HCF & LCM

medium importance~1 Q in Tier 117 formulas⚡ 12 shortcuts4 subtopics

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Subtopic 3 of 4·← Finding HCF & LCM (incl. fractions and decimals)·Two-step LCM/HCF cases (extra condition, N-digit bounds) →

Standard word problems (tiles, bells, groups, divisible numbers)

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Almost every CGL HCF/LCM question is a dressed-up version of one template:

Largest tile / rod / measure / greatest number that divides → HCF.
Bells/traffic lights together, same starting time again → LCM of intervals.
Least number divisible by each of a set → LCM; greatest such N-digit number = largest multiple of the LCM within range.
Least number leaving the SAME remainder r with each divisor → LCM + r.
Largest number leaving the SAME remainder r → HCF of (number − r); if remainders are not given, HCF of pairwise differences.
Least number leaving DIFFERENT remainders: try LCM − (divisor − remainder) when each remainder = divisor − constant.

Groups/military columns/planting trees in rows → HCF of the counts/dimensions.

📖 Notes 🧩 5 question types ⚡ Formulas 🎯 15 practice Q

Detailed notes

The template table — decide HCF or LCM in five seconds

Question wording	Tool
Greatest number that divides a, b, c (exactly, or with remainders)	HCF
Least number divisible by a, b, c	LCM
Largest tile / rod / measure / biggest equal group	HCF
Bells or lights ringing together, laps meeting again	LCM
Least number leaving remainder r with every divisor	LCM + r
Greatest number leaving remainder r with every divisor	HCF of (number − r)
Remainder = divisor − c in every case	LCM − c (least value)

Why the two remainder rules work

If N leaves remainder r with every divisor, then N − r is exactly divisible by all of them.

The smallest positive value of N − r is the LCM, so the least such N is LCM + r.
The greatest number dividing all the (number − r) values is their HCF. Example: least number leaving remainder 5 with 12, 15, 20 → LCM $= 540$ → answer 545.

When the remainders differ

If each remainder is exactly c less than its divisor, then N + c is divisible by every divisor, so the least N is LCM − c. Example: remainders 5, 8, 11 with divisors 6, 9, 12 — each is divisor − 1 → N $= 36 - 1 = 35$ . Check: $35 = 5 \times 6 + 5 = 3 \times 9 + 8 = 2 \times 12 + 11$ ✓.

Bells, lights and laps

Convert all intervals to the same unit, take the LCM, then:

next time together = start time + LCM;
how many times together within T: divide T by the LCM (add 1 if the moment of starting counts). Example: bells every 9, 12, 15 minutes, together at 8 a.m. → LCM $= 180$ min $= 3$ h → next together 11 a.m.

Tiles, rods, measures and groups

Largest square tile: side = HCF of the length and breadth (same units first); number of tiles = $\frac{L \times B}{\text{side}^2}$ .
Longest rod / biggest vessel: HCF of the dimensions or capacities; the count is total ÷ HCF.
Biggest equal groups (men and women split into identical teams): HCF of the counts. Example: containers of 144 L, 180 L, 240 L — largest vessel $= \gcd = 12$ L.

Units and classic traps

Convert everything first: 15 m 17 cm $= 1517$ cm; 2 hours $= 120$ minutes.
"Divides … leaving remainder" → subtract the remainders, then HCF. "Divisible by" → LCM. Mixing these up is the number-one error in this topic.
For LCM + r questions, other valid values are LCM·k + r — but the question asks for the least, so take k = 1.
A remainder must be smaller than its divisor; if your answer is not, you stopped one step early.

Quick revision

Divides with remainders → subtract remainders, take HCF.
Same remainder, least value → LCM + r; remainder = divisor − c → LCM − c.
Bells and laps: LCM, then add to the clock or count windows.
Tiles and rods: HCF first, then divide the area or the total.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Greatest number dividing with given remainders (HCF of differences)very common2 practice Q

How to spot it:

'Find the greatest number which divides a, b, c leaving remainders r1, r2, r3.'

\text{answer} = \gcd(a - r_1,\ b - r_2,\ c - r_3)

Subtract each remainder from its number: N − r is exactly divisible by the answer.
Take the HCF of the results.
Check the answer is bigger than every remainder.

Why: the unknown divisor divides N − r for every N, so it is a common factor — and the greatest one is the HCF.

Example: Find the greatest number which divides 85 and 72 leaving remainders 1 and 2 respectively.

$85 - 1 = 84$ , $72 - 2 = 70$ ; $\gcd(84, 70) = 14$ . Check: $85 = 6 \times 14 + 1$ , $72 = 5 \times 14 + 2$ .

Type 2: Least number leaving the same remainder r (LCM + r)very common2 practice Q

How to spot it:

'Find the least number which when divided by a, b, c leaves remainder r in each case.'

N = \text{LCM}(a, b, c) \times k + r \ \ (k = 1 \text{ for the least positive value})

Take the LCM of the divisors.
Add the remainder r.
If a bound appears (greater than 10 000, four-digit, …), fit N = LCM·k + r into the range.

Why: N − r must be a common multiple of all divisors; the least positive common multiple is the LCM.

Example: Find the least number which when divided by 8, 12, 15 and 20 leaves remainder 4 in each case.

LCM $(8, 12, 15, 20) = 120$ → least number $= 120 + 4 = 124$ . (244 also works but is not the least.)

Type 3: Remainder is (divisor − c) each time (LCM − c)common2 practice Q

How to spot it:

Remainders differ, but each one is a fixed amount less than its divisor: like 2, 3, 4 with divisors 3, 4, 5.

r_i = d_i - c \Rightarrow N = \text{LCM}(d_1, d_2, \dots) \times k - c

Confirm each divisor minus its remainder is the same number c.
N + c is divisible by every divisor → N = LCM − c for the least value.
Bigger values are LCM·k − c if a bound is given.

Why: adding c to N repairs every division to an exact one, so N + c is a common multiple.

Example: Find the least number which when divided by 6, 9 and 12 leaves remainders 5, 8 and 11 respectively.

Each remainder is divisor − 1 → N $= \text{LCM}(6, 9, 12) - 1 = 36 - 1 = 35$ . Check: $35 \bmod 6 = 5$ , $35 \bmod 9 = 8$ , $35 \bmod 12 = 11$ .

Type 4: Bells, lights and laps meeting again (LCM + clock)common2 practice Q

How to spot it:

Bells toll or lights change at different intervals, start together, and the question asks when they next coincide or how often in a window.

\text{gap} = \text{LCM}(\text{intervals}); \quad \#\text{times in } T = \left\lfloor \frac{T}{\text{LCM}} \right\rfloor \ (+1 \text{ if the start counts})

Convert every interval to the same unit.
LCM of the intervals = the gap between coincidences.
Add to the start time, or divide the window by the gap to count.
Decide whether the start counts — the wording tells you.

Why: two events with periods a and b coincide exactly at common multiples of a and b; the least one is the LCM.

Example: Three bells toll at intervals of 9, 12 and 15 minutes and toll together at 8 a.m. When do they next toll together?

LCM $(9, 12, 15) = 180$ minutes $= 3$ hours → 8 a.m. + 3 h $=$ 11 a.m.

Type 5: Largest tile, rod, measure or equal group (HCF)common3 practice Q

How to spot it:

Pave with the largest square tiles, cut planks of greatest equal length, fill containers with the biggest vessel, form biggest equal groups.

\text{side/length} = \gcd(\text{dimensions}); \quad \#\text{tiles} = \frac{L \times B}{h^2}, \quad \#\text{pieces} = \frac{\text{total length}}{h}

Convert all measurements to one unit.
HCF of the dimensions (or capacities / counts) is the answer's size.
Divide the total by the HCF to get how many tiles, pieces or groups.

Why: the tile or rod must fit each dimension a whole number of times — it is a common divisor, so the largest one is the HCF.

Example: What is the largest vessel which can fill containers of 144 L, 180 L and 240 L exactly, and how many times does it fill the 240 L one?

$\gcd(144, 180, 240) = 12$ L; $240 \div 12 = 20$ fills.

Formulas

Same remainder r

N = \text{LCM}(d_1, d_2, \dots) \times k + r

Different remainders, N ≡ −c

N = \text{LCM} \times k - c \text{ when } r_i = d_i - c

Largest tile count

\text{tiles} = \frac{L \times W}{h^2},\ h = \gcd(L, W)

Greatest n-digit multiple

\text{answer} = \underbrace{99\ldots9}_{n} - \left(\underbrace{99\ldots9}_{n} \bmod \text{LCM}\right)

Shortcut tricks

⚡ Subtract remainders, then HCF

'Same remainder' questions: HCF of (each given number minus its remainder). Remainder unknown: HCF of the pairwise differences.

Example: Find the greatest number which divides 85 and 72 leaving remainders 1 and 2 respectively.

$85 - 1 = 84$ , $72 - 2 = 70$ . $\gcd(84, 70) = 14$ .

⚡ LCM + r

'Least number leaving remainder r with each divisor': add r to the LCM.

Example: Find the least number which when divided by 12, 15, 20 and 27 leaves remainder 5 in each case.

LCM $= 540$ , answer $= 540 + 5 = 545$ .

⚡ Bells: add the LCM to the start time

Convert all intervals to the same unit, take the LCM, add to the given time.

Example: Three bells toll at intervals of 9, 12 and 15 minutes. They toll together at 8 a.m. When do they next toll together?

LCM $(9, 12, 15) = 180$ min $= 3$ h ⇒ 11 a.m.

Where students lose marks

Adding the remainder when the question asks for the greatest divisor (that needs HCF of differences, not LCM + r).
Forgetting to convert metres to centimetres (or minutes to seconds) before taking HCF/LCM.
For 'leaves remainder 2, 3, 4 with 3, 4, 5' type questions, seeing $r_i = d_i - 1$ and answering LCM $-1$ only if asked for the least; other values are LCM·k − 1.
Using HCF where 'divisible by all' demands LCM.

Practice sets — 15 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 10 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.