What is the relation between HCF and LCM?

For two numbers a and b: HCF(a,b) × LCM(a,b) = a × b. This formula works ONLY for two numbers.

How do you find HCF using the division method?

Divide the larger number by the smaller. Then divide the divisor by the remainder. Continue until remainder is 0. The last divisor is the HCF.

Three bells ring at intervals of 6, 8, and 12 minutes. When do they ring together again?

Find LCM(6, 8, 12) = 24. They ring together after 24 minutes.

Can HCF be greater than LCM?

No. HCF is always less than or equal to LCM. They are equal only when the two numbers are equal.

HCF and LCM — Formulas, Shortcuts, and Worked Examples | TNPSC, SSC, Bank

Q: What is the LCM of fractions?

LCM of fractions = LCM of numerators ÷ HCF of denominators.

Quick Concept Summary

HCF (Highest Common Factor), also called GCD (Greatest Common Divisor), is the largest number that exactly divides two or more given numbers. LCM (Least Common Multiple) is the smallest number that is exactly divisible by two or more given numbers. Together, these two concepts appear in almost every TNPSC and SSC aptitude paper.

The key relationship is: HCF × LCM = Product of the two numbers. This formula is a powerful shortcut — if you know any three of the four values, you can find the fourth instantly. For example, if two numbers are 12 and 18, HCF = 6, then LCM = (12 × 18) / 6 = 36. Always verify: 6 × 36 = 216 = 12 × 18. ✓

HCF and LCM appear in real-world problems too: finding the largest tile size to cover a floor without cutting, finding when recurring events (bells, signals, buses) coincide, and simplifying fractions. In exams, "word problems" involving HCF/LCM are common because they test whether you know which operation to apply — HCF for "largest" or "maximum sharing" problems, LCM for "when will they meet again" or "minimum quantity" problems.

Key Formulas

Formula	Description	Notes
HCF × LCM = a × b	Relation for two numbers a and b	Only valid for TWO numbers
HCF of fractions = HCF(numerators) / LCM(denominators)	For fractions p/q and r/s	Used in fraction simplification
LCM of fractions = LCM(numerators) / HCF(denominators)	For fractions p/q and r/s	Always ≥ each fraction
If a = H×m and b = H×n (where gcd(m,n)=1)	Express numbers in terms of HCF	m and n must be co-prime
LCM = H × m × n	Express LCM in terms of HCF	Where H = HCF, m and n co-prime
Number of common divisors of a and b = number of divisors of HCF(a,b)	Count of common factors	Useful in factor-counting problems

Worked Examples

Example 1: Finding HCF using the Euclidean Division Method

Problem: Find the HCF of 252 and 180.

Step 1: Divide the larger number by the smaller number.
        252 ÷ 180 = 1, remainder = 72

Step 2: Now divide 180 by the remainder 72.
        180 ÷ 72 = 2, remainder = 36

Step 3: Now divide 72 by the remainder 36.
        72 ÷ 36 = 2, remainder = 0

Step 4: When remainder = 0, the last divisor is the HCF.
        HCF(252, 180) = 36

Verification:
252 = 36 × 7 ✓
180 = 36 × 5 ✓
gcd(7, 5) = 1 (co-prime) ✓

Answer: HCF = 36

Example 2: Finding LCM using Prime Factorisation

Problem: Find the LCM of 36, 48, and 60.

Step 1: Prime factorisation of each number.
        36  = 2² × 3²
        48  = 2⁴ × 3
        60  = 2² × 3 × 5

Step 2: For LCM, take each prime factor at its highest power.
        Highest power of 2 = 2⁴ = 16
        Highest power of 3 = 3² = 9
        Highest power of 5 = 5¹ = 5

Step 3: LCM = 2⁴ × 3² × 5
             = 16 × 9 × 5
             = 720

Answer: LCM(36, 48, 60) = 720

Check: 720 ÷ 36 = 20 ✓
       720 ÷ 48 = 15 ✓
       720 ÷ 60 = 12 ✓

Example 3: Using the HCF × LCM = Product formula

Problem: The HCF of two numbers is 14 and their LCM is 168. If one number is 56, find the other number.

Step 1: Use the formula: HCF × LCM = First number × Second number
        14 × 168 = 56 × Second number

Step 2: Solve for the second number.
        Second number = (14 × 168) / 56
                      = 2352 / 56
                      = 42

Step 3: Verify.
        HCF(56, 42) = ?
        56 = 2³ × 7
        42 = 2 × 3 × 7
        HCF = 2 × 7 = 14 ✓

        LCM(56, 42) = 2³ × 3 × 7 = 8 × 3 × 7 = 168 ✓

Answer: The second number is 42.

Example 4: Bell problem (LCM application)

Problem: Three bells ring at intervals of 9, 12, and 15 minutes respectively. They all ring together at 8:00 AM. At what time will they next ring together?

Step 1: Find the LCM of 9, 12, and 15.
        9  = 3²
        12 = 2² × 3
        15 = 3 × 5

        LCM = highest powers: 2² × 3² × 5
            = 4 × 9 × 5
            = 180 minutes

Step 2: Convert 180 minutes to hours.
        180 minutes = 3 hours

Step 3: Add to starting time.
        8:00 AM + 3 hours = 11:00 AM

Answer: The bells will next ring together at 11:00 AM.

Example 5: HCF and LCM of fractions

Problem: Find the HCF and LCM of 2/3, 4/9, and 8/27.

Step 1: Identify numerators and denominators.
        Numerators:   2, 4, 8
        Denominators: 3, 9, 27

Step 2: HCF of fractions = HCF of numerators / LCM of denominators
        HCF of 2, 4, 8:
          2 = 2
          4 = 2²
          8 = 2³
          HCF = 2 (smallest power of common prime)

        LCM of 3, 9, 27:
          3  = 3
          9  = 3²
          27 = 3³
          LCM = 3³ = 27

        HCF of fractions = 2/27

Step 3: LCM of fractions = LCM of numerators / HCF of denominators
        LCM of 2, 4, 8 = 2³ = 8
        HCF of 3, 9, 27 = 3

        LCM of fractions = 8/3

Answer: HCF = 2/27 and LCM = 8/3

Verification: HCF × LCM = (2/27) × (8/3) = 16/81
              Product of first two fractions: (2/3) × (4/9) = 8/27 ≠ 16/81
              Note: The product formula does NOT apply to fractions directly.

Shortcuts and Tricks

Trick 1: HCF from co-prime decomposition

If numbers are a = H×m and b = H×n where gcd(m,n) = 1, then:

HCF = H
LCM = H × m × n

Example: Two numbers have HCF 6. If the numbers are 6m and 6n with gcd(m,n)=1, and LCM = 72, then 6×m×n = 72 → m×n = 12. Possible co-prime pairs: (1,12), (3,4). Numbers: (6,72) or (18,24).

Trick 2: "Largest square tile" problems → HCF

Whenever a question asks "largest size of tile / rod / group / team that divides perfectly into two dimensions," the answer is HCF.

Example: A floor 48m × 36m; largest square tile → HCF(48, 36) = 12m

Trick 3: "When do they meet / ring together" → LCM

Any problem about periodicity ("buses at intervals of X and Y minutes — when do they depart together?") → LCM.

Trick 4: Least number divisible by multiple numbers

"Find the least number divisible by 4, 6, 8, 10, 12" → LCM(4,6,8,10,12) = 120

If the question says "leaves remainder r in each case" → Answer = LCM + r

Trick 5: Finding numbers when sum and HCF are given

If sum of two numbers = S and HCF = H, let the numbers be Hm and Hn. Then Hm + Hn = S → H(m+n) = S → m+n = S/H. Find co-prime pairs (m, n) that sum to S/H.

Example: Sum = 84, HCF = 12. Then m+n = 7. Co-prime pairs summing to 7: (1,6), (2,5), (3,4). But check: gcd(3,4)=1 ✓, gcd(2,5)=1 ✓, gcd(1,6)=1 ✓. So possible pairs of numbers: (12,72), (24,60), (36,48).

TNPSC Exam Pattern

Frequency: 2–4 questions per TNPSC Group II/IIA Preliminary paper. HCF/LCM is one of the most consistently tested topics across all competitive exams.

Common Question Types: 1. Direct: Find HCF/LCM of 2–3 given numbers 2. Formula-based: Given product and HCF/LCM, find the other value 3. Word problem: Bell/timer/bus problems requiring LCM 4. Tile/maximum grouping problems requiring HCF 5. Fraction HCF/LCM (rare but has appeared in Bank PO) 6. "What is the smallest number divisible by X, Y, Z with remainder R?"

Common Traps:

Using HCF × LCM = product for three or more numbers (formula is invalid)
Confusing when to use HCF vs LCM in word problems (check: "maximum/largest" → HCF; "minimum/smallest/when together" → LCM)
Bell problems: forgetting that the bells first ring together at time 0 (start), and the question asks "after how long do they ring together again" vs "how many times including the first"
Fraction HCF: many candidates flip — HCF of fractions has HCF on top and LCM on bottom

Marks Weightage: Each HCF/LCM question carries 1–2 marks. Bell-type word problems often carry 2 marks each.

Practice Questions

Q1. Find the HCF of 84 and 126.

Q2. Find the LCM of 15, 25, and 30.

Q3. The LCM of two numbers is 560 and their HCF is 8. If one number is 80, find the other.

Q4. Three bells ring at intervals of 4, 6, and 8 minutes. They ring together at 9:00 AM. How many times will they ring together in the next 2 hours (including 9:00 AM)?

Q5. Find the greatest number that divides 210, 315, and 147 exactly.

Q6. Find the HCF of 5/6, 10/9, and 15/18.

Q7. Find the smallest number which when divided by 8, 12, and 16 leaves a remainder of 3 in each case.

Q8. The product of two numbers is 4800 and their HCF is 20. How many pairs of such numbers are possible?

Q9. Find the LCM of 22, 33, and 44.

Q10. A road has two sets of signal lights. One changes every 40 seconds and the other every 60 seconds. If they change simultaneously at 10:00 AM, at what time will they next change together?

Answers

Q	Answer	Key Working
Q1	42	126 = 84×1+42; 84 = 42×2+0. HCF = 42
Q2	150	15=3×5; 25=5²; 30=2×3×5. LCM = 2×3×5² = 150
Q3	56	Second = HCF×LCM/first = 8×560/80 = 4480/80 = 56
Q4	6 times	LCM(4,6,8)=24 min. In 120 min: 120÷24=5 times after start. Including 9:00 AM: 5+1 = 6 times
Q5	21	210=2×3×5×7; 315=3²×5×7; 147=3×7². HCF = 3×7 = 21
Q6	5/18	HCF(5,10,15)=5; LCM(6,9,18)=18. HCF of fractions = 5/18
Q7	51	LCM(8,12,16)=48. Answer = 48+3 = 51
Q8	2 pairs	4800=20²×12. Need m×n=12 where gcd(m,n)=1. Pairs: (1,12) and (3,4). So 2 pairs: (20,240) and (60,80)
Q9	132	22=2×11; 33=3×11; 44=2²×11. LCM=2²×3×11 = 132
Q10	10:02 AM	LCM(40,60)=120 sec = 2 minutes. 10:00 AM + 2 min = 10:02 AM