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aiMOOC

Prime factorisation and divisibility - Mathematics Year 8

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Introduction

An animated sieve illustrating the search for prime numbers. — A sieve classifies integers as prime or composite; it is not already a factor tree for the six numbers in this course. The first frame may show the unmarked starting grid.

Why do equal-pack problems and repeating-event problems use different selections from the same prime factors? Build complete factorisations, then decide whether the context needs a highest common factor or a lowest common multiple.

You will factorise six integers, solve three HCF/LCM pairs, correct an incomplete factor tree and explain two original contextual solutions. Start by comparing 360 = 36 × 10 with 360 = 8 × 45: their branches differ, but their final prime leaves should agree.

Build and check complete prime factorisations

A prime is an integer greater than 1 with exactly two positive factors: 1 and itself. A composite integer greater than 1 has additional positive factors. The number 1 is neither prime nor composite. The integer 2 is the only even prime; other even integers greater than 2 have 2 as a proper factor. Do not confuse a number's factors with its multiples.

Construct a complete factor tree. Each branch replaces a composite number by a product of two smaller positive integers. Continue until every leaf is prime. For 360, choose 36 × 10. Then 36 = 6 × 6, each 6 = 2 × 3, and 10 = 2 × 5. The final prime leaves are 2, 3, 2, 3, 2, 5. Group equal primes to write $360 = 2^{3} \times 3^{2} \times 5$ .

A different first split, 8 × 45, gives 8 = 2 × 2 × 2 and 45 = 3 × 3 × 5. The same multiplicities appear. Branch order is not part of the factorisation's value. For positive integers greater than 1, the completed prime factorisation is unique apart from order. That is why choosing an efficient first split does not change the final result.

Check by rebuilding the number. The powers mean repeated multiplication: 2³ = 8 and 3² = 9, so 8 × 9 × 5 = 360. Writing 2 × 3 × 5 = 30 loses repeated factors. Writing 2³ × 9 × 5 still has the correct value, but 9 is composite, so it is not yet a product of primes. Completeness and equality are separate checks.

Repeated division is an alternative representation. Divide 360 by 2 three times to reach 45, by 3 twice to reach 5, and by 5 to reach 1. Retain every divisor. The final 1 is a stopping value, not an extra prime factor. A factor tree makes the decomposition visible; a division ladder can be efficient when small primes divide repeatedly.

The sieve animation asks which integers survive elimination of composite multiples. Use it to recognise candidate primes, not as proof that a particular factor tree has the right number of repeated leaves. A quick divisibility check can suggest a branch: an even integer is divisible by 2, a digit sum divisible by 3 indicates divisibility by 3, and an integer ending in 0 or 5 is divisible by 5. These tests help choose a method; the completed product still needs checking.

Choose prime powers for HCF and LCM

HCF and LCM answer different questions. The highest common factor is the greatest positive integer dividing both given integers. The lowest common multiple is the smallest positive integer divisible by both. A shared factor is no larger than either positive integer; a common multiple is at least as large as each. Those size checks often expose a swapped answer.

Worked pair: 72 and 90. Their prime decompositions are $72 = 2^{3} \times 3^{2}, 90 = 2 \times 3^{2} \times 5$ . For a common factor, choose only shared primes and the smaller multiplicity of each: one 2 and two 3 s. Thus the HCF is 2 × 9 = 18. Both 72 ÷ 18 = 4 and 90 ÷ 18 = 5 are integers.

For a common multiple, include every prime appearing in either number at its greater multiplicity. The LCM is 2³ × 3² × 5 = 360. Check 360 ÷ 72 = 5 and 360 ÷ 90 = 4. If a prime is absent from one factorisation, treat its exponent there as zero when comparing exponents. Its contribution to the HCF is then 1, while it may still contribute to the LCM.

Why these exponent choices work. A common factor cannot use more copies of a prime than either number contains. Taking the smallest exponents gives the greatest permitted shared product. A common multiple must contain enough copies for both numbers, so taking the largest exponents gives the least product meeting both demands. Multiplying the original numbers always gives a common multiple, but it need not be the lowest.

For positive integers a and b, $HCF (a, b) LCM (a, b) = a b$ . For 72 and 90, 18 × 360 = 6480 = 72 × 90. This is a useful independent relationship after finding one of the two quantities. It does not mean that an arbitrary factor and arbitrary multiple satisfy the identity.

Follow Corbettmaths’ original product-of-primes explanation. Its visible example uses 28, whereas our six-number set and HCF/LCM contexts are separate original tasks. Check both primality of leaves and the product when applying the method. Original: corbettmaths, Product of Primes - Corbettmaths. The written explanations and native tasks work without sound or the external player.

Choose a representation for the values, not a ritual. Listing factors or multiples can efficiently check a small pair. Prime powers make larger multiplicities clear and support both HCF and LCM without long lists. Explain why your chosen check is independent of the step that might contain the error.

Solve six numbers and three contextual pairs

Six complete factorisations. Attempt each number first, then compare with the original answer table. Every leaf must be prime and multiplying the powers must restore the starting integer.

Integer	Product of prime powers	Reconstruction
72	2³ × 3²	8 × 9 = 72
90	2 × 3² × 5	2 × 9 × 5 = 90
360	2³ × 3² × 5	8 × 9 × 5 = 360
84	2² × 3 × 7	4 × 3 × 7 = 84
150	2 × 3 × 5²	2 × 3 × 25 = 150
196	2² × 7²	4 × 49 = 196

Three pairs, two different selections per pair.

Pair	HCF	LCM
72 and 90	18	360
84 and 150	6	2100
150 and 196	2	14700

For 84 and 150, the shared primes are 2 and 3, each once, giving HCF 6. The LCM needs 2², 3, 5² and 7, giving 4 × 3 × 25 × 7 = 2100. For 150 and 196, only 2 is shared; LCM 2² × 3 × 5² × 7² = 14700. Check that each HCF divides both originals and each LCM is divisible by both. Product checks give 6 × 2100 = 84 × 150 and 2 × 14700 = 150 × 196.

Context: the greatest number of equal packs. An invented club has 84 red counters and 150 blue counters. It wants identical packs, using every counter with none left. The number of packs must divide both totals, so the greatest possible number is HCF 6. Each pack contains 14 red and 25 blue counters. Check six copies restore 84 and 150. The HCF counts packs here; it is not the total number of counters in each pack.

Context: the first shared repeat time. Two model signals begin together and repeat every 72 seconds and 90 seconds. The first later shared start is after LCM 360 seconds, or 6 minutes. That is five 72-second intervals and four 90-second intervals. Using HCF 18 seconds would not even complete one repeat of either signal. These are invented exact timing models, not measurements of real devices.

Error clinic and transfer. A learner stops 360 at 8 × 45, calls both leaves prime, and selects only one copy of every prime for the LCM. Continue both composite branches, restore multiplicities and use the context to select smaller shared or larger required exponents. As an unfamiliar check, 48 = 2⁴ × 3 and 180 = 2² × 3² × 5 give HCF 12 and LCM 720. Verify the divisions and the product identity. Retain all six factorisations, all three pairs and both contextual interpretations; after a genuine break, keep an actual later retry rather than claiming lasting mastery from one attempt.

Interactive Tasks

Quiz: Test Your Knowledge

Why is 1 not prime? (It has only one positive factor) (!It is an even number) (!It has three prime factors) (!It is divisible by every integer)

Which prime-power expression equals 360? (2 cubed times 3 squared times 5) (!2 times 3 times 5) (!2 squared times 3 cubed times 5) (!2 cubed times 3 times 5)

Why is 8 times 45 not a completed prime factorisation? (Both leaves are composite) (!Its product is not 360) (!Every tree must start with 36) (!A factor tree cannot have two branches)

What is the HCF of 72 and 90? (18) (!360) (!9) (!72)

What is the LCM of 72 and 90? (360) (!18) (!6480) (!90)

Which exponent selection gives the HCF? (Smallest exponents shared by both numbers) (!Largest exponents from either number) (!Add all exponents in both numbers) (!Keep each prime exactly once)

What is the HCF of 84 and 150? (6) (!2100) (!14) (!25)

What is the LCM of 150 and 196? (14700) (!2) (!29400) (!7350)

How many identical packs use all 84 red and 150 blue counters at most? (6 packs) (!39 packs) (!84 packs) (!150 packs)

When do the model 72-second and 90-second signals first repeat together? (360 seconds) (!18 seconds) (!162 seconds) (!90 seconds)

Memory Game

Prime	Integer greater than one with exactly two positive factors
Composite	Integer greater than one with extra positive factors
Factorisation	Writing a number as a product of factors
Exponent	Number of repeated equal prime factors in a power
HCF	Greatest positive integer dividing both given integers
LCM	Smallest positive integer divisible by both given integers

Drag and Drop

18	HCF of 72 and 90
360	LCM of 72 and 90
6	HCF of 84 and 150
2100	LCM of 84 and 150
14700	LCM of 150 and 196

Crossword Puzzle

PRIME	Integer greater than one with exactly two positive factors
COMPOSITE	Integer greater than one that is not prime
FACTOR	Integer dividing the stated integer exactly
MULTIPLE	Product of an integer with a positive whole-number count here
EXPONENT	Repeated-factor count in index notation
DIVISIBILITY	Property of being divided exactly without a remainder

LearningApps

Optional extra practice. Choose an appropriate activity with your teacher. If the external collection is unavailable, use the native activities and independent tasks in this course.

Cloze Text

Open Tasks

Easy

Finish a tree: Factorise 360 using 36 times 10; identify all prime leaves and rebuild the original.
Compare two branches: Repeat with 8 times 45 and explain why the multiplicities agree.
Complete the six-number set: Factorise 72, 90, 360, 84, 150 and 196 in prime-power notation.
Classify the leaves: Explain why 1, 8 and 45 cannot be prime leaves in the completed examples.

Standard

Solve all three pairs: Find HCF and LCM for 72/90, 84/150 and 150/196 with exponent reasons.
Check independently: Use exact divisions and HCF times LCM equals the original product for each pair.
Build equal packs: Use all 84 red and 150 blue counters in the greatest number of identical packs; give contents.
Find the repeat time: Interpret 72-second and 90-second signals starting together; include seconds and minutes.

Advanced

Transfer to a new pair: Find HCF and LCM for 48 and 180, explaining an efficient representation and two checks.
Correct a plausible mistake: Repair an incomplete 360 tree and the rule that every prime should appear only once.
Defend the exponent rules: Explain absent primes as exponent zero and why HCF and LCM choose different powers.
Retrieve later: Keep a genuinely later pair and contextual decision, including the first error and correction if needed.

Learning Check

Complete prime products: Give all six factorizations and reconstruction checks.
First pair with reasons: Find HCF and LCM of 72 and 90.
Remaining pairs: Solve 84/150 and 150/196 completely.
Two contextual interpretations: Distinguish packs from repeated signals.
Unfamiliar pair and error: Use 48 and 180; correct composite leaves and missing repetitions.
Evidence and method choice: Retain completed work and an actual later response.

Worked checks and assessment guidance

72=2³×3²; 90=2×3²×5; 360=2³×3²×5; 84=2²×3×7; 150=2×3×5²; 196=2²×7². Reconstruct 72, 90, 360, 84, 150, 196 respectively; every final base is prime.
Shared minima give 2×3²=18; union maxima give 2³×3²×5=360. Check 72/18=4, 90/18=5, 360/72=5, 360/90=4 and 18×360=72×90.
84/150: HCF 6, LCM 2100; 150/196: HCF 2, LCM 14700. Factor bases and exponent minima/maxima explain the choice; exact divisions and products verify each result.
84 red/150 blue allow 6 identical packs, each 14 red and 25 blue, with no leftovers. Signals 72/90 seconds first repeat together at 360 seconds=6 minutes, not HCF 18 seconds.
48=2⁴×3 and 180=2²×3²×5, so HCF 12 and LCM 720. 12×720=48×180=8640. Leaves 8/45 need further splitting; dropping repeated primes changes the original product.
Keep six complete trees or ladders, three checked pairs, contextual units and a chosen efficient method. A sieve identifies primes but is not a pre-completed tree; a real later retry is separate evidence, not an invented performance claim.

Learning evidence

Retain six prime factorisations, three HCF/LCM pairs, exact reconstruction/divisibility checks and two original contextual interpretations. Keep any actual later response separately.

OER and Wikipedia

Department for Education: England mathematics programme of study, Key Stage 3, geometry,number,fluency,reasoning and problem solving. Year 8 allocation is editorial within KS3. All numerical tasks,models,diagrams and worked answers are original. Actual Commons sources retain their creator/licence credits below; original-publisher videos supplement the written teaching. Reviewed 12 November 2026.

Sieve of Eratosthenes animation.gif, SKopp at German Wikipedia, CC BY-SA 3.0. Unchanged original; caption and learning activities are original.

Dienes Material (Dyskalkulie).jpg, Rhetos, CC0. Unchanged original; caption and learning activities are original.