Introduction

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LCM and HCF – A Comprehensive Guide for Competitive Exams (JKSSB Social Forestry Worker & Similar Tests)

Introduction

In the quantitative aptitude section of most government recruitment examinations—including the JKSSB Social Forestry Worker test—questions on Least Common Multiple (LCM) and Highest Common Factor (HCF) appear regularly. Though the concepts are elementary, the way they are twisted in exam questions can test a candidate’s speed, accuracy, and logical reasoning. A strong grasp of LCM and HCF not only helps in direct questions but also underpins topics such as fractions, ratios, time‑and‑work, and problems involving remainders.

This article is designed to give you a deep, exam‑oriented understanding of LCM and HCF: from basic definitions to advanced shortcuts, from theory to numerous solved examples, and finally to a set of practice questions that mirror the pattern of JKSSB and similar exams. By the end, you should be able to solve any LCM‑HCF problem within a minute, using the most efficient method available.

1. Concept Explanation ### 1.1 What is a Factor?

A factor (or divisor) of a number n is an integer that divides n exactly, leaving no remainder. Example: Factors of 12 are 1, 2, 3, 4, 6, 12.

1.2 What is a Multiple?

A multiple of a number n is the product obtained when n is multiplied by any integer.

Example: First five multiples of 7 are 7, 14, 21, 28, 35.

1.3 Highest Common Factor (HCF) – also called GCD The HCF (Highest Common Factor) or GCD (Greatest Common Divisor) of two or more integers is the largest integer that divides each of them without leaving a remainder.

  • Notation: HCF(a, b) or gcd(a, b).
  • If one of the numbers is zero, HCF(a,0) = |a| (by convention).

1.4 Least Common Multiple (LCM)

The LCM (Least Common Multiple) of two or more integers is the smallest positive integer that is divisible by each of them.

  • Notation: LCM(a, b).
  • If any number is zero, LCM is conventionally taken as 0 (but exam questions avoid zero).

1.5 Fundamental Relationship For any two positive integers a and b:

\[

\boxed{\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b}

\]

This identity is extremely useful: if you know HCF and one number, you can instantly find the other number’s LCM, and vice‑versa.

1.6 Methods to Compute HCF and LCM

Method When to Use Steps
Prime Factorisation Small to medium numbers; when you need both HCF and LCM together 1. Write each number as product of prime powers.
2. HCF = product of lowest powers of common primes.
3. LCM = product of highest powers of all primes appearing.
Division (Euclidean) Algorithm Large numbers; especially when only HCF is needed 1. Divide the larger number by the smaller, keep remainder.
2. Replace dividend with divisor and divisor with remainder.
3. Repeat until remainder = 0.
4. Last non‑zero divisor = HCF.
5. LCM = (a × b) / HCF (if needed).
Short‑Cut / Formula When numbers are given in a special form (e.g., consecutive, multiples) Use known results: HCF of consecutive numbers = 1; LCM of first n natural numbers = product of primes ≤ n with highest powers; etc.
Listing Multiples / Factors Very small numbers (≤ 20) – mainly for concept building List out multiples/factors until a common one appears. Not recommended for exams due to time consumption.

2. Key Facts & Properties (Exam‑Centric)

# Fact / Property Explanation / Usage
1 HCF ≤ each number; LCM ≥ each number Helps in eliminating options in MCQs.
2 If HCF(a,b) = 1, numbers are co‑prime → LCM = a × b. Frequently appears in fraction simplification.
3 For any set of numbers, HCF divides LCM. Useful to check consistency of given data.
4 HCF of fractions = HCF of numerators / LCM of denominators. Direct formula for HCF of rational numbers.
5 LCM of fractions = LCM of numerators / HCF of denominators. Symmetric to point 4.
6 If a = b × k (one number is a multiple of the other): HCF = b, LCM = a. Saves time in “one number divides the other” questions.
7 HCF of (a, b, c) = HCF(HCF(a,b), c). Associative property – extend to any count.
8 LCM of (a, b, c) = LCM(LCM(a,b), c). Same as above.
9 Product property: a × b = HCF × LCM (already noted). Enables quick calculation when one of HCF/LCM is known.
10 Remainder‑based problems: If a number leaves the same remainder r when divided by several numbers, then the number = LCM(divisors) × k + r. Core for “find the smallest number leaving remainder r” questions.
11 Successive division: To find the smallest number that when divided by given divisors leaves respective remainders, use the Chinese Remainder Theorem concept (often simplified for two divisors). Appears in higher‑level questions but can be solved by trial with LCM.
12 HCF of consecutive numbers = 1; LCM of first n natural numbers = ∏ p^{⌊log_p n⌋} (product over primes ≤ n). Useful for “find LCM of 1,2,…,10” type questions.
13 HCF and LCM of decimals: Convert to integers by multiplying by appropriate power of 10, compute HCF/LCM, then divide back by same factor. Rare but appears in some exams.
14 If HCF = d and LCM = L, then the numbers can be expressed as d × x and d × y where x, y are co‑prime and xy = L/d. Helps in “find two numbers given HCF and LCM”.
15 Number of pairs (a,b) with given HCF = h and LCM = l is 2^{k-1}, where k = number of distinct prime factors of (l/h). Useful for counting problems.

3. Detailed Examples (Step‑by‑Step)

Example 1 – Basic HCF & LCM via Prime Factorisation Find HCF and LCM of 48 and 180.

  1. Prime factorisation:
  • 48 = 2⁴ × 3¹
  • 180 = 2² × 3² × 5¹
  1. HCF = product of lowest powers of common primes:
  • Common primes: 2, 3 → min powers: 2², 3¹ → HCF = 2² × 3¹ = 4 × 3 = 12.
  1. LCM = product of highest powers of all primes present:
  • Primes: 2, 3, 5 → max powers: 2⁴, 3², 5¹ → LCM = 2⁴ × 3² × 5¹ = 16 × 9 × 5 = 720.
  1. Verify: HCF × LCM = 12 × 720 = 8640 = 48 × 180. ✔

Answer: HCF = 12, LCM = 720.

Example 2 – Euclidean Algorithm for HCF

Find HCF of 274 and 119.

  • 274 ÷ 119 = 2 remainder 36 → (274,119) → (119,36)
  • 119 ÷ 36 = 3 remainder 11 → (36,11) – 36 ÷ 11 = 3 remainder 3 → (11,3)
  • 11 ÷ 3 = 3 remainder 2 → (3,2)
  • 3 ÷ 2 = 1 remainder 1 → (2,1)
  • 2 ÷ 1 = 2 remainder 0 → stop.

Last non‑zero divisor = 1. Hence HCF(274,119) = 1 (they are co‑prime).

LCM = (274 × 119) / 1 = 32606.

Example 3 – HCF & LCM of Fractions

Find HCF and LCM of the fractions ³⁄₄, ⁵⁄₆, ⁷⁄₈. – HCF of fractions = HCF(numerators) / LCM(denominators)

  • Numerators: 3,5,7 → HCF = 1 (no common factor)
  • Denominators: 4,6,8 → LCM = 24 (2³×3)

→ HCF = 1⁄24

  • LCM of fractions = LCM(numerators) / HCF(denominators)
  • Numerators: 3,5,7 → LCM = 3×5×7 = 105
  • Denominators: 4,6,8 → HCF = 2

→ LCM = 105⁄2 = 52.5 (or 105⁄2 as a fraction).

Answer: HCF = 1⁄24, LCM = 105⁄2.

Example 4 – Remainder Problem (Common Remainder)

Find the smallest number which when divided by 6, 9, 15 leaves a remainder 4 in each case.

Let the required number be N.

Then N – 4 is exactly divisible by 6, 9, and 15.

So N – 4 = LCM(6,9,15) × k, where k is a positive integer.

  • LCM(6,9,15): – 6 = 2×3
  • 9 = 3²
  • 15 = 3×5

→ LCM = 2¹ × 3² × 5¹ = 2 × 9 × 5 = 90

Thus N = 90k + 4. The smallest positive N occurs when k = 1 → N = 90 + 4 = 94.

Check: 94 ÷ 6 = 15 remainder 4; 94 ÷ 9 = 10 remainder 4; 94 ÷ 15 = 6 remainder 4. ✔

Example 5 – Finding Numbers from HCF and LCM

The HCF of two numbers is 12 and their LCM is 180. If one number is 36, find the other.

Using product property:

Let the unknown number be x.

HCF × LCM = product of numbers → 12 × 180 = 36 × x

=> 2160 = 36x → x = 2160 ÷ 36 = 60.

Check: HCF(36,60) = 12, LCM(36,60) = 180. ✔

Example 6 – LCM of First n Natural Numbers (Shortcut)

Find LCM of numbers from 1 to 10. List primes ≤10: 2,3,5,7.

  • Highest power of 2 ≤10 → 2³ = 8
  • Highest power of 3 ≤10 → 3² = 9
  • Highest power of 5 ≤10 → 5¹ = 5
  • Highest power of 7 ≤10 → 7¹ = 7

LCM = 2³ × 3² × 5 × 7 = 8 × 9 × 5 × 7 = 2520.

(Indeed 2520 is the smallest number divisible by 1…10.)

4. Exam‑Focused Points & Quick‑Tricks

Situation Trick / Formula Why it Saves Time
One number divides the other (e.g., 24 and 96) HCF = smaller number, LCM = larger number No factorisation needed.
Numbers are co‑prime LCM = product, HCF = 1 Directly from definition.
HCF known, LCM unknown (or vice‑versa) Use LCM = (a×b)/HCF or HCF = (a×b)/LCM One step after product known.
Find smallest number leaving same remainder r N = LCM(divisors) × k + r Set k=1 for smallest.
Find largest number ≤ N leaving same remainder r Nmax = LCM × ⌊(N−r)/LCM⌋ + r Uses floor division.
HCF of three numbers Compute HCF pairwise iteratively Reduces to two‑number HCF repeatedly.
LCM of three numbers Compute LCM pairwise iteratively Same as above.
Decimal numbers Multiply all by 10^d (d = max decimal places), solve, then divide back Avoids dealing with fractions directly.
Fractions HCF = HCF(num)/LCM(den), LCM = LCM(num)/HCF(den) Memorise the “swap” rule.
Counting pairs with given HCF & LCM If L/H = ∏ p_i^{e_i}, number of unordered pairs = 2^{k-1} where k = number of distinct primes in L/H Useful for combinatorial LCM‑HCF questions.
When LCM of consecutive numbers is asked LCM(1,…,n) = ∏_{p≤n} p^{⌊log_p n⌋} No need to list all numbers.
Check plausibility of options HCF ≤ each option ≤ LCM; product of any two options = HCF×LCM if they are the two numbers Quick elimination in MCQs.

Tip for JKSSB: The exam usually contains 2–4 direct LCM/HCF questions and 2–3 application‑based questions (remainders, fractions, time‑and‑work). Practising the above shortcuts will let you solve each in under 30 seconds.

5. Practice Questions (with Solutions) ### Set A – Straight Computation

  1. Find the HCF and LCM of 84 and 126. 2. HCF of 221 and 323 is?
  2. LCM of 15, 20, 25 is?
  3. If HCF(a,b)=9 and LCM(a,b)=450, and a=45, find b.
  4. The HCF of three numbers is 4 and their LCM is 360. If two of the numbers are 8 and 20, find the third.

Solutions

  1. Prime factors: 84 = 2²·3·7; 126 = 2·3²·7 → HCF = 2¹·3¹·7¹ = 42; LCM = 2²·3²·7¹ = 4·9·7 = 252.
  2. Use Euclidean: 323÷221=1 r102; 221÷102=2 r17; 102÷17=6 r0 → HCF = 17.
  3. 15=3·5; 20=2²·5; 25=5² → LCM = 2²·3·5² = 4·3·25 = 300.
  4. Product = HCF×LCM = 9×450 = 4050; b = 4050/45 = 90.
  5. Let third = x. HCF(8,20,x)=4 → each number divisible by 4 → write 8=4·2, 20=4·5, x=4·y. LCM(8,20,x)=360 → LCM(4·2,4·5,4·y)=4·LCM(2,5,y)=360 → LCM(2,5,y)=90. LCM of 2 and 5 is 10, so we need LCM(10,y)=90 → y must be a divisor of 90 that makes LCM 90 → y=9 works (LCM(10,9)=90). Thus x=4·9=36.

Set B – Remainder & Application

  1. Find the least number which when divided by 5, 7, 9 leaves remainder 2 in each case.
  2. What is the greatest 4‑digit number divisible by 12, 15, and 18?
  3. Three bells toll at intervals of 6, 8, and 12 seconds. If they toll together at 9:00 am, after how many minutes will they toll together again?
  4. The HCF of two numbers is 13 and their LCM is 1170. If one number is 65, find the other.
  5. Find the HCF of the fractions ⁴⁄₉, ⁵⁄₁₂, ⁷⁄₁₈.

Solutions

  1. N‑2 must be divisible by 5,7,9 → LCM(5,7,9)=5·7·9=315 (since they’re pairwise co‑prime). Smallest N = 315+2 = 317.
  2. LCM(12,15,18): 12=2²·3, 15=3·5, 18=2·3² → LCM = 2²·3²·5 = 4·9·5=180. Greatest 4‑digit number ≤9999 divisible by 180: floor(9999/180)=55 → 55·180=9900. Answer = 9900.
  3. LCM of intervals = LCM(6,8,12): 6=2·3, 8=2³, 12=2²·3 → LCM = 2³·3 = 8·3=24 seconds. 24 seconds = 0.4 min → they toll together every 24 seconds. After 9:00 am, next toll at 9:00:24 am. In minutes, 24 sec = 0.4 min (or 2/5 minute).
  4. Product = HCF×LCM = 13×1170 = 15210. Other number = 15210/65 = 234. (Check: HCF(65,234)=13, LCM=1170.)
  5. HCF of fractions = HCF(numerators)/LCM(denominators). Numerators: 4,5,7 → HCF = 1. Denominators: 9,12,18 → LCM = 36 (9=3²,12=2²·3,18=2·3² → LCM = 2²·3²=4·9=36). Hence HCF = 1⁄36.

Set C – Higher‑Order / Conceptual 11. If the HCF of two numbers is 1 and their LCM is 210, how many unordered pairs (a,b) of positive integers satisfy this?

  1. The product of two numbers is 5040 and their HCF is 12. Find their LCM. 13. Find the smallest number which when divided by 8, 12, 16 leaves remainders 3, 7, 11 respectively.
  2. Three containers contain 40 L, 60 L, and 80 L of milk. What is the maximum capacity of a container that can measure the milk in each container exactly an integer number of times? 15. The LCM of two numbers is 4 times their HCF. If the sum of the numbers is 60, find the numbers.

Solutions

  1. HCF=1 → numbers are co‑prime. LCM = a×b = 210. So we need unordered co‑prime factor pairs of 210. Prime factorisation of 210 = 2·3·5·7 (four distinct primes). Number of ways to split the set of primes into two subsets (order doesn’t matter) = 2^{k-1} where k=4 → 2^{3}=8. However, we must exclude the case where one subset is empty (which would give a=1, b=210, still valid because HCF(1,210)=1). Actually empty subset corresponds to a=1, which is allowed. So all 2^{k-1}=8 pairs are valid. List: (1,210),(2,105),(3,70),(5,42),(6,35),(7,30),(10,21),(14,15). Answer = 8 pairs.
  1. LCM = (product)/HCF = 5040/12 = 420.
  1. Observe that each remainder is divisor‑minus‑5: 8‑5=3, 12‑5=7, 16‑5=11. So if we add 5 to the desired number N, it becomes exactly divisible by 8,12,16. Thus N+5 = LCM(8,12,16). LCM: 8=2³,12=2²·3,16=2⁴ → LCM = 2⁴·3 = 16·3=48. Hence N = 48‑5 = 43. Check: 43 mod 8 = 3, mod 12 = 7, mod 16 = 11.
  1. The required capacity is the HCF of the three volumes (largest measure that divides each exactly). HCF(40,60,80):
  • HCF(40,60)=20, HCF(20,80)=20 → HCF = 20 L.

Answer = 20 L.

  1. Let HCF = h, then LCM = 4h. Let the numbers be hx and hy where x,y are co‑prime. Then:
  • HCF = h (by construction).
  • LCM = h·xy = 4h → xy = 4.

Since x,y are co‑prime positive integers with product 4, the only possibilities are (1,4) or (2,2). But (2,2) gives HCF=2h, not h (since they share factor 2). So we must take (x,y) = (1,4) or (4,1). Thus numbers are h·1 = h and h·4 = 4h. Their sum = h + 4h = 5h = 60 → h = 12. Hence numbers are 12 and 48. Check: HCF(12,48)=12, LCM=48, indeed LCM = 4·HCF. Answer = 12 and 48.

6. FAQs (Frequently Asked Questions)

Q1. Is it always necessary to compute both HCF and LCM?

A. No. Many questions ask for only one of them. Identify what is required first; if only HCF is needed, use Euclidean algorithm (fast). If only LCM is needed, you can compute HCF first and then use LCM = (a×b)/HCF, which often saves time.

Q2. Can HCF be greater than LCM?

A. For two positive integers, HCF ≤ each number ≤ LCM, so HCF can never exceed LCM. The only case where HCF = LCM is when the two numbers are equal.

Q3. How to handle decimals in HCF/LCM problems? A. Convert all decimals to integers by multiplying each by the same power of 10 (the maximum number of decimal places among them). Compute HCF/LCM on the integers, then divide the result by the same factor.

Q4. What is the fastest way to find LCM of a set of numbers like 2,4,8,16?

A. Recognise that they are powers of 2. The LCM is the highest power present → 16. In general, if numbers are in a geometric progression with integer ratio, the LCM is the largest term.

Q5. Are there any tricks for HCF of three or more numbers without doing pairwise steps?

A. You can use the prime factorisation method for all numbers at once: write each number as product of primes, then take the minimum exponent for each prime across all numbers. This is essentially the same as pairwise but done in one go.

Q6. In remainder problems, why do we add the remainder to the number before taking LCM?

A. If N leaves remainder r when divided by d, then N−r is divisible by d. Making N−r a common multiple of all divisors ensures N satisfies all remainder conditions simultaneously.

Q7. Can HCF or LCM be zero? A. HCF of (0,0) is undefined (or taken as 0 by some conventions), but HCF(0,a)=|a|. LCM involving zero is conventionally 0 because 0 is a multiple of every integer. Exam questions avoid zero to prevent ambiguity.

Q8. How to verify an answer quickly? A. Use the product property: HCF×LCM should equal the product of the two numbers (if you have two numbers). For more than two numbers, check that each original number divides the LCM and that the HCF divides each number.

Q9. Are LCM and HCF concepts used elsewhere in the JKSSB syllabus?

A. Yes. They appear in:

  • Simplification of fractions (HCF for reducing).
  • Time‑and‑work problems (LCM for combined cycles).
  • Problems on synchronising events (bells, traffic lights).
  • Number system questions (finding smallest/largest numbers with given remainders).
  • Ratio and proportion (when scaling quantities).

Q10. Is there any shortcut to find LCM of first n natural numbers without prime factorisation?

A. For small n (≤20) you can memorise:

LCM(1–10)=2520, LCM(1–12)=27720, LCM(1–15)=360360, LCM(1–20)=232792560.

These values often appear directly in exams; memorising them saves time.

7. Final Revision Checklist

Before the exam, run through this quick mental checklist:

  • [ ] Definitions clear: HCF = greatest divisor, LCM = smallest multiple.
  • [ ] Relationship: HCF × LCM = product of two numbers (remember it).
  • [ ] Prime factorisation method mastered for both HCF and LCM.
  • [ ] Euclidean algorithm ready for large numbers (especially when only HCF needed).
  • [ ] Fractions: HCF = HCF(num)/LCM(den); LCM = LCM(num)/HCF(den).
  • [ ] Remainder problems: N = LCM(divisors)·k + r.
  • [ ] Application: bells, measuring containers, synchronising events.
  • [ ] Special cases: one number divides the other → HCF = smaller, LCM = larger; co‑prime → LCM = product.
  • [ ] Product property used to find missing number when HCF/LCM known.
  • [ ] Quick values: LCM(1–10)=2520, LCM(1–12)=27720, etc. – [ ] Time management: aim ≤45 seconds per LCM/HCF question.

If you can tick all boxes confidently, you are well‑prepared to tackle any LCM‑HCF question that appears in the JKSSB Social Forestry Worker exam or similar tests.

—

Good luck, and may your calculations be swift and accurate!

Editorial Team

Editorial Team

Founder & Content Creator at EduFrugal

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