Standard word problems (tiles, bells, groups, divisible numbers) — notes, question types & practice · HCF & LCM

01

Overview

Every word problem in this topic is a dressed-up HCF or LCM question. The fastest first step is to name the family: 'greatest number that divides' means HCF, 'least number divisible by' means LCM. For bells that toll together every 6, 8 and 12 minutes, the gap is $\text{LCM} = 24$ minutes.

02

The template table

Question wording	Tool
Greatest number that divides a, b, c	HCF
Least number divisible by a, b, c	LCM
Largest tile, rod or biggest equal group	HCF
Bells or lights together again	LCM
Least number, remainder r with every divisor	LCM + r
Greatest number, remainder r with every divisor	HCF of (number - r)

Tip: 'Greatest' pairs with HCF; 'least' pairs with LCM. Fix the family first, then the arithmetic is short.

03

Divides with remainders

'Find the greatest number which divides 70 and 125 leaving remainders 5 and 8.' The divisor must divide $70 - 5 = 65$ and $125 - 8 = 117$ , so it divides their HCF. $\gcd(65, 117) = 13$ . Check: $70 = 5 \times 13 + 5$ and $125 = 9 \times 13 + 8$ .

Rule: Subtract each remainder from its number, then take the HCF. The answer must exceed every remainder.

04

Same remainder with every divisor

If N leaves remainder r with every divisor, then $N - r$ is divisible by all of them. The least such N is LCM + r. For divisors 9, 12, 15 with remainder 5: $\text{LCM} = 180$ , so $N = 185$ .

05

Remainders that differ by a constant

When each remainder is the same amount c below its divisor, add c instead. For divisors 6, 9, 12 with remainders 5, 8, 11: each remainder is divisor $- 1$ , so $N + 1$ is a common multiple. $N = \text{LCM}(6, 9, 12) - 1 = 36 - 1 = 35$ . Check: $35 \bmod 6 = 5$ , $35 \bmod 9 = 8$ , $35 \bmod 12 = 11$ .

Watch: Each remainder must be smaller than its divisor. A remainder bigger than its divisor means the working stopped one step early.

06

Bells, lights and laps

Convert every interval to one unit, then take the LCM. Buses leaving every 10, 15 and 20 minutes together at 6 a.m. meet again after $\text{LCM} = 60$ minutes, at 7 a.m. To count meetings inside a window, divide the window by the LCM; add 1 only if the starting moment itself counts.

07

Tiles, rods, vessels and groups

The largest square tile has side $\gcd(L, W)$ in the same units; the tile count is $\dfrac{L \times W}{h^2}$ . The longest rod cut from planks of 42 cm and 49 cm has length $\gcd = 7$ cm, giving $6 + 7 = 13$ pieces. Equal groups of people: the group size is the HCF of the counts.

Example: A hall is 8.4 m by 5.6 m, so 840 cm by 560 cm. The largest square tile has side $\gcd(840, 560) = 280$ cm, and $\dfrac{840 \times 560}{280^2} = 6$ tiles.

08

Units first

Convert everything before computing: 2 m 40 cm is 240 cm, and 3 hours is 180 minutes. Mixed units are the number-one error here, because the HCF or LCM changes with the unit.

09

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Greatest number dividing with given remainders

How to spot it:

The question asks for the greatest number which divides two or three numbers leaving stated remainders.

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

Method

Subtract each remainder from its number; the divisor divides each difference.
Take the HCF of the differences.
Check the answer exceeds every remainder.

Why it works:

The unknown divisor divides each N minus r, so it is a common factor, and the greatest one is the HCF.

Try this

Find the greatest number which divides 70 and 125 leaving remainders 5 and 8.

Show solution

$70 - 5 = 65$ and $125 - 8 = 117$ .
$\gcd(65, 117) = 13$ .
Check: $70 = 5 \times 13 + 5$ , $125 = 9 \times 13 + 8$ .

Answer

13

Type 2very common2 practice Q

Least number leaving the same remainder (LCM + r)

How to spot it:

The question asks for the least number which when divided by several divisors leaves the same remainder each time.

N = \text{LCM}(a, b, c) \times k + r

Method

Take the LCM of the divisors.
Add the remainder r for the least value.
With a bound (four-digit, above 1000), pick k so that LCM x k + r fits.

Why it works:

N minus r must be a common multiple of all divisors, and the least positive one is the LCM.

Try this

Find the least number which when divided by 9, 12 and 15 leaves remainder 5 in each case.

Show solution

$\text{LCM}(9, 12, 15) = 180$ .
Least number $= 180 + 5 = 185$ .
Check: $185 \bmod 9 = 185 \bmod 12 = 185 \bmod 15 = 5$ .

Answer

185

Type 3common2 practice Q

Remainder is divisor minus c each time

How to spot it:

The remainders differ, but each one sits a fixed amount below its own divisor.

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

Method

Subtract each remainder from its divisor; confirm the same value c every time.
Then N + c is divisible by every divisor.
The least N is LCM minus c.

Why it works:

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

Try this

Find the least number which when divided by 5, 6 and 8 leaves remainders 4, 5 and 7.

Show solution

$5 - 4 = 1$ , $6 - 5 = 1$ , $8 - 7 = 1$ , so $c = 1$ .
$\text{LCM}(5, 6, 8) = 120$ .
$N = 120 - 1 = 119$ .

Answer

119

Type 4common2 practice Q

Bells, lights and laps meeting again

How to spot it:

Events repeat at different intervals, start together, and the question asks when they next coincide or how often.

\text{gap} = \text{LCM}(\text{intervals})

Method

Convert every interval to the same unit.
The LCM of the intervals is the gap between coincidences.
Add the gap to the start time, or divide the window by it to count meetings.
Check the wording for whether the starting moment counts.

Why it works:

Two repeating events coincide exactly at common multiples of their periods, and the first one is the LCM.

Try this

Three bells toll every 6, 8 and 12 minutes and toll together at 7 a.m. When do they next toll together?

Show solution

$\text{LCM}(6, 8, 12) = 24$ minutes.
$7$ a.m. $+ 24$ minutes.
$= 7$ : $24$ a.m.

Answer

7:24 a.m.

Type 5common3 practice Q

Largest tile, rod, vessel or equal group

How to spot it:

The question asks for the largest square tile, longest equal rod, biggest vessel, or biggest identical groups.

\text{size} = \gcd(\text{dimensions}); \quad \#\text{tiles} = \dfrac{L \times W}{h^2}

Method

Convert all measurements to one unit.
The HCF of the dimensions or counts gives the size.
Divide the total by the HCF for the count of tiles, pieces or groups.

Why it works:

The tile or rod must fit each dimension a whole number of times, so it is a common divisor; the largest is the HCF.

Try this

A hall is 8.4 m long and 5.6 m wide. Find the largest square tile that paves it exactly, and the number of tiles.

Show solution

Metres to centimetres: $840$ and $560$ .
Side $= \gcd(840, 560) = 280$ cm $= 2.8$ m.
Tiles $= \dfrac{840 \times 560}{280^2} = 6$ .

Answer

2.8 m side, 6 tiles

10

Formula sheet

Same remainder r

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

Remainder is divisor minus c

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

Divides with remainders

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

Largest tile count

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

11

Shortcuts that save time

⚡ Subtract remainders, then HCF

For 'greatest number dividing a and b leaving remainders r1 and r2', the answer is the HCF of (a - r1) and (b - r2).

Example

Find the greatest number which divides 1000 and 750 leaving remainders 4 and 6.

Show solution

$1000 - 4 = 996$ and $750 - 6 = 744$ .
$\gcd(996, 744) = 12$ .
Check: $1000 \bmod 12 = 4$ and $750 \bmod 12 = 6$ .

Answer

12

⚡ LCM plus r

For 'least number leaving remainder r with each divisor', add r to the LCM of the divisors.

Example

Find the least number which when divided by 12, 16 and 24 leaves remainder 8 in each case.

Show solution

$\text{LCM}(12, 16, 24) = 48$ .
Least number $= 48 + 8 = 56$ .
Check: $56 \bmod 12 = 56 \bmod 16 = 56 \bmod 24 = 8$ .

Answer

56

⚡ Bells: add the LCM to the clock

Convert all intervals to one unit, take the LCM, and add it to the given start time.

Example

Four lights blink every 12, 15, 18 and 30 seconds and blink together at noon. When do they next blink together?

Show solution

$\text{LCM}(12, 15, 18, 30) = 180$ seconds.
$180$ seconds $= 3$ minutes.
Noon $+ 3$ minutes $=$ 12:03.

Answer

12:03

12

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Adding the remainder in a 'greatest number that divides' question.

That family needs the HCF of the numbers minus their remainders.

Mistake 02

Mixing metres with centimetres, or minutes with seconds.

Convert all measurements to one unit before the HCF or LCM.

Mistake 03

Reporting LCM + r when a larger value was demanded by a bound.

Fit LCM times k plus r into the stated range; the least uses k = 1.

Mistake 04

Using the HCF where the question says 'divisible by all'.

'Divisible by all' is an LCM question; 'divides all' is HCF.

Mistake 05

Counting the starting moment in a bells question without checking the wording.

Divide the window by the LCM; add 1 only if the start counts.

13

Quick revision

Read this the night before the exam.

'Greatest that divides' $\to$ HCF; 'least divisible by' $\to$ LCM.
Divides with remainders: subtract remainders, take HCF.
Same remainder r, least value: LCM $+ r$ .
Each remainder $=$ divisor $- c$ : least value LCM $- c$ .
Bells and laps: LCM of intervals, then add to the clock.
Tiles and rods: HCF of dimensions; count $=$ area $\div h^2$ or total $\div h$ .
Convert units before anything else.

14

Practice: 15 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 10 questions

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