Prime factorisation is a structural x-ray of a whole number. It reveals which prime building blocks are present and how many copies of each are available. From that structure, Highest Common Factor, Lowest Common Multiple, square and cube conditions, and repeated-cycle problems become easier to reason about.
This twenty-fifth Secondary 4 Mathematics Learning Guide develops prime factors, HCF, LCM and divisibility as one number-structure system. It belongs to the Secondary Mathematics Hub and S1–S4 Capability Map.
Current syllabus connection: prime factors, HCF, LCM, roots and numerical structure remain part of the Number and Algebra foundation in the 2026 O-Level Mathematics 4052 and 2027 SEC G3 Mathematics K310 progression.
Prime numbers are multiplicative building blocks
A prime number has exactly two positive factors: 1 and itself. Composite whole numbers greater than 1 can be written as products of primes.
For example, 360 can be decomposed as:
360 = 36×10 = 2³×3²×5.
The final prime factorisation is independent of the route used to reach it. A factor tree may split 360 as 36×10 or 18×20, but the same prime powers should remain.
Worked Example 1 | Prime factorise efficiently
Write 756 as a product of prime factors.
756=2×378=2²×189=2²×3×63=2²×3³×7.
756 = 2²×3³×7.
Check by multiplication: 4×27×7=756.
HCF keeps only factors shared by all numbers
For two numbers written in prime-power form, the HCF uses only prime factors common to both, taking the smaller exponent for each common prime.
Suppose A=2⁴×3²×5 and B=2³×3×7. The common primes are 2 and 3. Their smaller exponents are 3 and 1, so HCF=2³×3=24.
Worked Example 2 | HCF from prime factors
Find the HCF of 360 and 504.
360=2³×3²×5.
504=2³×3²×7.
HCF=2³×3²=72.
Check: 360÷72=5 and 504÷72=7.
LCM contains enough prime factors to cover every number
The LCM takes every prime factor appearing in either number, using the larger exponent needed.
For 360=2³×3²×5 and 504=2³×3²×7:
LCM=2³×3²×5×7=2520.
Both 360 and 504 divide 2520 exactly.
HCF and LCM answer different kinds of questions
| Question structure | Likely tool |
|---|---|
| Largest equal group size, greatest identical piece length | HCF |
| First time repeated cycles coincide, smallest common batch size | LCM |
| Common factor shared by all quantities | HCF |
| Smallest number divisible by several quantities | LCM |
Worked Example 3 | Largest identical grouping
A school has 84 red cards and 126 blue cards. It wants to make the greatest possible number of identical packs using all cards, with each pack containing the same number of red cards and the same number of blue cards. How many packs can be made?
The number of packs must divide both 84 and 126. Find HCF:
84=2²×3×7.
126=2×3²×7.
HCF=2×3×7=42 packs.
Each pack contains 84/42=2 red cards and 126/42=3 blue cards.
Worked Example 4 | Repeating cycles
One signal flashes every 18 seconds and another every 24 seconds. They flash together now. When will they next flash together?
18=2×3² and 24=2³×3.
LCM=2³×3²=72 seconds.
The problem asks for the smallest positive time that is a whole number of both 18-second and 24-second cycles.
Perfect squares require even prime exponents
A positive whole number is a perfect square when every exponent in its prime factorisation is even. For example, 2⁴×3²×5⁶ is a perfect square.
This follows because squaring doubles every exponent.
Worked Example 5 | Smallest multiplier to make a square
Find the smallest positive integer k such that 180k is a perfect square.
180=2²×3²×5.
The exponents of 2 and 3 are already even. The exponent of 5 is odd, so multiply by one more factor 5:
k=5, because 180×5=900=30².
Perfect cubes require exponents that are multiples of three
Cubing triples every prime exponent. Therefore a perfect cube has exponents divisible by 3.
Worked Example 6 | Smallest multiplier to make a cube
Find the smallest positive integer k such that 108k is a perfect cube.
108=2²×3³. The exponent of 3 is already a multiple of 3. The exponent of 2 needs one more factor 2 to reach 3.
k=2, because 216=6³.
Divisibility is often faster than full division
- Divisible by 2: last digit even.
- Divisible by 3: digit sum divisible by 3.
- Divisible by 5: last digit 0 or 5.
- Divisible by 9: digit sum divisible by 9.
- Divisible by 10: last digit 0.
These tests can screen factors before a longer prime decomposition.
Worked Example 7 | Missing digit and divisibility
The number 47a2 is divisible by 9. Find the possible digit a.
Digit sum=4+7+a+2=13+a. For divisibility by 9, 13+a must be a multiple of 9. With a from 0 to 9, the only possibility is 18.
13+a=18, so a=5.
HCF and LCM have a useful two-number check
For positive integers a and b:
HCF(a,b)×LCM(a,b)=a×b.
For 360 and 504, 72×2520=181440 and 360×504=181440. This identity is a powerful verification tool.
Worked Example 8 | Recover an unknown number
Two positive integers have HCF 12 and LCM 420. One integer is 60. Find the other.
Use HCF×LCM=product of the two integers:
12×420=60×n.
n=84. Check HCF(60,84)=12 and LCM(60,84)=420.
Common failure modes
| Error | Cause | Repair |
|---|---|---|
| Uses largest exponents for HCF | HCF and LCM rules reversed | HCF asks what both numbers definitely share |
| Uses only common primes for LCM | LCM seen as overlap instead of coverage | Include every prime needed by either number |
| Chooses HCF for repeated cycles | Question language not classified | Ask for first common future multiple |
| Calls a number square when one exponent is odd | Prime-power structure not checked | Inspect every exponent |
| Finds a common factor but not the highest | Process stops early | Prime factorise or continue factor comparison |
Independent practice
- Prime factorise 540.
- Find the HCF of 168 and 252.
- Find the LCM of 72 and 90.
- Two alarms repeat every 28 minutes and 42 minutes. They sound together now. When do they next coincide?
- Find the smallest k such that 75k is a perfect square.
- Find the smallest k such that 250k is a perfect cube.
Explained answers
1. 540=54×10=2²×3³×5.
2. 168=2³×3×7 and 252=2²×3²×7, so HCF=84.
3. 72=2³×3² and 90=2×3²×5, so LCM=2³×3²×5=360.
4. LCM(28,42)=84 minutes.
5. 75=3×5², so multiply by 3. k=3.
6. 250=2×5³. The exponent of 2 must rise from 1 to 3, so multiply by 2². k=4.
Final thought
Prime factors reveal how a number is built. HCF asks what several numbers share. LCM asks what one number must contain to cover several cycles or divisibility demands.
Decompose first. Then choose whether the problem needs shared structure or complete coverage.
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