FN4, HN4 - Highest common factors and lowest common multiples | Eduqas GCSE Mathematics

Find the highest common factor and lowest common multiple using lists and prime factors. You will explain why the two methods use different prime powers, and decide which one solves a grouping or repeating-cycle problem.

Highest common factor

When two repeating schedules start together, when will they next coincide? To answer that later, first distinguish a number that divides both amounts from one that both amounts divide into.

The factors of $18$ include $1,2,3,6$ ; these are also factors of $30$ . A common factor divides each of the numbers exactly. We often need the greatest one.

Highest common factor

The highest common factor (HCF) of two or more positive integers is the largest positive integer that divides every one of them exactly.

Listing factors is a reliable starting method for small numbers. Write complete lists so that a missed factor cannot hide the highest shared value.

Worked example 1: HCF from factor lists

Question. Find the HCF of $18$ and $30$ .

Plan. List both sets of factors, identify their common factors, then select the largest.

\begin{aligned} 18:&\quad1,2,3,6,9,18\\ 30:&\quad1,2,3,5,6,10,15,30\\ \text{common factors}:&\quad1,2,3,6 \end{aligned}

Answer. The HCF is $6$ .

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

For larger numbers, compare their prime products. A common factor can use only primes found in both numbers. For each shared prime, use the smaller number of copies: the factor must divide the number with fewer copies too.

Worked example 2: HCF from prime factors

Question. Find the HCF of $84$ and $126$ .

Plan. Write prime factorisations and take the shared prime powers with the smaller indices.

\begin{aligned} 84&=2^2\times3\times7\\ 126&=2\times3^2\times7\\ \operatorname{HCF}(84,126)&=2\times3\times7=42 \end{aligned}

Only one $2$ can be shared because $126$ has just one. Only one $3$ can be shared because $84$ has just one. Both have one $7$ .

Answer. $42$ .

Check. $84\div42=2$ and $126\div42=3$ . Those remaining factors share no factor greater than $1$ , so there is no additional common factor to include.

If the numbers have no shared prime, their HCF is $1$ . Such numbers are called coprime; they need not both be prime. For example, $8$ and $9$ are coprime.

Watch out

Taking the larger prime powers would give $2^2\times3^2\times7=252$ for the second example. That cannot be the HCF: it does not divide either number, and exceeds both of them.

Find a common divisor of a fresh pair using prime factors.

Lowest common multiple

A common multiple is a number that each starting number divides into exactly.

Multiples of $6$ are $6,12,18,24,\ldots$ . Multiples of $8$ are $8,16,24,32,\ldots$ . The first positive value in both lists is $24$ .

Lowest common multiple

The lowest common multiple (LCM) of two or more positive integers is their smallest positive common multiple. Zero is excluded, even though it is a multiple of every integer.

For small numbers, list positive multiples until the lists first meet. Continue far enough for the larger starting number to appear in its own list too.

Worked example 3: LCM from multiple lists

Question. Find the LCM of $6$ and $8$ .

Plan. List positive multiples in increasing order and find the first shared value.

\begin{aligned} 6:&\quad6,12,18,24,30,\ldots\\ 8:&\quad8,16,24,32,\ldots \end{aligned}

Answer. The LCM is $24$ .

Check. $24\div6=4$ and $24\div8=3$ . The smaller multiples $8$ and $16$ of the larger number are not multiples of $6$ .

A prime-factor method avoids long lists. A common multiple must contain enough copies of each prime to include either original number. Use every prime that appears, with its larger index. Shared copies can serve both numbers, so they are not counted twice.

Worked example 4: LCM from prime powers

Question. Find the LCM of $84$ and $126$ .

Plan. Compare the prime factorisations and include the larger power of each prime.

\begin{aligned} 84&=2^2\times3\times7\\ 126&=2\times3^2\times7\\ \operatorname{LCM}(84,126)&=2^2\times3^2\times7\\ &=252 \end{aligned}

Two copies of $2$ are needed to include $84$ ; two copies of $3$ are needed to include $126$ . One copy of $7$ serves both.

Answer. $252$ .

Check. $252=3\times84=2\times126$ . Removing any prime factor from the selected product would leave too few copies for at least one starting number, so this common multiple is the lowest.

If one number already divides the other, the larger number is their LCM: the LCM of $6$ and $18$ is $18$ . Multiplying the starting numbers always gives a common multiple, but it may be larger than necessary: $6\times8=48$ , while their LCM is $24$ .

Key idea

HCF: shared primes, smaller powers. LCM: all primes present, larger powers. The first must divide each number; the second must be divisible by each number.

Find the smallest common multiple of a new pair.

Choosing a method

Choose HCF or LCM by deciding what the unknown number must do, rather than searching for one keyword.

If the unknown divides each amount into equal whole groups, it is a common factor. If it is reached after whole numbers of both repeating cycles, it is a common multiple. The requested largest or first value then selects HCF or LCM.

Situation	What must be true?	Method
Greatest number of identical packs using everything	Number of packs divides each total	HCF
Largest equal lengths cut from two lengths, with no waste	Chosen length divides both original lengths	HCF
First repeat together after a shared start	Elapsed time is a multiple of both intervals	LCM

First use the grouping interpretation. “Identical” means every pack has the same number of apples and the same number of pears; it does not mean apples and pears must occur in equal numbers within a pack.

Worked example 5: the greatest number of identical packs

Question. A stall has $42$ apples and $63$ pears. It makes the greatest possible number of identical packs using all the fruit. How many packs can it make, and what is in each pack?

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

\begin{aligned} 42&=2\times3\times7\\ 63&=3^2\times7\\ \operatorname{HCF}(42,63)&=3\times7=21\\ 42\div21&=2 &&\text{apples per pack}\\ 63\div21&=3 &&\text{pears per pack} \end{aligned}

Answer. $21$ packs, each containing $2$ apples and $3$ pears.

Check. $21\times2=42$ and $21\times3=63$ , so all fruit is used. The HCF makes this the greatest possible number of packs.

Now change from dividing fixed totals to waiting for two repeating events. The unknown must contain whole cycles of both lengths.

Worked example 6: the next repeat together

Question. One bus service leaves every $12$ minutes and another every $18$ minutes. Both leave at $10{:}00$ . At what time do they next leave together?

Plan. Find the LCM of the intervals, then add that elapsed time to $10{:}00$ .

\begin{aligned} 12:&\quad12,24,36,48,\ldots\\ 18:&\quad18,36,54,\ldots\\ \operatorname{LCM}(12,18)&=36\text{ minutes} \end{aligned}

Answer. $10{:}36$ .

Check. In $36$ minutes the first service completes $3$ intervals and the second completes $2$ . At $18$ minutes only the second repeats, so that earlier time does not work.

The opening schedule question therefore needs a common multiple, because the elapsed time must contain a whole number of each interval. The shared start itself is time zero; “next” asks for the first positive common time.

Watch out

The HCF of $12$ and $18$ is $6$ , but neither bus has completed an interval after $6$ minutes. A divisor of the intervals cannot answer this repeating-cycle question.

Use the same decision with different intervals.

Quick Recap

The HCF divides every starting number: use shared primes with their smaller powers.
The LCM is divisible by every starting number: use all primes present with their larger powers.
Equal groups with no leftovers point to a common factor; repeating cycles point to a common multiple.

Two schedules next coincide after their first positive common multiple of time intervals.

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FN4, HN4 - Highest common factors and lowest common multiples

Highest common factor

Lowest common multiple

Choosing a method

Quick Recap

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