HCF and LCM

HCF (Highest Common Factor) is the largest number that divides all given numbers exactly. LCM (Lowest Common Multiple) is the smallest number that all given numbers divide into exactly. Find them using prime factorisation: HCF = product of common prime factors with the LOWEST powers; LCM = product of all prime factors with the HIGHEST powers. Most important formula: for any two numbers, HCF × LCM = Product of the two numbers. Also, HCF of fractions = HCF of numerators / LCM of denominators, and LCM of fractions = LCM of numerators / HCF of denominators. HCF is always a factor of LCM.

Two students stare at the same word problem. One reads "largest size" and instantly writes "HCF"; the other reads it and freezes. The difference isn't talent — it's a small set of keyword reflexes that turn HCF and LCM word problems into one-line decisions. This lesson hands you those reflexes for SSC CHSL.

Definition: The HCF (Highest Common Factor, also called GCD) of two or more numbers is the largest positive integer that divides each of them exactly.

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

The Two Mental Pictures

If you remember just one image per concept, you'll never confuse them again:

• HCF = "splitting into the biggest equal groups". You have several quantities; you want to break each down into equal pieces, and you want those pieces to be as large as possible. Whenever the problem speaks of dividing, cutting, distributing equally, or finding "the largest box / largest piece / largest length" — HCF.

• LCM = "coming back together at the same moment". You have several cycles repeating at different intervals; you want to know when they will all synchronise again. Whenever the problem speaks of bells ringing together, lights blinking together, traffic signals matching, or finding "the smallest / least number that is a multiple of each" — LCM.

Keyword Reflex Table

If the question uses any of these words, your hand should already be writing the answer:

• HCF triggers: "largest", "greatest", "maximum", "biggest", "cuts into equal pieces", "divides", "distributes equally so nothing is left over", "longest tape/rope/length".
• LCM triggers: "least", "smallest", "minimum", "together again", "ring at the same time", "simultaneously", "smallest number divisible by".

Classic HCF Problems

Cutting equal pieces

Question: Three ropes are 60 m, 84 m and 132 m long. They are to be cut into pieces of equal length, each piece being as long as possible. Find the length of each piece.

Solution:

Step 1: "As long as possible, equal pieces from several ropes" → HCF.
Step 2: HCF(60, 84, 132). Prime factorise:
60 = 2² × 3 × 5; 84 = 2² × 3 × 7; 132 = 2² × 3 × 11.
Common factors: 2² × 3 = 12.
Step 3: Each piece = 12 m. Number of pieces = 60/12 + 84/12 + 132/12 = 5 + 7 + 11 = 23 pieces.

Conclusion: Each piece is 12 m long; you get 23 pieces in total.

Distributing items equally

Question: 84 chocolates, 126 toffees and 168 biscuits are to be packed into identical gift boxes such that each box contains the same number of each item and no item is left over. Find the maximum number of boxes.

Solution:

Step 1: "Maximum number of identical boxes, no item left over" → HCF.
Step 2: HCF(84, 126, 168) = 42 (you can verify: 84 = 42×2, 126 = 42×3, 168 = 42×4).

Conclusion: 42 boxes, each with 2 chocolates, 3 toffees, 4 biscuits.

Classic LCM Problems

Bells, lights and signals

Question: Three bells toll at intervals of 8, 12 and 18 minutes respectively. They tolled together at 9:00 a.m. When do they next toll together?

Solution:

Step 1: "Toll together again" → LCM.
Step 2: LCM(8, 12, 18). Prime: 8 = 2³, 12 = 2² × 3, 18 = 2 × 3². Take the highest power of each prime: 2³ × 3² = 8 × 9 = 72.
Step 3: 72 minutes after 9:00 a.m. = 10:12 a.m.

Conclusion: They next ring together at 10:12 a.m.

Smallest number with the same remainder

Definition: A number that leaves the same remainder r when divided by a, b, c is of the form LCM(a, b, c) × k + r, where k is any non-negative integer.

Question: Find the smallest number that leaves remainder 4 when divided by 6, 8 and 12.

Solution:

Step 1: "Smallest, same remainder" → LCM + remainder.
Step 2: LCM(6, 8, 12) = 24. Add the remainder: 24 + 4 = 28.

Conclusion: 28. Check: 28 ÷ 6 = 4 rem 4; 28 ÷ 8 = 3 rem 4; 28 ÷ 12 = 2 rem 4. ✓

Smallest number with different remainders that share a pattern

If the divisors leave different remainders r₁, r₂, r₃ but the gap d_i = (divisor_i − r_i) is the same for all, the required smallest number is LCM(a, b, c) − d. This pattern is a CHSL favourite — recognise it by checking that (divisor − remainder) is constant.

The Master Formula

For any two positive integers a and b:

HCF(a, b) × LCM(a, b) = a × b

This lets you find one if you have the other. Useful when the problem hands you, say, HCF = 12 and the two numbers as 36 and 60 — instantly LCM = (36×60)/12 = 180.

For three or more numbers this product-rule does not extend directly (be careful — SSC sometimes plants a trap there). It only works for two numbers at a time.

Why It Matters

HCF/LCM word problems appear in every SSC CHSL Tier-I paper, usually as 1–3 questions in the Quant section. They are pure speed: if you spot the keyword in 5 seconds, the rest is arithmetic. Speed here also feeds into ratios, time-and-work, time-distance, and clock problems — every chapter that secretly leans on LCM under the hood.

Why it matters: A confident HCF/LCM aspirant gains 2–4 marks per shift and finishes each problem in under 45 seconds, leaving more time for the trickier Data Interpretation block later.

Real-world example: The Mumbai BEST bus depot uses a slightly more complex LCM idea to schedule four bus routes that share the same junction. If the routes have headways of 6, 9, 15 and 20 minutes, the simultaneous arrival happens once every LCM(6, 9, 15, 20) = 180 minutes, i.e. once every 3 hours. The timetabling team uses exactly this calculation when planning shared-platform usage.

Common Misconceptions

Common misconception: "HCF is for division, LCM is for multiplication." Not quite — both can be computed from prime factorisation. Use them based on the word problem's intent, not on the arithmetic operation.

Common misconception: "Bigger numbers means LCM." No — the biggest piece comes from HCF, the smallest synchronisation interval comes from LCM. The size of the original numbers doesn't tell you which to use.

Common misconception: "HCF × LCM = product of all numbers, even for three or more." Only for two numbers. For three or more, this product rule fails in general.

Common misconception: "If the problem says 'minimum', use HCF; if 'maximum', use LCM." It's the opposite. Minimum → LCM (smallest common multiple); Maximum → HCF (largest common factor).

Q: Three bells ring at intervals of 9, 12 and 15 minutes. If they ring together at 10:00 am, when will they next ring together? Solution: We need the LCM of 9, 12, 15. Prime factors: 9 = 3², 12 = 2²×3, 15 = 3×5. LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180 minutes = 3 hours. So they ring together again at 10:00 am + 3 hours = 1:00 pm. Answer: 1:00 pm.

HCF and LCM is a guaranteed 1–2 question area in SSC CHSL — "largest number that divides", "smallest number divisible by", bell/traffic-light timing, and fraction problems. One core identity and two fraction formulas cover almost everything.

Core relations

• Product identity: HCF × LCM = product of the two numbers. (Only for two numbers.)
• HCF of fractions = HCF(numerators) / LCM(denominators).
• LCM of fractions = LCM(numerators) / HCF(denominators).
• HCF always divides the numbers; LCM is always a multiple of them; HCF ≤ each number ≤ LCM.

Quantity	Method
HCF	product of lowest powers of common primes
LCM	product of highest powers of all primes
Two numbers	HCF × LCM = a × b

Remainder patterns

• Greatest number dividing a, b, c leaving the same remainder r → HCF of the differences (a−b), (b−c).
• Least number leaving remainder r with each divisor → LCM of divisors + r.
• Bells toll together again after LCM of the intervals.

Exam Tricks & Tips

• 🎯 With two numbers, if HCF and LCM are given, get the missing number instantly from HCF × LCM = a × b.
• 🎯 "Same remainder" problems → HCF of successive differences (the remainder cancels out).
• 🎯 LCM of the numbers is always divisible by their HCF — use it to reject impossible option pairs.
• 🎯 For "greatest number that divides a, b, c leaving remainders p, q, r" → HCF of (a−p), (b−q), (c−r).
• 🎯 To add/compare fractions fast, take LCM of denominators as common denominator.
• ❌ The product identity fails for three or more numbers — never write HCF×LCM = a×b×c.

Expected exam pattern

"Find the greatest number dividing 258 and 323 leaving remainders 2 and 3", or "bells ring at 4,6,8,10 sec; when together again?". Usually 30–40 seconds.

Quick recap

Use prime factorisation for HCF (lowest powers) and LCM (highest powers); apply HCF×LCM=a×b for two numbers; and the fraction formulas HCF=HCF(num)/LCM(den), LCM=LCM(num)/HCF(den).

Cover the answer, recall, then check. 11 cards on HCF & LCM for SSC CHSL.

Q1. State the product identity for two numbers.
A1. HCF × LCM = product of the two numbers (a × b). Valid only for two numbers.

Q2. LCM of two numbers is 84, HCF is 6, one number is 12. Find the other.
A2. Other = (HCF×LCM)/12 = (6×84)/12 = 504/12 = 42.

Q3. Formula for HCF of fractions?
A3. HCF(numerators) / LCM(denominators).

Q4. Formula for LCM of fractions?
A4. LCM(numerators) / HCF(denominators).

Q5. HCF of 2/3, 4/9, 6/15?
A5. HCF(2,4,6)/LCM(3,9,15) = 2/45.

Q6. How do you find HCF by prime factorisation?
A6. Multiply the lowest powers of the common prime factors.

Q7. How do you find LCM by prime factorisation?
A7. Multiply the highest powers of all prime factors that appear.

Q8. Greatest number that divides 70 and 125 leaving remainders 5 and 8?
A8. HCF of (70−5)=65 and (125−8)=117 → HCF(65,117)=13.

Q9. Least number which when divided by 6, 8, 12 leaves remainder 2 each time?
A9. LCM(6,8,12) + 2 = 24 + 2 = 26.

Q10. Bells toll at intervals 4, 6, 8, 10 s together at start; after how long together again?
A10. LCM(4,6,8,10) = 120 s = 2 minutes.

Q11. Does HCF×LCM = product work for three numbers?
A11. No — the identity holds only for two numbers.