Common Mistakes with HCF and LCM in GCSE Maths

It is frustrating to finish a calculation neatly and discover that you answered the wrong question. Common mistakes with HCF and LCM in GCSE Maths usually come from confusing factors with multiples, losing repeated prime factors, or choosing the wrong method in a word problem. The fix is to decide what your answer must do before you calculate it.

Treat each error as a specific check: identify the misunderstanding, then practise the correct method.

Your HCF and LCM checklist

• HCF means highest common factor: the largest positive whole number that divides exactly into every given number.
• LCM means lowest common multiple: the smallest positive whole number that every given number divides into exactly.
• Prime factorisation preserves every factor: repeated primes must stay in your working.
• Context determines the method: equal groups often suggest HCF; repeating events meeting again often suggest LCM.
• Your answer needs checking: does it divide the original numbers, or is it divisible by them?

These ideas belong together, which is why MathsGenie's Prime Factors, HCF and LCM lesson covers them as one topic.

Common mistakes with HCF and LCM in GCSE Maths

Confusing factors with multiples

The mistake: our imagined student finds a number larger than both starting numbers and labels it the HCF.

The correct method: ask which direction the division goes. A factor divides into a number; a multiple is divisible by that number.

For $12$ and $18$, list the positive factors:

$$12:1,2,3,4,6,12$$ $$18:1,2,3,6,9,18$$

The largest shared factor is $6$, so:

$$\mathrm{HCF}(12,18)=6$$

Now list positive multiples:

$$12:12,24,36,48,\dots$$ $$18:18,36,54,\dots$$

The first shared multiple is $36$, so:

$$\mathrm{LCM}(12,18)=36$$

Check: for positive whole numbers, the HCF cannot exceed the smaller number. The LCM cannot be smaller than the larger number. These checks catch many swapped answers.

Stopping a factor tree too early

The mistake: the student writes $60=6\times 10$ and treats that as a prime factorisation.

The correct method: neither $6$ nor $10$ is prime. Continue splitting every composite factor until all the end numbers are prime:

$$60=6\times 10=(2\times 3)\times (2\times 5)$$

Therefore:

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

Multiply back to check that no factor disappeared. Also remember that $1$ is not prime, so it should not appear as a prime factor at the end of a branch.

The shape of the factor tree does not matter. Starting with a different factor pair must still produce the same prime factors, including repetitions.

Losing repeated primes when finding the HCF

The mistake: the student sees that two numbers share the primes $2$ and $3$, then uses each only once.

Consider:

$$48=2^4\times 3,72=2^3\times 3^2$$

Both numbers contain three copies of $2$ and one copy of $3$. Therefore:

$$\mathrm{HCF}(48,72)=2^3\times 3=24$$

The correct method: for the HCF, use only shared primes and take the smaller power of each.

If you use a Venn diagram, place matching prime factors in the overlap one pair at a time. Multiply the factors in the overlap to find the HCF. Do not add them.

Multiplying both numbers to find the LCM

The mistake: the student automatically multiplies the two original numbers.

Why it happens: their product is certainly a common multiple, but it is not necessarily the lowest one. Shared factors can be counted unnecessarily.

Using the same factorisations, the LCM needs enough copies of each prime to contain either original number:

$$\mathrm{LCM}(48,72)=2^4\times 3^2=144$$

The correct method: use every prime that appears and take the larger power of each.

Check both divisions:

$$144÷48=3,144÷72=2$$

In a Venn diagram, multiply all the factors across the entire diagram, counting each entry once. The overlap is already shared, so do not count it twice.

Use the MathsGenie lesson and Question Bank to practise choosing smaller powers for HCF and larger powers for LCM while the distinction is fresh.

A stick-figure student replaces a huge bag of past mistakes with a small notebook of things to check next time.

Choosing a method from one word

The mistake: the student spots “largest” and chooses HCF without considering what the answer represents.

The correct method: describe the job of the unknown number.

Suppose a club has $48$ pencils and $72$ stickers. It wants the greatest possible number of identical packs, with no items left over. The number of packs must divide both quantities, so use HCF:

$$Number\; of\; packs=24$$

Each pack contains:

$$48÷24=2pencils,72÷24=3stickers$$

Now suppose two lights flash every $48$ seconds and $72$ seconds. They flash together now. The next shared flashing time must be a multiple of both intervals, so use LCM:

$$Time\; until\; they\; next\; flash\; together=144seconds$$

The numbers have not changed. The relationship has. Always finish by answering the actual question, including units where needed.

Make your working useful to the examiner

A correct method should be visible, not hidden in your head. Write the prime factorisations clearly, label whether you are finding HCF or LCM, and show the multiplication that produces your answer.

For questions involving three numbers, extend the same rule carefully: an HCF prime must occur in all three, while the LCM must include enough copies to accommodate each number.

HCF and LCM sit within the wider number content of GCSE Maths revision. When you later meet them in topic-based exam booklets, keep using the same checking routine rather than relying on how familiar a question looks.

Exam-style practice: find the error before it happens

Try these independently, explaining your choice of method before checking the answers.

• Find the HCF and LCM of $18$ and $30$.
• A student says the LCM of $8$ and $12$ is $96$ because $8\times 12=96$. Explain the mistake and give the correct answer.
• A teacher shares $36$ red counters and $54$ blue counters into the greatest possible number of identical groups. Find the number of groups and the contents of each group.

Answers and checks

For the first question:

$$18=2\times 3^2,30=2\times 3\times 5$$

Therefore:

$$\mathrm{HCF}=2\times 3=6,\mathrm{LCM}=2\times 3^2\times 5=90$$

For the second, $96$ is a common multiple, but not the lowest. Since $8=2^3$ and $12=2^2\times 3$, the LCM is $2^3\times 3=24$.

For the third, use HCF because the group count must divide both quantities. The HCF is $18$, giving $18$ groups containing $2$ red counters and $3$ blue counters each.

Turn each mistake into your next check

The goal is to choose the method yourself, even when a question looks different. Turn your errors into checks, then use MathsGenie's Lessons and Question Bank to practise the mistake you most need to fix.