LCM & HCF — SSC CGL Study Notes
Everything the chapter tests — in plain words, with one example under every rule and the traps SSC likes to set.
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LCM and HCF is one of the shortest chapters in SSC CGL Quant and one of the most reliable for marks. Almost every question is one of about a dozen patterns. These notes cover each pattern once, in simple language, with the rule stated in general letters, one worked example, a shortcut and the trap that costs marks.
Topics covered: factors and multiples · finding HCF and LCM · the product rule · fractions and decimals · prime-power forms · bells, lights and runners · least and greatest numbers with remainders · measuring rods and cutting · numbers in a ratio · pairs from a product or a sum · perfect squares · polynomials · traps.
1. Factors, multiples, HCF and LCM
Concept
A factor of a number divides it exactly (3 is a factor of 12). A multiple is the number times a whole number (12, 24, 36 are multiples of 12).
HCF (Highest Common Factor) of some numbers = the largest number that divides all of them.
LCM (Least Common Multiple) = the smallest number that all of them divide.
Two numbers are co-prime when their HCF is 1 (8 and 15). They need not be prime themselves.
Example 12 and 18. Factors of 12: 1, 2, 3, 4, 6, 12. Factors of 18: 1, 2, 3, 6, 9, 18. Common: 1, 2, 3, 6 → HCF = 6. Multiples of 12: 12, 24, 36, … and of 18: 18, 36, … → LCM = 36.
Shortcut If one number divides the other, the smaller one is the HCF and the larger one is the LCM. HCF(8, 24) = 8, LCM(8, 24) = 24.
2. Finding the HCF
Method A — prime factorisation
Rule Write each number as a product of primes. The HCF takes every common prime at its lowest power.
Example 360 = 23·32·5 and 84 = 22·3·7. Common primes 2 and 3, lowest powers 22 and 31. HCF = 4 × 3 = 12.
Method B — Euclid's division (best for big numbers)
Rule Divide the larger by the smaller. Then divide the divisor by the remainder. Repeat until the remainder is 0. The last divisor is the HCF.
Example HCF(1794, 2346): 2346 = 1794 × 1 + 552; 1794 = 552 × 3 + 138; 552 = 138 × 4 + 0. HCF = 138.
Shortcut Any common factor of two numbers also divides their difference. So HCF(a, b) = HCF(b, a − b). For 2346 and 1794 the difference is 552, and you can start from 552 straight away.
3. Finding the LCM
Method A — prime factorisation
Rule The LCM takes every prime that appears in any number, at its highest power.
Example 36 = 22·32, 54 = 2·33, 90 = 2·32·5. Highest powers: 22, 33, 5. LCM = 4 × 27 × 5 = 540.
Method B — the division ladder
2 | 36 54 90
3 | 18 27 45
3 | 6 9 15
| 2 3 5 LCM = 2×3×3 × 2×3×5 = 540
Divide the whole row by any prime that divides at least two of the numbers; carry the others down unchanged. Stop when no prime divides two of them. Multiply all the divisors and the bottom row.
Shortcut A number that divides another member of the set can be dropped before finding the LCM. LCM(15, 25, 75, 90) = LCM(75, 90) = 450, because 15 and 25 both divide 75.
Trap In the ladder, forgetting the numbers left on the bottom row is the commonest slip — every one of them must be multiplied in.
4. The four relations you must know
| Relation | What it says | Use it when |
|---|
| Product rule | For two numbers a, b: a × b = HCF × LCM | Any two of {a, b, HCF, LCM} given → find the third |
| HCF divides LCM | LCM is always a multiple of HCF | "Which of these can / cannot be the HCF (or LCM)" |
| Sandwich | HCF ≤ smaller ≤ larger ≤ LCM; all equal only when a = b | True / false statements |
| Co-prime split | If HCF = h, the numbers are ha and hb with a, b co-prime; LCM = hab | Ratio questions, "how many pairs", "which pair" |
Example HCF 12, LCM 180, one number 36. Other = 12 × 180 ÷ 36 = 60. Check: HCF(36, 60) = 12 ✓.
Example Two numbers have HCF 6 and LCM 180. All pairs: ab = 180 ÷ 6 = 30, co-prime splits (1, 30), (2, 15), (3, 10), (5, 6) → pairs (6, 180), (12, 90), (18, 60), (30, 36). Four pairs, and every one has product 1080.
Trap The product rule is for two numbers only. For 2, 2, 2: HCF × LCM = 4, product = 8. Never apply it to three numbers.
Trap A stated HCF and LCM can be impossible: HCF 8 with LCM 84 is out, because 84 ÷ 8 is not whole.
5. Fractions and decimals
Rule (fractions in lowest terms)
LCM of fractions = LCM of numerators ÷ HCF of denominators
HCF of fractions = HCF of numerators ÷ LCM of denominators
Mixed numbers → improper fractions first. Unreduced fractions → reduce first, or the rule gives a wrong answer.
Example LCM(3/4, 9/10, 15/8) = LCM(3, 9, 15) ÷ HCF(4, 10, 8) = 45 ÷ 2 = 45/2. HCF of the same = HCF(3, 9, 15) ÷ LCM(4, 10, 8) = 3/40.
Trap 4/6, 6/9, 10/15 are all 2/3, so their HCF is 2/3. Applying the rule without reducing gives 2/90 — wrong.
Rule (decimals) Pad every decimal to the same number of places, drop the point, find the LCM or HCF of the whole numbers, then put back the same number of places.
Example LCM(0.4, 0.06, 0.9): 0.40, 0.06, 0.90 → 40, 6, 90 → LCM 360 → 3.60 = 3.6. HCF(1.5, 0.75, 2.25) → 150, 75, 225 → 75 → 0.75.
Shortcut LCM ÷ HCF of decimals equals LCM ÷ HCF of the padded whole numbers — the decimal shift cancels.
6. Prime-power forms and scaling
Rule When numbers are given as 2a·3b·…, work prime by prime: HCF takes the lower power, LCM the higher. If an exponent is unknown, the HCF or LCM pins it down.
Example A = 23·5m, B = 25·52, LCM = 25·54. Prime 5: max(m, 2) = 4 → m = 4.
Rule (scaling) HCF(ka, kb) = k × HCF(a, b) and LCM(ka, kb) = k × LCM(a, b). Multiplying both numbers by k multiplies both HCF and LCM by k. Multiplying only ONE number by k changes the LCM only by the part of k the LCM does not already have.
Example LCM(a, b) = 84 → LCM(3a, 3b) = 252. But with a = 24, b = 40 (LCM 120), LCM(a, 3b) = LCM(24, 120) = 120 — the 3 was already inside 24.
Shortcut HCF(a + b, b) = HCF(a, b). Adding one number to the other never changes the HCF — that is why Euclid's method works.
7. Bells, lights and runners (LCM)
Rule Things that repeat every p, q, r seconds happen together every LCM(p, q, r) seconds. Convert all intervals to ONE unit first.
| Question | Answer |
|---|
| After how long do they next coincide? | LCM |
| At what time do they coincide for the k-th time after the start? | start + k × LCM |
| How many times in a period T (after the start)? | T ÷ LCM; add 1 if the start is counted |
| Runners: when together at the start again? | LCM of lap times; laps by a runner = LCM ÷ own lap time |
Example Signals change every 40 s, 1 min, 1 min 20 s → 40, 60, 80 s → LCM 240 s = 4 min. In 1 hour they coincide 60 ÷ 4 = 15 times after the start, 16 if the start is counted.
Trap "How many times in 15 minutes, both ends included" and "how many MORE times" differ by exactly one. Read which one is asked.
8. Least and greatest numbers with remainders (LCM)
| Condition | Form of the number | Example |
|---|
| Divisible by a, b, c | L × k (L = LCM) | Least: L. Least four-digit: first L×k ≥ 1000 |
| Leaves the SAME remainder r on a, b, c | L × k + r | Remainder 5 on 6, 9, 12: 36k + 5 → 41, 77, 113 … |
| Divisor − remainder is the SAME d for each | L × k − d | Remainders 1, 3, 5 on 4, 6, 8 (gap 3): 24k − 3 → 21, 45, 69 … |
| Plus one more condition (divisible by 7, or remainder on 7) | Walk the family L × k + r and test each | Multiple of 11 with remainder 4 on 6, 8, 9: 72k + 4 → 76, 148, 220 |
Least to add / subtract Find the remainder r of N on L. Subtract r to reach the multiple below; add L − r to reach the multiple above. The two always add up to L.
Example 4763 and the LCM of 8, 9, 12 (= 72): 4763 = 72 × 66 + 11. Subtract 11 or add 61.
Shortcut Greatest n-digit multiple of L = (largest n-digit number) − its remainder on L. Greatest three-digit multiple of 180: 999 − 99 = 900.
Trap "Least number leaving remainder 5 on 6, 9, 12" is 5 itself (quotient 0), so SSC asks for the least three-digit such number. Also, a remainder must be smaller than every divisor.
9. Measuring rods, cutting and sharing (HCF)
Rule "Greatest length that measures all of them exactly", "longest equal pieces with nothing wasted", "greatest number of identical packs with nothing left" — all three are the HCF. The number of placements / pieces / items per pack = each quantity ÷ HCF.
Example Ropes 84 cm, 126 cm, 210 cm. HCF = 42 cm. Pieces: 2 + 3 + 5 = 10.
Example 96 apples, 144 oranges, 216 mangoes → HCF 24 baskets, each with 4 + 6 + 9 = 19 fruits.
Example (with leftovers) 130 roses and 175 lilies, and 4 roses and 7 lilies are left over → HCF(126, 168) = 42 bouquets. The answer must be bigger than each leftover.
Shortcut Largest square tile for a floor a × b has side HCF(a, b); tiles needed = (a ÷ HCF) × (b ÷ HCF).
Trap Convert metres-and-centimetres to centimetres before taking the HCF. 3 m 60 cm is 360, not 3.6 and not 36.
10. Greatest divisor with remainders (HCF)
| Type | Method | Example |
|---|
| Same remainder, NOT stated | HCF of the differences of the numbers | 1237, 1849, 2665: differences 612, 816, 1428 → HCF 204. Remainder = 1237 mod 204 = 13 |
| Same remainder r, stated | Subtract r from each, then HCF | Remainder 5 on 317, 629, 1097 → HCF(312, 624, 1092) = 156 |
| Different remainders, stated | Subtract each its own remainder, then HCF | Remainders 4 and 7 from 1096 and 1375 → HCF(1092, 1368) = 12 |
Why differences work If a = nq₁ + r and b = nq₂ + r, then a − b = n(q₁ − q₂). The same remainder cancels, so n divides every difference.
Trap The divisor must exceed every remainder. HCF 8 with a stated remainder of 8 is impossible.
Shortcut Every divisor giving the same remainder is a factor of that HCF. "How many such divisors exceed 20?" → count the factors of the HCF above 20 (and above the remainder).
11. Numbers in a ratio; pairs from a product or a sum
Rule (ratio) Numbers in the ratio a : b : c (no common factor) are ha, hb, hc. Then HCF = h and LCM = h × LCM(a, b, c). Fractional ratios: multiply by the LCM of the denominators, then reduce.
Example Ratio 3 : 4 : 5, HCF 14 → numbers 42, 56, 70, LCM = 14 × 60 = 840.
Example Ratio 1/2 : 1/3 : 1/4 → × 12 → 6 : 4 : 3. LCM 72 → h = 72 ÷ 12 = 6 → numbers 36, 24, 18, HCF 6.
Rule (product given) Product P and HCF h: write ha, hb with ab = P ÷ h², a and b co-prime. Each co-prime split of P ÷ h² is one pair. The LCM is fixed: P ÷ h.
Rule (sum given) Sum S and HCF h: a + b = S ÷ h, a and b co-prime. Each co-prime split is one pair; the LCM of a pair is hab.
Example Sum 144, HCF 12 → a + b = 12 → co-prime splits (1, 11), (5, 7) → pairs (12, 132) and (60, 84). LCMs 132 and 420.
Rule (reciprocals) 1/x + 1/y = (x + y) ÷ (xy) and xy = HCF × LCM. Sum 64, HCF 8, LCM 120 → 64 ÷ 960 = 1/15.
Trap Reduce the ratio first. With 6 : 9 unreduced, h is no longer the HCF; 6 : 9 = 2 : 3.
12. Perfect squares, n-digit numbers, polynomials
Least perfect square / cube divisible by a set Take the LCM. A square needs every prime power EVEN, a cube needs every power a multiple of 3. Multiply by whatever is missing.
Example LCM(6, 15, 20) = 60 = 22·3·5. Powers of 3 and 5 are odd → × 15 → least square 900. Least cube divisible by 12 and 18: LCM 36 = 22·32 → × 6 → 216.
Counting multiples in a range Multiples of L from 1 to N: ⌊N ÷ L⌋. Four-digit multiples of 60: ⌊9999/60⌋ − ⌊999/60⌋ = 166 − 16 = 150.
Polynomials Factorise fully. HCF = common factors at lowest powers; LCM = every factor at its highest power. HCF × LCM = product of the two expressions, exactly as for numbers.
Example x² − 4 = (x − 2)(x + 2) and x² − x − 2 = (x − 2)(x + 1). HCF = x − 2, LCM = (x − 2)(x + 2)(x + 1). If the HCF of x² + 5x + 6 and x² + ax + 8 is x + 2, put x = −2 in the second: 4 − 2a + 8 = 0 → a = 6.
Letters HCF(n, n + 1) = 1; HCF(n, n + 2) = 2 for even n; LCM(n, n + 1) = n(n + 1); HCF(2n + 1, n) = 1; HCF(6n + 4, 4n + 2) = 2.
13. Mind map and traps
Mind map
Want the GREATEST that divides / measures / packs → HCF
Want the LEAST that is divisible / repeats together / fits all → LCM
Same remainder not stated → HCF of differences | stated → subtract, then HCF
Same remainder on division → L·k + r | constant gap d → L·k − d
Two given, third wanted (two numbers) → a·b = HCF·LCM
Ratio a : b → ha, hb; HCF = h; LCM = hab
Fractions → LCM(tops)/HCF(bottoms), HCF(tops)/LCM(bottoms)
Decimals → pad, drop the point, restore the point
The seven traps
- Product rule used on three numbers.
- Fractions not reduced before the fraction rule.
- Mixed units (m and cm, min and s) not converted before the LCM/HCF.
- Counting the starting toll — "how many times" vs "how many MORE times".
- Remainder larger than a divisor — impossible, the question is telling you something.
- Forgetting the bottom row in the division ladder.
- Ratio not in lowest terms before setting h as the HCF.
FAQs
What is the fastest way to find the HCF of two large numbers?
Euclid's method: divide the larger by the smaller, then the divisor by the remainder, and repeat. The last non-zero remainder is the HCF.
Is HCF × LCM = product true for three numbers?
No — only for two. 2, 4, 8 have HCF 2, LCM 8, product 64.
How do I handle "same remainder" questions?
Remainder not stated: HCF of the differences. Remainder stated: subtract it from each number, then HCF. The divisor must be larger than the remainder.
What is the LCM of fractions?
LCM of numerators ÷ HCF of denominators, with every fraction in lowest terms. HCF of fractions is HCF of numerators ÷ LCM of denominators.
How many questions does SSC CGL ask from this chapter?
Usually 1–2 in Tier I and 1–2 in Tier II — bells, remainders, measuring rods and ratio pairs. Short chapter, high return.