1.02c - Highest factors and lowest multiples | OCR GCSE Mathematics

Common factors help you share or cut exactly; common multiples help you match repeating quantities. You will find the highest common factor and lowest common multiple using lists and prime factorisations, then choose between them in context.

Common factors and multiples

Two bells ring every eighteen seconds and every thirty seconds, starting together. When will they next ring together? The answer must be a multiple of both intervals, rather than a factor of them.

For smaller numbers, lists make the difference visible. The factors of $12$ are $1,2,3,4,6,12$ ; its positive multiples begin $12,24,36,48,\ldots$ . A factor divides into the number. A multiple contains whole copies of it.

HCF and LCM

The highest common factor (HCF) of positive integers is the greatest positive integer that divides every one of them exactly. The lowest common multiple (LCM) is the smallest positive integer that is a multiple of every one of them. Zero is a common multiple, but is excluded by “positive” in the LCM definition.

For an HCF, list all positive factors and choose the greatest shared value. For an LCM, list increasing positive multiples until you find the first shared value. Do not simply choose the largest number written down.

Worked example 1: comparing twelve and eighteen

Question. Find the HCF and LCM of $12$ and $18$ by listing.

Plan. Compare factor lists for the HCF, and increasing multiple lists for the LCM.

\begin{aligned} \text{factors of }12&:1,2,3,4,6,12\\ \text{factors of }18&:1,2,3,6,9,18\\ \text{HCF}&=6 &&\text{greatest shared factor} \end{aligned}

\begin{aligned} \text{multiples of }12&:12,24,36,48,\ldots\\ \text{multiples of }18&:18,36,54,\ldots\\ \text{LCM}&=36 &&\text{first shared positive multiple} \end{aligned}

Answer. HCF $=6$ ; LCM $=36$ .

Check. $12\div6=2$ and $18\div6=3$ . Also, $36\div12=3$ and $36\div18=2$ . The complete factor lists and increasing multiples establish highest and lowest.

Keep twelve and change only the second number to twenty. Neither answer should be carried over without comparing again.

Worked example 2: comparing twelve and twenty

Question. Find the HCF and LCM of $12$ and $20$ by listing.

Plan. Keep the factors of twelve, list the new factors and extend the multiples as far as needed.

\begin{aligned} \text{factors of }12&:1,2,3,4,6,12\\ \text{factors of }20&:1,2,4,5,10,20\\ \text{HCF}&=4 &&\text{greatest shared factor} \end{aligned}

\begin{aligned} \text{multiples of }12&:12,24,36,48,60,\ldots\\ \text{multiples of }20&:20,40,60,\ldots\\ \text{LCM}&=60 &&\text{first shared positive multiple} \end{aligned}

Answer. HCF $=4$ ; LCM $=60$ .

Check. Four divides both numbers, and sixty is divisible by each. The lists show no larger common factor or smaller positive common multiple.

Two useful boundaries follow from the definitions. If the numbers have no common factor above one, their HCF is one. If one number divides the other, the smaller is the HCF and the larger is the LCM: for $6$ and $18$ , they are $6$ and $18$ . Equal positive numbers have that same number as both HCF and LCM.

Watch out

An LCM is a multiple of each original number, not necessarily their product. For twelve and eighteen, $216$ is a common multiple but the LCM is $36$ .

Change the second number once more and compare the lists.

Choosing prime powers

Prime factorisation makes the same comparison without writing long multiple lists.

Consider $84=2^2\times3\times7$ and $180=2^2\times3^2\times5$ . A common factor can use only primes present in both numbers, and no more copies than either number contains. It can therefore contain two twos and one three, but neither five nor seven.

A common multiple must contain enough prime factors to include each original number. It needs two twos, two threes, one five and one seven. Shared copies need not be counted twice: the same two twos can serve both originals.

Key idea

For the HCF, take each shared prime to the smaller power. For the LCM, take every prime present to the larger power. A missing prime contributes nothing to the HCF, but must still be included in the LCM if the other number needs it.

These choices explain why the answers are highest and lowest. Another copy in the HCF would fail to divide at least one original. Removing a needed copy from the LCM would stop it being a multiple of at least one original.

Worked example 3: shared and unshared primes

Question. Given $84=2^2\times3\times7$ and $180=2^2\times3^2\times5$ , find their HCF and LCM.

Plan. Use the common copies for the HCF and enough copies for either number for the LCM.

\begin{aligned} \text{HCF}&=2^2\times3 &&\text{shared primes, fewer copies}\\ &=4\times3=12\\ \text{LCM}&=2^2\times3^2\times5\times7 &&\text{all primes, enough copies}\\ &=4\times9\times35=1260 \end{aligned}

Answer. HCF $=12$ ; LCM $=1260$ .

Check. $84\div12=7$ , $180\div12=15$ ; $1260\div84=15$ , $1260\div180=7$ . Both answers have the required divisibility.

Keep eighty-four and change only the power of three in the second factorisation, from $3^2$ to $3^3$ . The second number becomes $540$ .

Worked example 4: one larger prime power

Question. Given $84=2^2\times3\times7$ and $540=2^2\times3^3\times5$ , find their HCF and LCM.

Plan. Reconsider the number of threes needed; all other prime comparisons stay the same.

\begin{aligned} \text{HCF}&=2^2\times3=12 &&\text{eighty-four still has one three}\\ \text{LCM}&=2^2\times3^3\times5\times7 &&\text{three threes now needed}\\ &=4\times27\times35=3780 \end{aligned}

Answer. HCF $=12$ ; LCM $=3780$ .

Check. $3780\div84=45$ , $3780\div540=7$ , while twelve still divides both originals. The LCM tripled when the second number gained a factor of three.

For two positive integers, there is a useful connection: their HCF multiplied by their LCM equals their product. Each prime's copies are split into the smaller and larger counts, so together they use exactly the copies in the two originals. Here, $12\times1260=15120=84\times180$ . This gives a check, or a way to recover one answer from the other. The statement is for two numbers; do not apply it unchanged to three.

Watch out

For the LCM of eighty-four and one hundred and eighty, including only shared primes gives $36$ , which is not divisible by either original. Include the unshared five and seven too.

Change only the second number's power of three again.

Sharing and repeating cycles

Choose the method by asking what the required number must do: divide each given quantity exactly, or be reached by repeated copies of each quantity.

If lengths are cut into equal pieces without waste, the piece length must divide each original length, so factors matter. If events repeat and must coincide, the elapsed time must be a multiple of each interval. The words “greatest” and “next” then select the HCF or LCM.

Requirement	Relationship	Choice
Longest equal pieces, no waste	Piece length divides each length	HCF
Greatest number of identical packs, nothing left	Number of packs divides each item count	HCF
Next simultaneous repeat	Elapsed time is a positive multiple of each interval	LCM

For packs, “identical” means each pack has the same number of each kind of item. The HCF gives the number of packs; divide the original counts by it to find each pack's contents.

Worked example 5: the longest equal pieces

Question. Ropes of lengths $18\,\mathrm m$ and $30\,\mathrm m$ are cut into equal whole-metre lengths with no waste. Find the greatest possible piece length.

Plan. Find the HCF because the piece length must divide both rope lengths.

\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{HCF}&=6 &&\text{greatest shared divisor} \end{aligned}

Answer. Each piece can be $6\,\mathrm m$ long.

Check. The ropes make $18\div6=3$ pieces and $30\div6=5$ pieces, with no waste. No greater common factor appears in the complete lists.

Keep the numbers eighteen and thirty, but change the requirement from dividing quantities to matching repeats. This reverses which relationship we need.

Worked example 6: the next simultaneous ring

Question. Two bells ring every $18$ seconds and every $30$ seconds. They ring together now. After how many seconds will they next ring together?

Plan. Find the first positive time that is a multiple of both intervals.

\begin{aligned} 18&=2\times3^2,\qquad30=2\times3\times5 &&\text{prime factorisations}\\ \text{LCM}&=2\times3^2\times5 &&\text{enough copies for both}\\ &=90 \end{aligned}

Answer. They next ring together after $90$ seconds.

Check. This is five intervals of eighteen seconds and three intervals of thirty seconds. The earlier positive thirty-second times, $30$ and $60$ , are not multiples of eighteen.

The opening bells therefore meet after ninety seconds. The answer six would divide both intervals, but would not be a ringing time for either bell after the start.

Watch out

Do not choose HCF just because the problem contains two whole numbers. State the relationship first: a piece length divides; a repeat time is a multiple. Include the context in the answer so its meaning is clear.

Find the next coincidence with new intervals.

Quick Recap

The HCF is the greatest shared positive factor; the LCM is the smallest shared positive multiple.
Lists suit smaller numbers. Prime factorisations compare repeated prime copies efficiently.
Choose shared primes to smaller powers for the HCF; every required prime to its larger power for the LCM.
Equal sharing or cutting uses divisors; coinciding repeats use multiples.

The bells next ring together after $90$ seconds, the LCM of eighteen and thirty.

Last Accessed

Number

Algebra

Ratio and proportion

Geometry and measures

Graphs and functions

Probability

Statistics

1.02c - Highest factors and lowest multiples

Common factors and multiples

Choosing prime powers

Sharing and repeating cycles

Quick Recap

Learn faster with Nova