HCF & LCM: definitions and core relations — notes, question types & practice · HCF & LCM

01

Overview

The HCF (highest common factor) is the largest number that divides each given number exactly. The LCM (lowest common multiple) is the smallest number that each given number divides exactly. For 12 and 18 the common factors are 1, 2, 3 and 6, so the HCF is 6; their multiples first meet at 36, so the LCM is 36.

02

The size order

The HCF is never bigger than the smallest number. The LCM is never smaller than the largest number. So HCF $\le$ every number $\le$ LCM, for numbers of 1 or more. This order kills wrong options at sight.

Rule: HCF $\le$ smallest number, and LCM $\ge$ largest number. An option breaking this order is wrong.

03

The product relation

For exactly two numbers, HCF times LCM equals the product of the two numbers. Check on 12 and 18: $6 \times 36 = 216 = 12 \times 18$ . The rule fails for three numbers, so never use it there. One common use is finding a missing number: other number $= \dfrac{\text{HCF} \times \text{LCM}}{\text{given number}}$ .

Example: HCF 15, LCM 225, one number 45. The other is $\dfrac{15 \times 225}{45} = 75$ . Check: $\gcd(45, 75) = 15$ and $\text{LCM} = 225$ .

04

Numbers in a ratio

Numbers in ratio $a : b$ (parts co-prime) with HCF $h$ are $ha$ and $hb$ . Then:

LCM $= h \times a \times b$
sum $= h(a + b)$
difference $= h(b - a)$

Ratio 3 : 4 with HCF 6 gives 18 and 24, LCM $6 \times 12 = 72$ , sum $6 \times 7 = 42$ . Working backwards from an LCM: ratio 4 : 9 and LCM 252 give $h = \dfrac{252}{4 \times 9} = 7$ , so the numbers are 28 and 63.

Tip: Ratio and LCM given: divide the LCM by the product of the ratio parts. That quotient is the HCF.

05

Co-prime and close numbers

Co-prime numbers have HCF 1, so their LCM is simply their product. Consecutive integers are always co-prime. The HCF also divides every difference of the numbers: two numbers 20 apart can never have HCF 12, because 12 does not divide 20.

One more standard shape: the HCF is $h$ , and the other two factors of the LCM are $m$ and $n$ (co-prime). The numbers are $hm$ and $hn$ , and the LCM is $hmn$ . HCF 12 with other factors 5 and 7 gives numbers 60 and 84, and LCM $12 \times 35 = 420$ .

Watch: The LCM is always a multiple of the HCF. An LCM option that is not a multiple of the given HCF is wrong at sight.

06

Counting pairs

A question may ask how many pairs of numbers have a given product $P$ and HCF $h$ . Write the numbers as $ha$ and $hb$ with co-prime $a$ and $b$ , so $ab = \dfrac{P}{h^2}$ . The answer is the number of co-prime factor pairs of that quotient. The pair $(1, \text{quotient})$ counts as one pair. For product 216 with HCF 6: $\dfrac{216}{36} = 6$ , and 6 has co-prime pairs $(1, 6)$ and $(2, 3)$ , so 2 pairs exist: 6 with 36, and 12 with 18.

07

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

Find the HCF or LCM of small numbers directly

How to spot it:

The question asks plainly for the HCF or LCM of two or three small numbers, with no story around it.

\text{HCF} = \text{common primes, lowest powers}; \quad \text{LCM} = \text{all primes, highest powers}

Method

Factorise each number into primes.
HCF: multiply the primes common to all numbers, at the lowest power of each.
LCM: multiply every prime that appears, at the highest power of each.
Sanity check: HCF at most the smallest number, LCM at least the largest.

Why it works:

A common divisor cannot use more of any prime than the poorest number holds; a common multiple needs the richest stock of every prime.

Try this

Find the LCM of 12, 15 and 20.

Show solution

$12 = 2^2 \times 3$ , $15 = 3 \times 5$ , $20 = 2^2 \times 5$ .
Take the highest power of every prime: $2^2$ , $3$ , $5$ .
LCM $= 4 \times 3 \times 5 = 60$ .

Answer

60

Type 2very common2 practice Q

HCF x LCM = product (two numbers)

How to spot it:

Any two of product, HCF, LCM and one number are given, and a third value is asked.

\text{HCF} \times \text{LCM} = a \times b \ \Rightarrow\ b = \dfrac{\text{HCF} \times \text{LCM}}{a}

Method

Confirm exactly two numbers are involved.
Write the product relation and fill in the three known values.
Divide to get the fourth value.
Check: the HCF must divide the LCM, and the HCF must divide both numbers.

Why it works:

Writing $a = ha'$ and $b = hb'$ with co-prime parts, the LCM is $ha'b'$ , so HCF times LCM equals $ab$ .

Try this

The HCF of two numbers is 12, their LCM is 240, and one number is 48. Find the other number.

Show solution

Other $= \dfrac{12 \times 240}{48}$ .
$= \dfrac{2880}{48} = 60$ .
Check: $\gcd(48, 60) = 12$ and $\text{LCM} = 240$ .

Answer

60

Type 3very common2 practice Q

Numbers in a given ratio, HCF or LCM known

How to spot it:

Two numbers are in a ratio and the HCF or LCM is given; a sum, difference or the numbers themselves are asked.

\text{numbers} = hm,\ hn; \quad \text{LCM} = hmn, \quad \text{sum} = h(m + n)

Method

Cancel the ratio to co-prime parts $m : n$ first.
HCF given: the numbers are $hm$ and $hn$ .
LCM given: $h = \dfrac{\text{LCM}}{mn}$ , then the numbers are $hm$ and $hn$ .
Answer what is asked: sum, difference, product or the numbers.

Why it works:

The ratio fixes only the shape of the pair; the HCF is the scale factor that stretches it.

Try this

Two numbers are in the ratio 5 : 7 and their LCM is 700. Find their sum.

Show solution

$h = \dfrac{700}{5 \times 7} = \dfrac{700}{35} = 20$ .
Numbers: $20 \times 5 = 100$ and $20 \times 7 = 140$ .
Sum $= 100 + 140 = 240$ .

Answer

240

Type 4common2 practice Q

HCF given with the other factors of the LCM

How to spot it:

The HCF is given together with the two leftover factors of the LCM, and the numbers are asked.

\text{numbers} = hm,\ hn \ \ (m, n \text{ co-prime}); \quad \text{LCM} = hmn

Method

The numbers are HCF times the first other factor, and HCF times the second.
The two other factors must be co-prime; otherwise the data is impossible.
If asked, the LCM is HCF times the product of the two factors.

Why it works:

The LCM equals the HCF times the leftover co-prime parts, by the structure of the two ideas.

Try this

The HCF of two numbers is 17 and the other two factors of their LCM are 11 and 16. Find the greater number.

Show solution

Numbers: $17 \times 11 = 187$ and $17 \times 16 = 272$ .
$\gcd(11, 16) = 1$ , so the data is valid.
Greater number $= 272$ .

Answer

272

Type 5common2 practice Q

Co-prime, consecutive and property questions

How to spot it:

The word co-prime or consecutive appears, or a pair must be tested for being co-prime.

\gcd(a, b) = 1 \Rightarrow \text{LCM}(a, b) = ab

Method

Co-prime with LCM given: divide the LCM by the given number to get the other.
Testing a pair for co-prime: check for any common factor above 1.
Difference checks: a common factor must divide the difference of the pair.

Why it works:

With no common prime between them, nothing cancels, so the LCM is the plain product.

Try this

The LCM of two co-prime numbers is 312 and one of them is 13. Find the other number.

Show solution

Co-prime, so LCM $=$ product.
Other $= \dfrac{312}{13} = 24$ .
Check: $\gcd(13, 24) = 1$ .

Answer

24

08

Formula sheet

Product relation (two numbers)

\text{HCF} \times \text{LCM} = a \times b

Other number

b = \dfrac{\text{HCF} \times \text{LCM}}{a}

Co-prime case

\gcd(a, b) = 1 \Rightarrow \text{LCM}(a, b) = ab

HCF divides every difference

\gcd(a, b) \mid (a - b)

LCM is a multiple of the HCF

\gcd(a, b) \mid \text{LCM}(a, b)

Ratio pair

\text{numbers} = hm,\ hn; \quad \text{LCM} = hmn

09

Shortcuts that save time

⚡ Ratio split

Numbers in ratio a : b (co-prime parts) with HCF h are ha and hb. LCM = h times a times b; sum = h times (a plus b).

Example

Two numbers are in the ratio 3 : 8 and their HCF is 9. Find their LCM.

Show solution

Numbers: $9 \times 3 = 27$ and $9 \times 8 = 72$ .
LCM $= 9 \times 3 \times 8$ .
$= 216$ .

Answer

216

⚡ Product divided by HCF

Given any two of product, HCF, LCM, the product relation settles the third in one division.

Example

The product of two numbers is 2940 and their HCF is 14. Find the LCM.

Show solution

$\text{HCF} \times \text{LCM} = \text{product}$ .
$\text{LCM} = \dfrac{2940}{14}$ .
$= 210$ .

Answer

210

⚡ Co-prime pairs from the quotient

For pair-count questions, divide the product by the square of the HCF, then count co-prime factor pairs of the result.

Example

The product of two numbers is 216 and their HCF is 6. How many such pairs exist?

Show solution

$ab = \dfrac{216}{6^2} = 6$ .
Co-prime pairs of 6: $(1, 6)$ and $(2, 3)$ .
So 2 pairs: 6 with 36, and 12 with 18.

Answer

2

10

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using HCF x LCM = product for three numbers.

The relation holds for exactly two numbers only.

Mistake 02

Forgetting the co-prime pair (1, quotient) when counting pairs.

That pair is always valid; include it in the count.

Mistake 03

Reading 'greatest number that divides' as an LCM question.

'Greatest that divides' is HCF; 'least that is divisible by' is LCM.

Mistake 04

Taking the HCF of a ratio pair to be the ratio itself.

The HCF is the scale factor h; the ratio only gives the co-prime parts.

Mistake 05

Reporting an LCM that is not a multiple of the HCF.

The LCM is always a multiple of the HCF; recheck the arithmetic.

11

Quick revision

Read this the night before the exam.

Two numbers: $\text{HCF} \times \text{LCM} = ab$ . Three numbers: no such rule.
Co-prime numbers: HCF 1, LCM product. Consecutive integers are co-prime.
Ratio $a : b$ with HCF $h$ : numbers $ha$ , $hb$ ; LCM $hab$ ; sum $h(a + b)$ .
HCF divides every difference; LCM is a multiple of the HCF.
Other factors $m$ , $n$ of the LCM: numbers $hm$ , $hn$ ; LCM $hmn$ .
Pair count: co-prime factor pairs of product $\div h^2$ .

12

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 6 min · wrong answers go to your mistake notebook automatically.