C1.1; E1.1 - Highest common factor and lowest common multiple | Cambridge IGCSE International Mathematics

Two lights flash together, then repeat every 12 seconds and every 20 seconds. When will they next flash together? Common factors and common multiples in C1.1 and E1.1 help you choose between dividing quantities into equal groups and bringing repeating events together.

Common factors

If 18 and 30 objects are each split into 6 equal groups, neither leaves a remainder: 6 divides both totals. This makes 6 a common factor. The words “highest common factor” ask for the greatest possible divisor shared by the numbers.

Highest common factor

A common factor divides each of the given positive integers exactly. Their highest common factor, or HCF, is the greatest of these common factors. For example, the HCF of 8 and 12 is 4, not 2, because 4 is the greater shared divisor.

For small numbers, list all factors and select the greatest common entry. A prime factorisation is more efficient for larger numbers. Match repeated copies of each prime: a factor shared by both numbers cannot use more copies than either number contains.

Worked example 1: HCF from lists

Question. Find the HCF of 18 and 30.

Plan. Make complete factor lists, then select the largest shared value.

\begin{aligned} \text{factors of 18: }&1,2,3,6,9,18\\ \text{factors of 30: }&1,2,3,5,6,10,15,30\\ \text{common factors: }&1,2,3,6 \end{aligned}

Answer. The HCF is 6.

Check. $18\div6=3$ and $30\div6=5$ . The complete lists contain no greater shared factor.

The next pair has longer factor lists. Change the representation to prime products while keeping the same decision: take only what both numbers contain.

Worked example 2: HCF from prime products

Question. Find the HCF of 48 and 180.

Plan. Factorise both numbers and keep the common copies of primes.

\begin{aligned} 48&=2\times2\times2\times2\times3=2^4\times3\\ 180&=2\times2\times3\times3\times5=2^2\times3^2\times5\\ \mathrm{HCF}&=2\times2\times3=12 \end{aligned}

Both products contain two copies of 2 and one copy of 3. A third 2 would not divide 180; a second 3 would not divide 48; 5 is absent from 48.

Answer. 12.

Check. $48\div12=4$ , $180\div12=15$ . The remaining factors 4 and 15 have no common factor above 1, so no extra shared factor was missed.

In index form, this means taking the smaller power of each common prime. If no prime is shared, the HCF is 1; for example, 8 and 9 have HCF 1.

Watch out

Do not include a prime just because it appears somewhere. A factor of both numbers must fit inside both prime products. HCF uses only shared primes, with the smaller available number of copies.

Use either lists or prime products for the next pair.

Common multiples

A common multiple goes in the opposite direction: both given numbers must divide it. For example, 24 contains four lots of 6 and three lots of 8. We seek the first positive total that both counting patterns reach.

Lowest common multiple

A common multiple is a multiple of each given number. The lowest common multiple, or LCM, is their smallest positive common multiple. Zero is a common multiple, but it is excluded by “positive”.

Use lists for small numbers. With prime products, the LCM must contain enough copies of every prime to accommodate either original number. Choose the larger power of each prime appearing in either product, including primes that appear in only one.

Worked example 3: LCM from lists

Question. Find the LCM of 6 and 8.

Plan. List positive multiples until the first shared value appears.

\begin{aligned} \text{multiples of 6: }&6,12,18,24,30,\ldots\\ \text{multiples of 8: }&8,16,24,32,\ldots \end{aligned}

Answer. 24.

Check. $24\div6=4$ and $24\div8=3$ . No smaller positive multiple of 8, namely 8 or 16, is divisible by 6.

Now use the earlier larger pair. The factorisations stay the same, but the question has changed from what fits inside both numbers to what can contain both.

Worked example 4: LCM from prime products

Question. Find the LCM of 48 and 180.

Plan. Include the greatest required number of copies of each prime.

\begin{aligned} 48&=2^4\times3\\ 180&=2^2\times3^2\times5\\ \mathrm{LCM}&=2^4\times3^2\times5\\ &=16\times9\times5=720 \end{aligned}

Four copies of 2 are needed for 48, two copies of 3 for 180, and one copy of 5 for 180. Repeating any prime more often would make an unnecessary larger multiple.

Answer. 720.

Check. $720\div48=15$ and $720\div180=4$ . Removing any selected prime copy would stop at least one division being exact.

Multiplying the two numbers always gives a common multiple, but it may be larger than necessary. Here $48\times180=8640$ , so it would not answer a question asking for the lowest.

Watch out

HCF and LCM are different choices. For 48 and 180, HCF uses $2^2\times3$ ; LCM uses $2^4\times3^2\times5$ . For positive integers, the HCF cannot exceed the smaller number, and the LCM cannot be below the larger number.

Check divisibility and whether your answer is the smallest suitable positive value.

Choosing a common measure

The key decision in a word problem is whether the answer divides the given quantities or is divisible by them. Equal grouping needs a divisor. Repeating cycles need a shared total number of steps or units of time.

Situation	What the answer must do	Choice
Greatest number of identical packs, using all objects	Divide every object total	HCF
Longest equal pieces cut from several lengths	Divide every length	HCF
First time repeating events coincide again	Be a multiple of every interval	LCM

The question's conditions matter. “Largest” on its own is not a method: ask what must divide what. For a cycle problem, “together now” establishes a common starting time, and “next” excludes time zero.

Worked example 5: making identical packs

Question. A club has 42 red badges and 30 blue badges. It makes the greatest possible number of identical packs, using every badge. Find the number of packs and the number of each colour in a pack.

Plan. Find the HCF for the number of packs, then divide each stock by it.

\begin{aligned} 42&=2\times3\times7\\ 30&=2\times3\times5\\ \mathrm{HCF}&=2\times3=6\\ 42\div6&=7\qquad 30\div6=5 \end{aligned}

Answer. 6 packs, each with 7 red badges and 5 blue badges.

Check. $6\times7=42$ and $6\times5=30$ , with nothing left. A greater number of packs cannot divide both totals.

The next problem still asks for a shared quantity, but the given numbers are intervals to repeat, not totals to divide. This changes HCF to LCM.

Worked example 6: synchronising cycles

Question. Two bells ring together at 10:00. One rings every 18 minutes and the other every 24 minutes. When do they next ring together?

Plan. Find the LCM of the two intervals, then add it to the starting time.

\begin{aligned} 18&=2\times3^2\\ 24&=2^3\times3\\ \mathrm{LCM}&=2^3\times3^2=72\text{ minutes}\\ 72\text{ minutes}&=1\text{ hour }12\text{ minutes} \end{aligned}

Answer. 11:12.

Check. $72\div18=4$ and $72\div24=3$ : each bell has completed a whole number of intervals. The prime selection gives the first positive shared time.

For the opening lights, $12=2^2\times3$ and $20=2^2\times5$ . Their LCM is $2^2\times3\times5=60$ , so they next flash together after 60 seconds.

Problem-solving

Write a short sentence before calculating: “The answer divides both totals” or “Both intervals divide the answer.” It makes the HCF/LCM choice visible and gives you a useful final check.

Make the same decision for a new pair of cycles.

Quick Recap

The HCF is the greatest divisor shared by the numbers.
The LCM is the smallest positive multiple shared by them.
Prime products use smaller shared powers for HCF, and larger required powers for LCM.
Decide whether the answer divides the given numbers or is divisible by them.

The lights in the opening question next flash together after 60 seconds.

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C1.1; E1.1 - Highest common factor and lowest common multiple

Common factors

Common multiples

Choosing a common measure

Quick Recap

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