HCF LCM and Choosing the Right Model

What you will learn

Use exact divisibility to decide whether a problem needs a common factor or a common multiple. By the end you should be able to:

• Find the HCF and LCM of two or three positive integers and justify every prime power chosen
• Identify what the unknown quantity must divide, or what it must be divisible by
• Solve largest identical-kit and next-common-event problems with stated conditions
• Use and verify the product identity for two positive integers
• Reduce one specified-remainder problem to exact divisibility

This lesson follows Factors Divisibility and Prime Factorisation. All numbers for which we find HCF and LCM here are positive integers: 1, 2, 3, … . We do not apply these procedures to fractions, negative numbers or zero. HCF means highest common factor, also called greatest common divisor or GCD. LCM means least common multiple; we always mean the least positive common multiple.

1 Retrieve factors before choosing a method

Try these three short checks before reading the feedback.

1. Write 60 as a product of prime powers.
2. List the positive common factors of 8 and 12.
3. In 35 = 5 × 7, explain the relationship between 5 and 35.

Feedback:

1. 60 = 2² × 3 × 5. Multiply back: 4 × 3 × 5 = 60. Every factor in the prime list is prime.
2. The factors of 8 are 1, 2, 4, 8; those of 12 are 1, 2, 3, 4, 6, 12. The common factors are 1, 2, 4.
3. 5 is a factor of 35 because 35 ÷ 5 = 7 with remainder 0. Equivalently, 35 is a multiple of 5 because 35 = 5 × 7.

If you listed multiples instead of factors, first complete the sentence “this number divides the given total exactly.” If a prime factorisation does not multiply back to the original number, repair it using the previous lesson. We now use that knowledge rather than repeat the divisibility tests.

2 Two different directions of divisibility

A common factor divides every given number exactly. The HCF is the greatest positive number with that property. Since 1 divides every positive integer, the HCF is at least 1. If no prime factor is shared by all the numbers, the HCF is 1, not 0.

A common multiple is divisible by every given number. The LCM is the smallest positive number with that property. Although 0 is divisible by every positive integer, it is excluded from our LCM definition. The product of the given numbers is a common multiple, but it may contain unnecessary repeated prime factors and need not be the least one.

Read the direction before calculating:

• HCF: each given number ÷ the answer must be a whole number
• LCM: the answer ÷ each given number must be a whole number

Worked example 1 Meaning through lists and prime factors

Find the HCF and LCM of 18 and 30.

The positive factors of 18 are 1, 2, 3, 6, 9, 18. The positive factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30. Their common factors are 1, 2, 3, 6, so HCF = 6.

List positive multiples in increasing order:

• 18: 18, 36, 54, 72, 90, …
• 30: 30, 60, 90, …

The first meeting is 90, so LCM = 90.

Now use 18 = 2 × 3² and 30 = 2 × 3 × 5. For a common factor we can take one 2 and one 3: both numbers contain them. A second 3 fails because 30 does not contain it, and 5 fails because 18 does not contain it. Thus HCF = 2 × 3 = 6.

For a common multiple we must provide one 2, two 3s and one 5, so that the requirements of both numbers are met. Those factors give LCM = 2 × 3² × 5 = 90. There is no need to count the shared 2 and 3 twice.

Check: 18 ÷ 6 = 3 and 30 ÷ 6 = 5; 90 ÷ 18 = 5 and 90 ÷ 30 = 3. These checks confirm divisibility. The full lists or the prime-factor reasoning establish “greatest” and “least”. The product 18 × 30 = 540 is a common multiple, but it is not the LCM.

3 Why the exponent table works

A prime exponent records how many copies of that prime occur. In an exponent table, write 0 when the prime is absent. This means no copy is available or required from that number; its contribution to a product is 1.

For the HCF, a chosen prime power must fit inside every number. Its exponent cannot exceed any entry in that prime's column. The largest exponent that still fits is therefore the smallest column entry. Do this for every prime. A minimum of 0 means that prime is omitted. If all minima are zero, the HCF is 1.

For the LCM, the number we build must contain enough copies for every given number. Its exponent must reach the largest entry in each prime column. Choosing exactly that largest entry gives enough without adding unnecessary factors. Every common multiple must contain these required powers, so it is a multiple of the number built this way. That is why this construction gives the least positive common multiple.

The same reasoning works for two numbers or for three. We are not choosing the smaller or larger original number: we compare the exponents separately for each prime.

Worked example 2 A three-number exponent table

Find the HCF and LCM of 72, 90 and 120.

72 = 2³ × 3²; 90 = 2 × 3² × 5; 120 = 2³ × 3 × 5.

Number	Exponent of 2	Exponent of 3	Exponent of 5
72	3	2	0
90	1	2	1
120	3	1	1
Smallest in column	1	1	0
Largest in column	3	2	1

HCF = 2¹ × 3¹ = 6. The 5 is omitted because 72 contains no 5. LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360.

Check the factors: 72 ÷ 6 = 12; 90 ÷ 6 = 15; 120 ÷ 6 = 20. Check the multiples: 360 ÷ 72 = 5; 360 ÷ 90 = 4; 360 ÷ 120 = 3.

Error check: 30 is a common factor of 90 and 120, but it does not divide 72. “Common” must include every given number. Ignoring a zero entry changes the problem.

Two quick size checks help catch mistakes: the HCF cannot exceed the smallest given positive integer, and the LCM cannot be less than the largest. These checks can reject a wrong answer, but cannot by themselves prove that an answer is correct.

4 Let the unknown quantity choose the model

Before using either abbreviation, name the unknown and complete a divisibility sentence.

1. If the unknown must divide each given total exactly, it is a common factor. If the greatest such value is wanted, use HCF.
2. If the unknown must be divisible by every given size or interval, it is a common multiple. If the least positive such value is wanted, use LCM.

Do not choose by a single word such as “groups” or “maximum”. A greatest kit count divides the available totals. A least total that can be arranged into two specified group sizes must be divisible by those sizes. Also put measurements in the same unit before comparing them.

Worked example 3 The greatest number of identical kits

All 84 notebooks and 126 pens must be used to make the greatest possible number of identical kits, with nothing left over. Every kit must contain the same number of notebooks and the same number of pens. Find the number of kits and the contents of each.

Let the kit count be the unknown. It must divide both 84 and 126 exactly. We therefore need their HCF. 84 = 2² × 3 × 7 and 126 = 2 × 3² × 7. HCF = 2 × 3 × 7 = 42 kits.

Each kit has 84 ÷ 42 = 2 notebooks and 126 ÷ 42 = 3 pens. Check: 42 × 2 = 84 and 42 × 3 = 126, so all items are used and every kit matches.

Error check: 42 is the number of kits, not the number of items in each kit. Fewer kits could contain more items each, but the question asks for the greatest kit count. The LCM does not divide the available totals and cannot answer that question.

Worked example 4 The next common event

Three signals sound at fixed intervals of 8, 12 and 18 seconds, without skipping any signal. They sound together now. After how many positive seconds will all three next sound together?

The unknown is elapsed time from now. For each signal to sound then, this time must contain a whole number of its intervals. We need the least positive common multiple of 8, 12 and 18. 8 = 2³; 12 = 2² × 3; 18 = 2 × 3². LCM = 2³ × 3² = 8 × 9 = 72 seconds.

Check: 72 ÷ 8 = 9 intervals; 72 ÷ 12 = 6 intervals; 72 ÷ 18 = 4 intervals. The maximum prime-exponent requirements show that no smaller positive elapsed time works for all three.

Error check: 0 is the coincidence happening now, not the next one. Adding or averaging the intervals does not enforce divisibility by all three. The shared starting instant and fixed, unskipped intervals are essential conditions for this model.

5 A product identity for exactly two positive integers

For positive integers a and b: HCF(a, b) × LCM(a, b) = a × b.

Here is why. Focus on one prime. Suppose it occurs r times in a and s times in b. The HCF takes the smaller of r and s, and the LCM takes the larger. Between them they count r + s copies, exactly as the product a × b does. In symbols, min(r, s) + max(r, s) = r + s; “min” and “max” mean smaller and larger. This is true for every prime, including a prime absent from one number, whose count is 0. The two products therefore contain exactly the same prime factors.

For the pair 18 and 30, this gives 6 × 90 = 540 = 18 × 30. If the two numbers are co-prime, their HCF is 1 and their LCM equals their product. That conclusion depends on the HCF being 1; it is not a rule to multiply any two numbers to find their LCM.

Worked example 5 Find a missing number and verify it

Two positive integers have HCF 12 and LCM 180. One number is 36. Find the other.

Their product is 12 × 180 = 2160. Divide by the known number: Other number = 2160 ÷ 36 = 60.

Now verify both original conditions. 36 = 2² × 3² and 60 = 2² × 3 × 5. Therefore HCF = 2² × 3 = 12 and LCM = 2² × 3² × 5 = 180. Both match the question. The product formula produces a candidate; checking its HCF and LCM guards against inconsistent given data or arithmetic errors.

Do not extend the identity to three numbers. For 4, 6 and 8, HCF = 2 and LCM = 24. Their product is 4 × 6 × 8 = 192, whereas HCF × LCM = 2 × 24 = 48. With three prime counts, taking only the smallest and largest does not in general count all three.

6 A specified remainder changes what must divide exactly

For a positive divisor, division with remainder means: Dividend = divisor × whole-number quotient + remainder, with 0 ≤ remainder < divisor.

If two stated dividends leave the same specified remainder, subtract that remainder from each. The divisor must then divide both adjusted numbers exactly. In the type used below, both adjusted numbers are positive. When the greatest divisor is wanted, find their HCF, then check that the stated remainder is smaller than the candidate divisor and substitute back. Subtracting is a way to expose exact divisibility, not permission to ignore the remainder condition.

Worked example 6 The greatest divisor leaving remainder 2

Find the greatest positive integer that, when used to divide 98 and 146, leaves remainder 2 in each case.

1. Subtract the remainder: 98 − 2 = 96 and 146 − 2 = 144.
2. Any valid divisor must divide both 96 and 144. Factorise: 96 = 2⁵ × 3 and 144 = 2⁴ × 3².
3. HCF = 2⁴ × 3 = 16 × 3 = 48, so no valid divisor can exceed 48.
4. Verify the candidate: 98 = 48 × 2 + 2 and 146 = 48 × 3 + 2.
5. The remainder is valid because 0 ≤ 2 < 48. Thus 48 is the required greatest divisor.

Error check: HCF(98, 146) = 2 answers a different question about exact division. Dividing by 2 leaves remainder 0 here, and a remainder of 2 would be invalid with divisor 2. We have solved a greatest-divisor question; we have not introduced a general “LCM plus remainder” rule.

7 Independent practice

Attempt all twelve questions before reading the key. Choose one option and write the divisibility condition or calculation that supports it. This is untimed learning practice, with no negative marking or pass cutoff; it is not a full RRB CBT simulation.

1. What is the HCF of 24 and 40?

A. 4 B. 8 C. 16 D. 120

2. What is the LCM of 15 and 20?

A. 5 B. 30 C. 60 D. 300

3. What is the HCF of 18, 30 and 42?

A. 6 B. 3 C. 18 D. 126

4. What is the LCM of 6, 10 and 15?

A. 1 B. 15 C. 60 D. 30

5. Ropes of 96 cm and 144 cm are cut into pieces of the same positive whole-centimetre length, with no waste. What is the greatest possible piece length?

A. 24 cm B. 48 cm C. 72 cm D. 288 cm

6. All 60 pencils and 84 erasers are to be used in the greatest possible number of identical kits. Each kit must have the same number of pencils and the same number of erasers. Which result is correct?

A. 6 kits; 10 pencils and 14 erasers each B. 12 kits; 7 pencils and 5 erasers each C. 12 kits; 5 pencils and 7 erasers each D. 24 kits; 5 pencils and 7 erasers each

7. Two alarms sound at fixed intervals of 9 minutes and 15 minutes, without skipping any alarm. They sound together now. After how many positive minutes will they next sound together?

A. 45 minutes B. 3 minutes C. 24 minutes D. 135 minutes

8. What is the least positive number of beads that can be packed with none left over either into groups of 12 or, as a separate choice, into groups of 18?

A. 6 B. 24 C. 216 D. 36

9. Two positive integers have HCF 6 and LCM 210. One integer is 30. What is the other integer?

A. 7 B. 36 C. 42 D. 70

10. What is the greatest positive integer that, when used to divide 111 and 147, leaves remainder 3 in each case?

A. 36 B. 3 C. 48 D. 108

11. A learner writes 24 = 2³ × 3 and 36 = 2² × 3², then chooses the larger exponent of each prime and reports HCF = 72. Which correction is right?

A. 72 is the HCF because it uses all prime factors B. 72 is the LCM; the HCF is 12 C. The HCF is 6 because repeated primes must be ignored D. The HCF is 0 because the exponents differ

12. A learner claims that HCF × LCM always equals the product of three positive integers. Which calculation correctly disproves the claim?

A. For 4, 6, 8: HCF = 2, LCM = 96, and both products are 192 B. For 4, 6, 8: HCF = 4, LCM = 24, and both products are 96 C. For 4, 6, 8: HCF = 2, LCM = 24, and both products are 48 D. For 4, 6, 8: HCF × LCM = 2 × 24 = 48, but 4 × 6 × 8 = 192

8 Explained answer key

1. B. 24 = 2³ × 3 and 40 = 2³ × 5. The three copies of 2 are common to both, so HCF = 2³ = 8. Check: 24 ÷ 8 = 3 and 40 ÷ 8 = 5. The common factor 4 is not the greatest; 16 does not divide 24 or 40.

2. C. 15 = 3 × 5 and 20 = 2² × 5. The least common multiple needs 2², 3 and one copy of 5: 4 × 3 × 5 = 60. Check: 60 ÷ 15 = 4 and 60 ÷ 20 = 3. Although 300 is also a common multiple, it is not the least.

3. A. 18 = 2 × 3², 30 = 2 × 3 × 5 and 42 = 2 × 3 × 7. Only one 2 and one 3 are available in every number, so HCF = 2 × 3 = 6. Check: 18 ÷ 6 = 3, 30 ÷ 6 = 5 and 42 ÷ 6 = 7.

4. D. 6 = 2 × 3, 10 = 2 × 5 and 15 = 3 × 5. Taking each required prime once gives LCM = 2 × 3 × 5 = 30. It contains the prime factors needed by each number. Check: 30 ÷ 6 = 5, 30 ÷ 10 = 3 and 30 ÷ 15 = 2. Taking only primes shared by all three would find the HCF, which is 1.

5. B. The piece length must divide both rope lengths exactly, so use HCF. 96 = 2⁵ × 3 and 144 = 2⁴ × 3², giving HCF = 2⁴ × 3 = 48 cm. The ropes give 96 ÷ 48 = 2 and 144 ÷ 48 = 3 pieces. Their LCM, 288 cm, is longer than either rope and cannot be cut from either one.

6. C. The kit count must divide 60 and 84. Since 60 = 2² × 3 × 5 and 84 = 2² × 3 × 7, the greatest count is HCF = 12. Each kit contains 60 ÷ 12 = 5 pencils and 84 ÷ 12 = 7 erasers. Check: 12 × 5 = 60 and 12 × 7 = 84. 6 kits are possible but do not give the greatest count.

7. A. The elapsed time must be a positive multiple of both 9 and 15. Since 9 = 3² and 15 = 3 × 5, LCM = 3² × 5 = 45 minutes. This is 45 ÷ 9 = 5 intervals of the first alarm and 45 ÷ 15 = 3 intervals of the second. The current time, 0 minutes, is excluded by “next”.

8. D. The unknown is the total number of beads. It must be divisible by 12 and by 18, so find their LCM. 12 = 2² × 3 and 18 = 2 × 3² give LCM = 2² × 3² = 36. Then 36 ÷ 12 = 3 groups or 36 ÷ 18 = 2 groups. The word “groups” alone does not mean HCF: here the unknown total must be a multiple of each given group size.

9. C. For two positive integers, the product is HCF × LCM = 6 × 210 = 1260. The other number is 1260 ÷ 30 = 42. Verify the original data: 30 = 2 × 3 × 5 and 42 = 2 × 3 × 7, so HCF = 2 × 3 = 6 and LCM = 2 × 3 × 5 × 7 = 210. The product calculation and both checks agree.

10. A. Remove the specified remainder: 111 − 3 = 108 and 147 − 3 = 144. A valid divisor must divide both. Since 108 = 2² × 3³ and 144 = 2⁴ × 3², HCF = 2² × 3² = 36. Verify: 111 = 36 × 3 + 3 and 147 = 36 × 4 + 3, with 0 ≤ 3 < 36. The divisor 3 cannot leave remainder 3 because a remainder must be smaller than its divisor.

11. B. A common divisor cannot contain more copies of a prime than either number supplies. Use the smaller exponents: HCF = 2² × 3 = 12. Check: 24 ÷ 12 = 2 and 36 ÷ 12 = 3. The larger exponents give LCM = 2³ × 3² = 72, which is divisible by 24 and 36. As 72 exceeds both positive numbers, it cannot divide either of them.

12. D. 4 = 2², 6 = 2 × 3 and 8 = 2³. Their HCF is 2 and LCM is 2³ × 3 = 24. Hence HCF × LCM = 48, whereas their product is 192. One valid counterexample disproves an “always” claim. For three exponents, adding only the smallest and largest does not generally count all three; the two-number proof cannot be extended in that way.

9 Use an error to choose what to revisit

• Confused factor and multiple, or obtained HCF 0: revisit section 2
• Chose the wrong prime powers or ignored an absent prime: revisit section 3
• Chose a method only from a word such as “groups”: revisit section 4 and name the unknown
• Used the product identity with three numbers or did not verify a missing number: revisit section 5
• Used the original dividends or forgot remainder < divisor: revisit section 6

Cover the solution, retry a missed question, and explain why your chosen answer divides the totals or is divisible by the intervals. Then check the result in the original situation. A correct numerical answer without the right divisibility model is worth revisiting too.

Sources and next step

The teaching, examples and questions are original. The mathematical reference is NCERT Class VI Mathematics Chapter 3 Playing with Numbers on IIT Kanpur SATHEE: §3.7 Highest Common Factor, including Exercise 3.6; §3.8 Lowest Common Multiple; and §3.9 Some Problems on HCF and LCM, including Exercise 3.7. These sections support the definitions, prime-factor methods and exact-measure/common-multiple models. The two-number product identity above is justified here with our own prime-count argument; it is not presented as a proof quoted from that chapter.

The full HTML sections were inspected on 30 September 2026. The mirror does not establish a verified NCERT edition/reprint year or printed-page mapping, so section headings are used instead of guessed pages. Its tables contain transcription issues; source exercises and diagrams have not been copied. The ambiguous “least number leaving a remainder” wording in §3.9 Example 14 is not used in this lesson.

Next, Fractions Decimals and Exact Calculation uses common factors to simplify fractions and common multiples to choose shared denominators.

Analogy

Cutting shorter pieces and building a shared length

Imagine strips of lengths 18 cm and 30 cm. There are two different jobs.

For the cutting job, cut each strip into pieces of the same greatest possible whole-centimetre length, with no waste. The piece length must divide both 18 and 30, so it is 6 cm. The strips produce 3 and 5 pieces.

For the building job, you have enough whole 18 cm strips and enough whole 30 cm strips. Make one line using only 18 cm strips and another using only 30 cm strips. Find the shortest positive length both lines can have, without cutting any strip. This shared length must be a multiple of 18 and 30, so it is 90 cm: 5 strips of 18 cm and 3 strips of 30 cm.

HCF answers what can fit into both given totals. LCM answers what can contain an exact number of each given unit. The comparison is about exact lengths; the kit and signal problems still require you to name their own unknown quantities and conditions.

Quick reference

Meaning and prime powers

Use positive integers throughout this lesson.

• HCF is the greatest positive common divisor. It divides every given number. Choose the smallest exponent of each prime across all numbers; an absent prime has count 0. If no prime is shared by all, HCF = 1.
• LCM is the least positive common multiple. Every given number divides it. Choose the largest exponent of every prime that occurs in any number.
• HCF ≤ the smallest given number; LCM ≥ the largest given number. These are checks, not complete proofs.

Model choice

Name the unknown before choosing a method.

• Greatest exact piece size or greatest identical-kit count: the unknown divides each total → HCF
• Least total accommodating each given group size: every group size divides the unknown → LCM
• Next coincidence after a shared start, with fixed unskipped intervals: every interval divides the positive elapsed time → LCM

Use common units and check the result in the original situation.

Two-number identity

For two positive integers a, b only: HCF(a, b) × LCM(a, b) = a × b. Smaller prime count + larger prime count = their total count. A missing number equals (HCF × LCM) ÷ the known number; verify both HCF and LCM afterward. Co-prime pairs have HCF 1, so their LCM is their product. Do not apply the two-number identity to three numbers.

Specified common remainder

For the type taught here, subtract the stated remainder from both dividends, find the HCF of the two positive adjusted numbers, and check the candidate in the original divisions. Always require 0 ≤ remainder < divisor. Finding a common factor is not enough if the remainder condition fails.