Lowest Common Multiple

Finding the smallest number that is a multiple of two or more given numbers, and using it to solve problems

Before you start

You can already write a number as a product of its prime factors in index notation, for example $60 = 2^{2} \times 3 \times 5$ , and find a highest common factor (HCF) from prime factors. This lesson uses both.

Common Multiples and the LCM

The multiples of 12 are $12 \times 1$ , $12 \times 2$ , $12 \times 3$ and so on. A number that is a multiple of each of two or more numbers is a common multiple of them.

Example 1: Listing multiples

Question: Find the lowest common multiple of 12 and 18.

Multiples of 12: 12, 24, 36, 48, 60, 72, 84, 96, 108, ...

Multiples of 18: 18, 36, 54, 72, 90, 108, ...

Common multiples of 12 and 18: 36, 72, 108, ...

Answer: The lowest of the common multiples is 36, so the LCM of 12 and 18 is 36.

Definition

The lowest common multiple (LCM) of two or more whole numbers is the smallest positive whole number that is a multiple of each of them.

Every common multiple is a multiple of the LCM: 36, 72, 108, ... are the multiples of 36.

Listing is slow when the numbers are large. Prime factorisation works for any numbers.

LCM by Prime Factorisation

A number is a multiple of $12 = 2^{2} \times 3$ only if its own prime factorisation contains $2^{2} \times 3$ . So the LCM must contain the whole prime factorisation of each number, and nothing more.

Rule: LCM by prime factorisation

Write each number as a product of its prime factors in index notation.
Take every prime factor that appears in any of the numbers.
For each one, take the highest power that appears.
Multiply these together.

Example 2: Two numbers

Question: Find the LCM of 12 and 18 by prime factorisation.

$12 = 2^{2} \times 3$

$18 = 2 \times 3^{2}$

The primes that appear are 2 and 3. The highest power of 2 is $2^{2}$ (in 12) and the highest power of 3 is $3^{2}$ (in 18).

LCM $= 2^{2} \times 3^{2} = 4 \times 9 = 36$

Answer: The LCM of 12 and 18 is 36, as in Example 1.

Example 3: Three numbers

Question: Find the LCM of 20, 35 and 42.

Write each number as a product of prime factors and line the primes up in a table.

Number	2	3	5	7
$20 = 2^{2} \times 5$	$2^{2}$	–	$5$	–
$35 = 5 \times 7$	–	–	$5$	$7$
$42 = 2 \times 3 \times 7$	$2$	$3$	–	$7$
LCM takes	$2^{2}$	$3$	$5$	$7$

Every prime that appears is taken, even 3, which is in only one of the numbers. Of $2^{2}$ and $2$ , the higher power $2^{2}$ is taken.

LCM $= 2^{2} \times 3 \times 5 \times 7 = 420$

Answer: The LCM of 20, 35 and 42 is 420.

LCM by Repeated Division

For two numbers, the LCM can also be found by repeated division, as the HCF can: the two numbers are divided together, one step below another, in a ladder.

Example 4: The ladder for two numbers

Question: Find the LCM of 12 and 18 by repeated division.

Divide both numbers by a prime factor they share, and write the results underneath. Repeat until the two numbers left have no common prime factor.

2 and 3 have no common prime factor, so stop. Multiply the divisors on the left by the two numbers left at the bottom.

LCM $= 2 \times 3 \times 2 \times 3 = 36$

Why it works: $12 = (2 \times 3) \times 2$ and $18 = (2 \times 3) \times 3$ . The divisors $2 \times 3$ are the part the two numbers share, and the LCM needs that part once, together with what each number has left at the bottom.

The divisors on their own give the HCF: $2 \times 3 = 6$ .

Answer: The LCM of 12 and 18 is 36.

Take care: for the LCM, this ladder works for two numbers only. Used in the same way on three numbers it can give a common multiple that is not the lowest, so for the LCM of three or more numbers use prime factorisation. For the HCF, repeated division works for three numbers too, because there only the divisors are multiplied.

HCF and LCM Side by Side

The HCF and the LCM come from the same prime factorisations, but they take different things from them.

	HCF	LCM
Which primes?	Only the primes common to all the numbers	Every prime that appears in any of the numbers
Which power?	The lowest power of each	The highest power of each

Example 5: Both from one table

Question: Find the HCF and the LCM of $2^{3} \times 3^{2} \times 5$ and $2^{2} \times 3^{4} \times 7$ . Leave your answers in index notation.

Number	2	3	5	7
First number	$2^{3}$	$3^{2}$	$5$	–
Second number	$2^{2}$	$3^{4}$	–	$7$
HCF takes	$2^{2}$	$3^{2}$	–	–
LCM takes	$2^{3}$	$3^{4}$	$5$	$7$

2 and 3 are common to both numbers, so the HCF takes the lower power of each. 5 and 7 are not common: the HCF leaves them out, but the LCM takes them.

HCF $= 2^{2} \times 3^{2}$

LCM $= 2^{3} \times 3^{4} \times 5 \times 7$

Answer: The HCF is $2^{2} \times 3^{2}$ and the LCM is $2^{3} \times 3^{4} \times 5 \times 7$ .

Word Problems

In a word problem, look for a number that must be a multiple of each number given, and the smallest one. That number is the LCM.

Example 6: Wheels

Question: The front wheel of a tricycle has a circumference of 60 cm and each rear wheel has a circumference of 45 cm. Point P on the front wheel and point Q on a rear wheel are both touching the ground. How far must the tricycle travel in a straight line before P and Q next touch the ground at the same time?

P touches the ground again after every 60 cm travelled, and Q after every 45 cm. The distance must be a multiple of both 60 and 45, and the smallest one: the LCM.

$60 = 2^{2} \times 3 \times 5$

$45 = 3^{2} \times 5$

LCM $= 2^{2} \times 3^{2} \times 5 = 180$

Answer: The tricycle must travel 180 cm. The front wheel then makes 3 full turns and each rear wheel 4.

Example 7: The smallest number that packs evenly

Question: Mr Tan can pack all his erasers into bags of 12, or into bags of 15, or into bags of 20, with none left over each time. What is the smallest number of erasers he could have?

The number of erasers is a multiple of 12, of 15 and of 20. The smallest such number is their LCM.

$12 = 2^{2} \times 3$ , $15 = 3 \times 5$ , $20 = 2^{2} \times 5$

LCM $= 2^{2} \times 3 \times 5 = 60$

Answer: The smallest number of erasers he could have is 60.

HCF or LCM?

Ask whether the answer must be a multiple of the numbers given or a factor of them.

LCM: things repeat (buses, bells, lights, turns of a wheel), or the size of each lot is given (bags of 12, packets of 8) and the total is not. You want when they next happen together, or the smallest total that makes whole lots. The LCM is never smaller than the largest number given.
HCF: the totals are given (60 red markers and 84 blue markers) and they are cut, shared or packed into equal groups with nothing left over. You want the greatest size of each part, or the greatest number of groups. The HCF is never larger than the smallest number given.

Common Mistakes

Taking only the common primes, or the lowest powers. That gives the HCF, not the LCM.
Multiplying the numbers together. $12 \times 18 = 216$ is a common multiple of 12 and 18, but the lowest is 36. The product of two numbers is their LCM only when they have no common prime factor.
Giving the LCM when the question asks for something else, such as a date, a time or a number of packets.

Practice

Work each question on paper before you open its solution.

Question 1

Find the LCM of 24 and 90 by prime factorisation.

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Solution:

$24 = 2^{3} \times 3$

$90 = 2 \times 3^{2} \times 5$

The highest powers are $2^{3}$ , $3^{2}$ and $5$ . The 5 is in 90 only, but it must still be taken.

LCM $= 2^{3} \times 3^{2} \times 5 = 8 \times 9 \times 5 = 360$

The LCM of 24 and 90 is 360.

Question 2

Use repeated division to find the LCM of 30 and 42. Write down their HCF as well.

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Solution:

5 and 7 have no common prime factor, so stop.

LCM $= 2 \times 3 \times 5 \times 7 = 210$

HCF $= 2 \times 3 = 6$

The LCM of 30 and 42 is 210 and their HCF is 6.

Question 3

To find the LCM of 40 and 60, Wei Ming writes: $40 = 2^{3} \times 5$ and $60 = 2^{2} \times 3 \times 5$ , so LCM $= 2^{2} \times 5 = 20$ . Explain his mistake and find the correct LCM.

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Solution:

His prime factorisations are right, but he took only the common primes, each with its lowest power. That gives the HCF.

A quick check: an LCM is never smaller than the largest number, here 60, so 20 cannot be right.

LCM $= 2^{3} \times 3 \times 5 = 120$

He found the HCF. The LCM of 40 and 60 is 120.

Question 4

Three numbers are $2^{2} \times 3^{3}$ , $2^{4} \times 5$ and $3 \times 5^{2} \times 7$ . Find (a) their LCM, in index notation, (b) their HCF.

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Solution:

(a) The primes that appear are 2, 3, 5 and 7. Their highest powers are $2^{4}$ , $3^{3}$ , $5^{2}$ and $7$ .

LCM $= 2^{4} \times 3^{3} \times 5^{2} \times 7$

(b) No prime is a factor of all three numbers: the third has no 2, the second has no 3, the first has no 5, and only the third has 7. So their only common factor is 1.

(a) The LCM is $2^{4} \times 3^{3} \times 5^{2} \times 7$ . (b) The HCF is 1.

Question 5

Decide whether each problem needs the HCF or the LCM, then solve it.

(a) A red light flashes every 18 seconds and a blue light flashes every 24 seconds. They have just flashed together. After how many seconds will they next flash together?

(b) A florist uses 18 roses and 24 orchids to make identical bouquets, with no flowers left over. What is the greatest number of bouquets she can make?

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Solution:

$18 = 2 \times 3^{2}$ and $24 = 2^{3} \times 3$

(a) The number of seconds must be a multiple of both 18 and 24, and the smallest one: the LCM. LCM $= 2^{3} \times 3^{2} = 72$

(b) The number of bouquets must be a factor of both 18 and 24, and the greatest one: the HCF. HCF $= 2 \times 3 = 6$

(a) The lights next flash together after 72 seconds. (b) She can make 6 bouquets, each with 3 roses and 4 orchids.

Question 6

Doris has a piano lesson every 6 days, a swimming lesson every 4 days and an art lesson every 8 days. She has all three lessons on 1 April. On what date will she next have all three on the same day?

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Solution:

The number of days must be a multiple of 6, of 4 and of 8, and the smallest one: the LCM.

$6 = 2 \times 3$ , $4 = 2^{2}$ , $8 = 2^{3}$

LCM $= 2^{3} \times 3 = 24$

24 days after 1 April is 25 April.

Doris next has all three lessons on 25 April.

Question 7

Paper plates are sold in packets of 8 and paper cups in packets of 10. For a class party, Mrs Lee wants to buy exactly as many plates as cups. What is the smallest number of packets of each that she must buy?

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Solution:

The number of plates is a multiple of 8 and the number of cups is a multiple of 10. The smallest number that is both is the LCM of 8 and 10.

$8 = 2^{3}$ , $10 = 2 \times 5$

LCM $= 2^{3} \times 5 = 40$

Plates: $40 \div 8 = 5$ packets. Cups: $40 \div 10 = 4$ packets.

She must buy 5 packets of plates and 4 packets of cups, which gives 40 of each.

Key Points

The LCM of two or more numbers is the smallest positive whole number that is a multiple of each of them.
By prime factorisation: take every prime that appears, each with its highest power, and multiply.
The HCF takes only the common primes, each with its lowest power.
By repeated division of two numbers: the LCM is the product of the divisors and the two numbers left at the bottom.
Use the LCM for things that next happen together and for the smallest number that packs evenly. Use the HCF for the greatest equal parts or groups.