Secondary Mathematics: Prime Factors, Divisibility, HCF and LCM

Secondary Mathematics · Worked Repair Guide 12

Whole numbers have internal structure. Some split into many factors; others split only into 1 and themselves. Prime factorisation exposes that structure in a form that can be reused for divisibility, highest common factor, lowest common multiple, square and cube conditions, fraction simplification and algebraic reasoning.

This guide develops one central habit: translate a whole-number problem into prime-factor structure before choosing the operation. HCF and LCM become far easier when the learner can see which prime powers are shared and which are required.

All examples are original teaching material. Number-theory extensions should be matched to the learner’s actual Secondary Mathematics route and school programme.

1. Factors divide exactly

A positive factor of 24 is a positive integer that divides 24 with no remainder. The factors are 1,2,3,4,6,8,12,24.

A multiple of 24 has the form 24n for an integer n. Factors divide into a number; multiples are generated from a number.

Entry check: 6 is a factor of 24 because 24÷6=4. The number 48 is a multiple of 24 because 48=24×2.

Confusing factors and multiples can reverse an entire HCF or LCM problem, so identify the direction first.

2. Prime numbers are the building blocks of positive integers

A prime number has exactly two positive factors: 1 and itself. Examples include 2,3,5,7,11 and 13.

The number 1 is not prime because it has only one positive factor. The number 2 is the only even prime.

A composite number greater than 1 can be decomposed into smaller positive integer factors. Repeating this process eventually reaches primes.

This is why prime factorisation acts like a structural fingerprint for positive integers.

3. Factor trees should end in primes

To factorise 84, one route is 84=12×7, then 12=3×4 and 4=2×2. Therefore 84=2²×3×7.

A different valid tree, such as 84=6×14, reaches the same prime factorisation after all composite branches are continued.

The order of factors may differ, but the prime powers are the same.

Do not stop a factor tree at 6, 10, 12 or another composite number. The endpoint must be prime factors.

4. Divisibility tests are structure shortcuts

A number is divisible by 2 if its last digit is even. It is divisible by 5 if its last digit is 0 or 5. It is divisible by 10 if its last digit is 0.

A number is divisible by 3 if the sum of its digits is divisible by 3, and divisible by 9 if the digit sum is divisible by 9.

For divisibility by 4, inspect the last two digits. For divisibility by 8, inspect the last three digits.

Worked check: 7,452 has digit sum 18, so it is divisible by 3 and 9. Its last two digits are 52, divisible by 4, so 7,452 is divisible by 4 as well.

5. Prime factorisation can confirm divisibility

If N contains at least the prime factors of d with at least the required powers, then d divides N.

For example, 360=2³×3²×5. Since 12=2²×3, the factor structure required for 12 is contained inside 360, so 12 divides 360.

But 16=2⁴ does not divide 360 because 360 contains only 2³.

This viewpoint becomes especially powerful in HCF, LCM, perfect-square and perfect-cube problems.

6. HCF keeps only what all numbers share

The highest common factor of two or more positive integers is the greatest positive integer dividing every number.

Let 72=2³×3² and 120=2³×3×5. The shared primes are 2 and 3. For each shared prime, take the smaller exponent: HCF=2³×3=24.

The word “common” is the key. A factor involving 5 cannot belong to the HCF because 72 has no factor 5.

Worked check: 24 divides both 72 and 120 exactly.

7. LCM includes everything needed by every number

The lowest common multiple is the smallest positive integer that is a multiple of every given number.

Using 72=2³×3² and 120=2³×3×5, take every prime appearing in either factorisation, with the largest required exponent: LCM=2³×3²×5=360.

The word “multiple” changes the logic. The LCM must contain enough prime factors to be divisible by both original numbers.

Worked check: 360÷72=5 and 360÷120=3.

8. HCF and LCM solve different stories

HCF is often used when identical largest groups or pieces must divide quantities exactly. LCM is often used when repeating cycles or schedules must coincide.

HCF example: 48 red cards and 60 blue cards are packed into the greatest possible number of identical packs with no leftovers. HCF(48,60)=12, so 12 packs can be formed, each with 4 red and 5 blue cards.

LCM example: one signal repeats every 12 seconds and another every 18 seconds. If they coincide now, they next coincide after LCM(12,18)=36 seconds.

The same two numbers can therefore produce different operations depending on the relationship described.

9. A useful identity links HCF and LCM for two positive integers

For two positive integers a and b:

HCF(a,b) × LCM(a,b) = a × b.

For 72 and 120, 24×360=8640 and 72×120=8640.

This identity can find a missing HCF or LCM when the other is known.

Do not use it mechanically for three or more numbers in the same simple form; the two-number identity is the direct version being used here.

10. Perfect squares have even prime exponents

A positive integer is a perfect square when every exponent in its prime factorisation is even.

For example 144=2⁴×3², so 144 is a square: 12².

Consider 180=2²×3²×5. To make the smallest perfect square multiple of 180, multiply by 5 so the exponent of 5 becomes 2. Thus 900=30².

The exponent condition comes from squaring: every prime exponent in n² is twice an integer.

11. Perfect cubes have exponents divisible by three

A positive integer is a perfect cube when every prime exponent is a multiple of 3.

216=2³×3³=6³.

For 72=2³×3², multiply by 3 to obtain 2³×3³=216, the smallest perfect-cube multiple of 72.

Again, the exponent condition is structural: cubing multiplies every prime exponent by 3.

12. Prime factors help simplify fractions completely

To simplify 84/126, factorise: 84=2²×3×7 and 126=2×3²×7.

The common factor is 2×3×7=42. Dividing numerator and denominator by 42 gives 2/3.

This also shows why the fraction is fully simplified: the remaining numerator 2 and denominator 3 share no prime factor.

For larger numbers, prime structure can be more reliable than repeated guessing of common divisors.

13. Algebraic terms also have factor structure

The same HCF idea extends to algebra. The HCF of 12x³y² and 18x²y⁵ is 6x²y².

For the numerical coefficients, HCF(12,18)=6. For x, take the smaller exponent 2. For y, take the smaller exponent 2.

Therefore 12x³y²+18x²y⁵ can be factorised as 6x²y²(2x+3y³).

This connects integer factorisation to algebraic factorisation: both seek common multiplicative structure.

14. Remainders expose near-divisibility

When 53 is divided by 7, 53=7×7+4. The remainder is 4.

The division algorithm expresses an integer N as N=dq+r with 0≤r<d for positive divisor d.

This structure supports divisibility: d divides N exactly when r=0.

It also supports cyclic reasoning. For example, numbers leaving remainder 2 when divided by 5 have forms 5q+2.

15. Capstone: combine factorisation, HCF and square structure

Let N=2⁵×3²×5³. Find the largest perfect square factor of N and the smallest positive integer by which N must be multiplied to become a perfect square.

For the largest square factor, take the greatest even exponent not exceeding each exponent: 2⁴×3²×5².

Its value is 16×9×25=3600=60².

To make N itself a square, each odd exponent needs one more copy of its prime. Exponents 5 and 3 are odd, so multiply by 2×5=10. Then N×10=2⁶×3²×5⁴, all even exponents.

This problem is difficult only if the learner sees the number as one large product rather than as a vector of prime exponents.

16. Independent practice

1. List all positive factors of 18.
2. State whether 29 is prime.
3. Write 72 as a product of prime powers.
4. Write 180 as a product of prime powers.
5. Use divisibility tests to decide whether 5,346 is divisible by 3.
6. Decide whether 5,346 is divisible by 9.
7. Find HCF(48,72).
8. Find LCM(48,72).
9. Find HCF(84,126).
10. Find LCM(12,30).
11. Two cycles repeat every 8 and 14 minutes. When do they next coincide?
12. Find the greatest number of identical packs from 36 red and 54 blue items with no leftovers.
13. Find the smallest integer multiplying 75 to make a perfect square.
14. Find the smallest integer multiplying 108 to make a perfect cube.
15. Simplify 90/150 completely.
16. Find the HCF of 18x⁴y² and 30x²y⁵.
17. If HCF(a,b)=6, LCM(a,b)=180 and a=30, find b.
18. Find the remainder when 83 is divided by 6.
19. Explain why 1 is not prime.
20. State the largest perfect-square factor of 2⁵×3³.

17. Worked answers

1. 1,2,3,6,9,18.

2. Yes. Its only positive factors are 1 and 29.

3. 2³×3².

4. 2²×3²×5.

5. Yes. Digit sum 18 is divisible by 3.

6. Yes. Digit sum 18 is divisible by 9.

7. 24.

8. 144.

9. 42.

10. 60.

11. 56 minutes. LCM(8,14)=56.

12. 18 packs. HCF(36,54)=18.

13. 3. 75=3×5²; multiply by 3 to obtain 3²×5².

14. 2. 108=2²×3³; multiply by 2 to obtain 2³×3³.

15. 3/5. HCF is 30.

16. 6x²y².

17. 36. Since 6×180=30b, b=36.

18. 5. 83=6×13+5.

19. It has only one positive factor, not exactly two.

20. 2⁴×3²=144. Take the greatest even exponent not exceeding each exponent.

18. Diagnose factor errors by asking what must be shared

Common failures include stopping a factor tree before primes, taking the larger exponent for HCF, taking the smaller exponent for LCM, confusing factors with multiples, or treating a square condition as “an even number of prime types” rather than even prime exponents.

A useful repair note names the job: “HCF keeps only shared primes at minimum powers”, “LCM includes all required primes at maximum powers”, or “perfect square requires every exponent even”.

Then change the numbers. Structural understanding should survive a new factor tree and a different set of prime exponents.

19. Continue through the BTT Mathematics library

Return to the BTT Mathematics Hub. Use Indices, Roots and Standard Form for exponent structure, Ratio, Percentage and the Correct Base for multiplicative comparison, and Signed Numbers, Brackets and Algebraic Structure for factorisation in algebra.

The BTT Mathematical Lab is the diagnostic route when integer-structure errors keep resurfacing.

20. Sources and scope

The numerical examples, packing contexts and cycle problems in this guide are original teaching material. Prime factorisation, divisibility, HCF and LCM are treated as mathematical structures rather than memorised recipes.

For the current Singapore Secondary curriculum doorway, see MOE: Curriculum for secondary schools. Match algebraic HCF, remainder notation and number-theory extensions to the learner’s actual subject level and school programme.