM = product for two numbers: other number $= \frac{23 \times 1449}{161}$ . (1 mark)
$1449 \div 161 = 9$ , so other number $= 23 \times 9 = 207$ . (½ mark)
Verify: $161 = 7 \times 23$ , $207 = 3^{2} \times 23$ ; HCF $= 23$ , LCM $= 3^{2} \times 7 \times 23 = 1449$ . (1 mark)
(ii) Write the numbers as $23 m$ and $23 n$ with m, n co-prime; then $23 mn = 1449$ , so $mn = 63$ . (1 mark)
Co-prime pairs with product 63: (1, 63) and (7, 9). [(3, 21) is rejected as HCF(3, 21) = 3.] So pairs are (23, 1449) and (161, 207). (1 mark)
(iii) Sums: $23 + 1449 = 1472$ and $161 + 207 = 368$ . (½ mark)

Answer: (i) 207 (ii) (23, 1449) and (161, 207) (iii) (161, 207), sum 368.

Example 5: Case study: sports day supplies · 4 marks · Case-basedHOTS / competency

Suppose a PE teacher is arranging drinks for sports day. Water bottles come only in cartons of 24, energy drinks only in cartons of 36 and juice packs only in cartons of 40.
(i) Write the prime factorisations of 24, 36 and 40. (1 mark)
(ii) What is the least number of each drink she must buy to have equal numbers of all three, and how many cartons of each is that? (1 mark)
(iii) Later she has 216 water bottles, 288 energy drinks and 360 juice packs. She makes identical kits, each with the same numbers of each drink, using all the drinks. Find the greatest number of kits and the contents of each kit. A student suggests making 90 kits instead. Is that possible? Justify. (2 marks)

Try it first, then show the solutionSolution

(i) $24 = 2^{3} \times 3$ , $36 = 2^{2} \times 3^{2}$ , $40 = 2^{3} \times 5$ . (1 mark)
(ii) Equal totals must be a common multiple of all three: LCM $= 2^{3} \times 3^{2} \times 5 = 360$ . Cartons: $360 \div 24 = 15$ , $360 \div 36 = 10$ and $360 \div 40 = 9$ . (1 mark)
(iii) $216 = 2^{3} \times 3^{3}$ , $288 = 2^{5} \times 3^{2}$ , $360 = 2^{3} \times 3^{2} \times 5$ . The number of kits must divide all three, so the greatest is HCF $= 2^{3} \times 3^{2} = 72$ . Each kit: $216 \div 72 = 3$ water bottles, $288 \div 72 = 4$ energy drinks and $360 \div 72 = 5$ juice packs. (1 mark)
90 kits is not possible: the number of kits must be a common factor of 216, 288 and 360, but $90 = 2 \times 3^{2} \times 5$ does not divide 216 (216 has no factor 5), so some drinks would be left over. Besides, 90 is more than the greatest possible number, 72. (1 mark)

Answer: (i) $2^{3} \times 3$ , $2^{2} \times 3^{2}$ , $2^{3} \times 5$ (ii) 360 of each: 15 cartons of water, 10 of energy drinks, 9 of juice (iii) 72 kits, each with 3 water bottles, 4 energy drinks and 5 juice packs; 90 kits is impossible because 90 does not divide 216.

Where students lose marks in HCF and LCM by Prime Factorisation

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

Mistake: Using the product rule for three numbers
The rule works prime by prime only when there are two powers to compare.
Wrong working
HCF(6, 10, 15) = 1, so LCM $= 6 \times 10 \times 15 = 900$ .
Right working
$6 = 2 \times 3$ , $10 = 2 \times 5$ , $15 = 3 \times 5$ ; LCM $= 2 \times 3 \times 5 = 30$ .
Do this: For three numbers, always find the LCM from the factorisations.
Usually costs: Full marks for the MCQ or the part.
Mistake: Swapping smallest and greatest powers
Students remember the rule as a pair but mix up which goes where.
Wrong working
HCF of $2^{3} \times 3$ and $2 \times 3^{2}$ $= 2^{3} \times 3^{2} = 72$ .
Right working
HCF $= 2 \times 3 = 6$ ; LCM $= 2^{3} \times 3^{2} = 72$ .
Do this: Check that your HCF divides both numbers: 72 does not divide 24, so it cannot be the HCF.
Usually costs: 1 to 2 marks.
Mistake: Including a prime that is not common in the HCF
A prime in only one number seems to have power 1, so students take it.
Wrong working
HCF of $2^{2} \times 3 \times 7$ and $2 \times 3^{2}$ $= 2 \times 3 \times 7 = 42$ .
Right working
7 is absent from the second number, so HCF $= 2 \times 3 = 6$ .
Do this: Treat a missing prime as power 0.
Usually costs: 1 mark.
Mistake: Leaving a composite number in the factorisation
Stopping the factor tree early at 4, 6 or 9.
Wrong working
$180 = 4 \times 5 \times 9$ , so the powers cannot be compared.
Right working
$180 = 2^{2} \times 3^{2} \times 5$ .
Do this: Every factor in the final line must be a prime, in power form.
Usually costs: ½ to 1 mark per factorisation.
Mistake: Forgetting the remainder adjustment
Students find the HCF or LCM of the given numbers directly.
Wrong working
Smallest number greater than 5 leaving remainder 5 with 28, 42, 63 is LCM $= 252$ .
Right working
$252 + 5 = 257$ .
Do this: Read the question again after getting the LCM or HCF and ask what must still be added or subtracted.
Usually costs: 1 to 2 marks.

Exam tips

Show each prime factorisation on its own line in power form. Examiners give step marks for these lines even if the final multiplication slips.
When numbers are already factorised, do not multiply them out; read off powers directly and multiply only at the end if a value is asked.
In MCQs, reject options quickly with checks: the HCF must divide every number and the LCM must be a multiple of each.
In word problems, write one line saying why you chose HCF or LCM, then finish with a sentence and units.
In assertion-reason, test claims about HCF × LCM on a pair of numbers and a triple before choosing.

Practice questions on HCF and LCM by Prime Factorisation

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

Q1 · 2 marks · Very short answerBoard level
Find the HCF and LCM of 120 and 252 by prime factorisation, and verify that HCF × LCM = product of the numbers.
Hint
Factorise both; 7 appears in only one number.
Answer
$120 = 2^{3} \times 3 \times 5$ , $252 = 2^{2} \times 3^{2} \times 7$ . HCF $= 2^{2} \times 3 = 12$ , LCM $= 2^{3} \times 3^{2} \times 5 \times 7 = 2520$ . $12 \times 2520 = 30240 = 120 \times 252$ .
Q2 · 1 mark · Assertion-reasonBoard level
Assertion (A): For the numbers 6, 10 and 15, HCF × LCM $= 6 \times 10 \times 15$ .
Reason (R): HCF × LCM = product of the numbers is true for every pair of numbers, but is not true in general for three numbers.
(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
Find the HCF and LCM of the three numbers from their factorisations and compare with the product.
Answer
(d) A is false but R is true. $6 = 2 \times 3$ , $10 = 2 \times 5$ , $15 = 3 \times 5$ , so HCF $= 1$ and LCM $= 30$ , giving HCF × LCM $= 30$ , while the product is 900. R is true: the product rule always holds for two numbers but not in general for three, as these numbers show.
Q3 · 2 marks · Very short answerBoard level
Check whether $1 2^{n}$ can end with the digit 0 for any natural number n. Give a reason.
Hint
A number ending in 0 is divisible by 10, so its prime factorisation must contain 5.
Answer
If $1 2^{n}$ ended in 0, it would be divisible by 10, so 5 would be one of its prime factors. (½ mark) But $1 2^{n} = (2^{2} \times 3)^{n} = 2^{2 n} \times 3^{n}$ , so its only primes are 2 and 3. (1 mark) By the uniqueness in the Fundamental Theorem of Arithmetic, 5 is not a factor, so $1 2^{n}$ can never end with 0. (½ mark)
Q4 · 3 marks · Short answerHarder board level
Find the greatest number that divides 383 and 605 leaving remainders 3 and 5 respectively.
Hint
Subtract the remainders first, then take the HCF. Check it exceeds both remainders.
Answer
$380 = 2^{2} \times 5 \times 19$ , $600 = 2^{3} \times 3 \times 5^{2}$ . HCF $= 2^{2} \times 5 = 20$ , which is greater than 3 and 5. Answer: 20.
Q5 · 3 marks · Short answerHarder board level
Find the least number that leaves remainder 7 when divided by each of 12, 18 and 30, and is also exactly divisible by 13.
Hint
Such numbers are 7 more than a common multiple of 12, 18 and 30. List them and test each for divisibility by 13.
Answer
$12 = 2^{2} \times 3$ , $18 = 2 \times 3^{2}$ , $30 = 2 \times 3 \times 5$ , so LCM $= 2^{2} \times 3^{2} \times 5 = 180$ . The numbers are $180 k + 7$ : 187, 367, 547, 727, 907, 1087, 1267, 1447, 1627, 1807, ... Dividing by 13 leaves remainders 5, 3, 1, 12, 10, 8, 6, 4, 2, 0, so the first one divisible by 13 is $1807 = 13 \times 139$ . Answer: 1807. (7 itself is ruled out, as 13 does not divide it.)
Q6 · 3 marks · Short answerHarder board level
For natural numbers x and y, the HCF of $2^{3} \times 3^{x} \times 5^{2}$ and $2^{y} \times 3^{4} \times 7$ is 24 and their LCM is $2^{4} \times 3^{4} \times 5^{2} \times 7$ . Find x and y.
Hint
Write 24 in prime form and compare smallest powers; use the LCM for greatest powers.
Answer
$24 = 2^{3} \times 3$ . Smaller of x and 4 is 1, so $x = 1$ . Greatest power of 2 in LCM is 4, and the first number has $2^{3}$ , so $y = 4$ (smaller of 3 and 4 is 3, matching the HCF). $x = 1$ , $y = 4$ .
Q7 · 4 marks · Long answerHOTS / competency
Three numbers have HCF 6 and LCM 72. Two of them are 12 and 18. Aman says the third is $\frac{6 \times 72}{12 \times 18} = 2$ . Find his error, and find all possible values of the third number.
Hint
The product rule is guaranteed only for two numbers. The third number must be a multiple of 6 that divides 72 and supplies the factor $2^{3}$ .
Answer
Error: HCF × LCM = product is guaranteed only for two numbers, not in general for three (also 2 is not a multiple of the HCF 6). $12 = 2^{2} \times 3$ , $18 = 2 \times 3^{2}$ , LCM(12, 18) $= 36 = 2^{2} \times 3^{2}$ . To get LCM $72 = 2^{3} \times 3^{2}$ , the third number c needs $2^{3}$ ; it must divide 72 and be a multiple of 6. So c is 24 or 72. Check: HCF(12, 18, 24) = 6 and HCF(12, 18, 72) = 6. Third number: 24 or 72.
Q8 · 5 marks · Long answerHOTS / competency
Three bells in a school toll at intervals of 24 minutes, 36 minutes and 60 minutes. They toll together at 8:00 a.m.
(i) Write the prime factorisations of 24, 36 and 60.
(ii) At what time will they next toll together?
(iii) How many times do they toll together from 8:00 a.m. to 8:00 p.m. the same day, counting both ends?
(iv) Ravi says that if the second bell is reset to toll every 45 minutes, the bells will meet at a different time. Is he right? Justify.
Hint
The bells toll together after every common multiple of the three intervals; the first such gap is the LCM.
Answer
(i) $24 = 2^{3} \times 3$ , $36 = 2^{2} \times 3^{2}$ , $60 = 2^{2} \times 3 \times 5$ . (1)
(ii) LCM $= 2^{3} \times 3^{2} \times 5 = 360$ minutes $= 6$ hours, so next at 2:00 p.m. (1½)
(iii) They toll together every 6 hours: 8:00 a.m., 2:00 p.m. and 8:00 p.m., that is 3 times. (1)
(iv) No. $45 = 3^{2} \times 5$ , so LCM(24, 45, 60) $= 2^{3} \times 3^{2} \times 5 = 360$ minutes again: the highest powers of 2, 3 and 5 are unchanged, so they still meet every 6 hours, next at 2:00 p.m. (1½)

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.HCF uses which primes and which powers?Show answer
Common primes only, each at its smallest power.
2.LCM uses which primes and which powers?Show answer
All primes present, each at its greatest power.
3.HCF × LCM = product holds for how many numbers?Show answer
Exactly two numbers.
4.HCF and LCM of two co-prime numbers a and b?Show answer
HCF = 1, LCM = ab.
5.If a divides b, what are HCF(a, b) and LCM(a, b)?Show answer
HCF = a, LCM = b.
6.Prime factorisation of 360?Show answer
$2^{3} \times 3^{2} \times 5$

HCF and LCM by Prime Factorisation: frequently asked questions

Do I have to use prime factorisation, or can I list factors?

When a question says by prime factorisation, you must show the factorisations to get the marks. Otherwise listing is allowed, but prime factorisation is faster and easier to check for larger numbers.

Why does HCF times LCM not equal the product for three numbers?

For two numbers, each prime's smaller and larger powers add up to its total power in the product. With three numbers, the middle power is counted in neither the HCF nor the LCM, so the totals stop matching.

Is Euclid's division algorithm for HCF in the syllabus?

No. Euclid's division lemma and algorithm are not in the 2026-27 syllabus, so use prime factorisation for HCF and LCM.

How do I know whether a word problem needs HCF or LCM?

If you are splitting quantities into the largest equal parts, use HCF. If things repeat and must meet again, or you want the least amount every size divides, use LCM.