Secondary 1 Mathematics Classroom | Chapter 1: Number Structure—Primes, Prime Factorisation, HCF, LCM, Squares, Cubes and Roots | G2/G3

SECONDARY 1 MATHEMATICS CLASSROOM · CHAPTER 1 · NUMBER STRUCTURE · G2/G3

Primes, Prime Factorisation, HCF, LCM, Squares, Cubes and Roots: Take the Number Apart Before You Decide What to Do With It

In this classroom, you will not begin by guessing whether a word problem needs HCF or LCM. You will begin by exposing the prime structure of the numbers and asking what the answer must do.

A whole number is not an isolated object. It has factors, multiples and a prime architecture. Prime factorisation reveals that architecture. Highest Common Factor finds the largest structure shared by several numbers. Lowest Common Multiple builds the smallest structure large enough to contain every required factor. Perfect squares and cubes appear when prime factors can be grouped into equal pairs or equal triples.

Classroom rule: decompose the number, identify the relationship, choose the required prime exponents, then rebuild and verify.

The current Secondary One G2 and G3 Mathematics syllabuses both include primes and prime factorisation, and the use of prime factorisation to find HCF, LCM, squares, cubes, square roots and cube roots. Schools may differ in sequence and pace, but this number-structure core is shared.

Official reference: MOE G2 and G3 Mathematics Syllabuses.

Navigate: factors and multiples · prime numbers · prime factorisation · HCF · LCM · squares and roots · cubes and roots · misconception clinic · guided practice · examination transfer · exit ticket.

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Featured Answer: What Is Number Structure?

Number structure is the study of how whole numbers are built and how they relate through factors, multiples and powers. Prime numbers act as the basic building blocks. Prime factorisation writes a composite number entirely in those building blocks. Once that representation is visible, HCF, LCM, square roots and cube roots can be found systematically.

For example:

360 = 2³ × 3² × 5.

This single line tells us that 360 is divisible by 2, 3 and 5; that it contains three factors of 2, two factors of 3 and one factor of 5; that it is not a perfect square because not every exponent is even; and that it is not a perfect cube because not every exponent is a multiple of 3.

The Simple Classroom Answer

Prime factorisation is the number’s blueprint. HCF keeps only what all numbers share. LCM keeps everything needed by any number. Square roots pair factors. Cube roots group them in threes.

• Factors divide a number exactly.
• Multiples are generated by multiplication.
• Primes have exactly two positive factors.
• Prime factorisation decomposes a number into prime factors.
• HCF uses the shared prime factors with the smaller available exponents.
• LCM uses every required prime factor with the larger required exponents.
• Perfect squares have even prime exponents.
• Perfect cubes have prime exponents divisible by 3.

How to Use This Classroom

1. Attempt every Your Turn question before opening its answer.
2. Write the prime factorisation instead of doing all reasoning mentally.
3. If a result is wrong, locate the first wrong factor, exponent or interpretation.
4. Repeat the idea with one changed number.
5. Explain why the method works before adding time pressure.
6. Close the page and return after a delay to test retrieval rather than recognition.

You need paper, a pen and a calculator for checking only. Most of this chapter should be understandable without asking the calculator to make the structural decisions.

1. Start the Lesson With Division That Leaves No Remainder

Teacher: Write 24 on the board. Ask which positive whole numbers divide 24 exactly.

The positive factors of 24 are:

1, 2, 3, 4, 6, 8, 12 and 24.

A factor divides exactly, leaving no remainder. Therefore 6 is a factor of 24 because 24 ÷ 6 = 4. The number 5 is not a factor of 24 because 24 ÷ 5 is not a whole number.

Factor question: does this number fit exactly inside the other number?

2. Find Factors in Pairs

Do not search randomly. Build factor pairs from the outside inward.

For 36:

• 1 × 36;
• 2 × 18;
• 3 × 12;
• 4 × 9;
• 6 × 6.

So the positive factors are 1, 2, 3, 4, 6, 9, 12, 18 and 36.

Once the smaller factor in a new pair would exceed the larger factor already reached, the search is complete.

Your Turn 1

List all positive factors of 40 using factor pairs.

Check your answer

Factor pairs are 1×40, 2×20, 4×10 and 5×8. The factors are 1, 2, 4, 5, 8, 10, 20 and 40.

3. Multiples Are Generated, Not Found Inside

Multiples of 6 are produced by multiplying 6 by whole numbers:

6, 12, 18, 24, 30, 36, …

There are infinitely many positive multiples. A factor list is finite; a multiple list continues.

Multiple question: can this number be generated by multiplying the given number by a whole number?

4. Direction Matters: Factor of and Multiple of Are Opposite Relationships

The number 6 is a factor of 42 because 42 ÷ 6 = 7. The number 42 is a multiple of 6 because 42 = 6 × 7.

But 42 is not a factor of 6. Read the direction of the sentence.

Statement	Test
a is a factor of b	b ÷ a is a whole number
b is a multiple of a	b = a × a whole number

5. Use Divisibility Tests as Search Tools

Divisibility tests help you identify likely factors quickly.

• Divisible by 2: final digit is even.
• Divisible by 3: digit sum is divisible by 3.
• Divisible by 5: final digit is 0 or 5.
• Divisible by 9: digit sum is divisible by 9.
• Divisible by 10: final digit is 0.

These tests do not replace prime factorisation. They help you choose the next prime divisor efficiently.

6. A Prime Number Has Exactly Two Positive Factors

A prime number greater than 1 has exactly two positive factors: 1 and itself.

The first few prime numbers are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, …

A prime is not “a number that cannot be divided”. It can be divided exactly by 1 and itself. The definition is about having exactly two positive factors.

7. The Number 1 Is Not Prime

The number 1 has only one positive factor: 1. A prime number needs exactly two positive factors, so 1 is not prime.

This distinction keeps prime factorisation unique. If 1 were treated as prime, extra factors of 1 could be inserted indefinitely without changing the number.

8. The Number 2 Is the Only Even Prime

The number 2 has exactly two positive factors: 1 and 2. Every other even positive whole number greater than 2 is divisible by 2 and therefore has at least three factors: 1, 2 and itself.

9. Composite Numbers Have More Than Two Positive Factors

A composite number greater than 1 can be decomposed into smaller positive factors. For example, 21 = 3 × 7, so 21 is composite.

The whole numbers greater than 1 are therefore either prime or composite.

10. Test Primality Systematically

To decide whether a small number is prime, test possible prime divisors up to its square root. If a composite number has a factor pair, at least one member of the pair is no greater than the square root.

To test 37, √37 is a little above 6. Test prime divisors 2, 3 and 5. The number 37 is divisible by none of them, so 37 is prime.

To test 51, the digit sum is 6, so 51 is divisible by 3. Therefore 51 = 3 × 17 and is composite.

11. Teacher Model 1: Prime or Composite?

Decide whether 91 is prime.

√91 is between 9 and 10, so test prime divisors 2, 3, 5 and 7.

• 91 is not even.
• Its digit sum 10 is not divisible by 3.
• It does not end in 0 or 5.
• 91 ÷ 7 = 13.

Therefore 91 = 7 × 13, so it is composite.

Your Turn 2

Classify each number as prime, composite or neither: 1, 2, 17, 27, 41, 57.

Answers

1 is neither prime nor composite. 2, 17 and 41 are prime. 27 = 3³ and 57 = 3×19, so both are composite.

12. Prime Factorisation Rewrites a Number Using Only Prime Factors

Prime factorisation decomposes a whole number greater than 1 into a product of primes.

For example:

60 = 2 × 2 × 3 × 5 = 2² × 3 × 5.

The final factors must all be prime. Writing 60 = 6 × 10 is a factorisation, but not yet a prime factorisation.

13. Use a Factor Tree to Keep Decomposing

Start with any valid factor pair. Continue splitting every composite branch until every endpoint is prime.

For 84:

• 84 = 7 × 12;
• 12 = 3 × 4;
• 4 = 2 × 2.

Therefore:

84 = 2² × 3 × 7.

Circle prime endpoints if that helps you see when decomposition is complete.

14. The Division-Ladder Method Repeats Prime Division

Another method is to divide repeatedly by the smallest convenient prime factor.

For 180:

• 180 ÷ 2 = 90;
• 90 ÷ 2 = 45;
• 45 ÷ 3 = 15;
• 15 ÷ 3 = 5;
• 5 ÷ 5 = 1.

So:

180 = 2² × 3² × 5.

Use the method that makes your steps easiest to verify.

15. Different Factor Trees Must Reach the Same Prime Blueprint

You might begin 84 as 7 × 12 or as 2 × 42. Both routes end at 2² × 3 × 7.

Every whole number greater than 1 has a unique prime factorisation apart from the order in which the prime factors are written. This means a different valid decomposition route should not change the final prime exponents.

16. Indices Compress Repeated Prime Factors

Instead of writing 2 × 2 × 2 × 3 × 3 × 5, write:

2³ × 3² × 5.

The exponent records how many copies of the prime factor are present.

Do not confuse 2³ with 2 × 3. The expression 2³ means 2 × 2 × 2 = 8.

17. Teacher Model 2: Prime Factorisation of 360

Divide systematically:

• 360 = 36 × 10;
• 36 = 2² × 3²;
• 10 = 2 × 5.

Combine equal prime bases:

360 = 2³ × 3² × 5.

Check by rebuilding: 8 × 9 × 5 = 360.

18. Rebuild the Number as a Compulsory Check

After writing a prime factorisation, multiply the prime powers back together.

If you claim 294 = 2 × 3 × 7², check:

2 × 3 × 49 = 294.

If the product does not return to the original number, the factorisation contains a missing or extra factor.

Your Turn 3

Write each number as a product of prime factors in index form:

1. 72
2. 200
3. 294
4. 1000

Worked answers

72 = 2³×3². 200 = 2³×5². 294 = 2×3×7². 1000 = 2³×5³.

19. HCF Is the Greatest Factor Shared by All the Numbers

The Highest Common Factor of two or more positive whole numbers is the greatest positive whole number that divides every one of them exactly.

For 12 and 18:

• factors of 12: 1, 2, 3, 4, 6, 12;
• factors of 18: 1, 2, 3, 6, 9, 18.

The common factors are 1, 2, 3 and 6. The highest is 6, so HCF(12,18) = 6.

20. Listing Factors Works for Small Numbers

For small inputs, factor lists can make HCF visible. But as numbers grow, prime factorisation is more systematic and easier to extend to three or more numbers.

21. Find HCF From Shared Prime Factors

Suppose:

48 = 2⁴ × 3

72 = 2³ × 3².

The primes shared by both numbers are 2 and 3. For each shared prime, use the smaller exponent available in all numbers:

HCF = 2³ × 3 = 24.

22. Why HCF Uses the Smaller Exponents

A common factor must fit inside every source number.

For the prime 2, the number 48 contains four copies but 72 contains only three. A common factor cannot demand four copies because 72 does not have that many. Therefore the shared structure is limited by the smaller exponent.

HCF asks: what is the largest prime structure that every number can supply?

23. Teacher Model 3: HCF of 84 and 126

Prime factorise:

84 = 2² × 3 × 7

126 = 2 × 3² × 7.

Take the common primes with smaller exponents:

HCF = 2 × 3 × 7 = 42.

Check: 84 ÷ 42 = 2 and 126 ÷ 42 = 3. The answer divides both numbers exactly.

24. HCF Appears in Equal-Grouping Problems

A teacher has 60 red cards and 84 blue cards. The cards are to be divided into the greatest possible number of identical groups, with no leftovers.

The number of groups must divide both 60 and 84. Therefore find HCF(60,84).

60 = 2² × 3 × 5

84 = 2² × 3 × 7.

HCF = 2² × 3 = 12.

There can be 12 identical groups. Each group has 5 red cards and 7 blue cards.

25. HCF Appears in Greatest-Exact-Length Problems

Two ribbons are 96 cm and 144 cm long. They are cut into equal pieces of the greatest possible whole-number length, with no waste.

The piece length must divide both ribbon lengths. Therefore:

HCF(96,144) = 48.

The greatest possible piece length is 48 cm.

26. HCF Has a Size Boundary

The HCF cannot exceed the smallest positive input because it must divide that number.

If you find HCF(18,30) = 90, the result is impossible before any detailed checking: 90 cannot divide 18.

Your Turn 4

1. Find HCF(72,120).
2. Find HCF(96,144,168).
3. 84 apples and 126 oranges are packed into the greatest possible number of identical packs with no leftovers. How many packs are made, and what is in each pack?

Worked answers

HCF(72,120)=24. HCF(96,144,168)=24. For 84 and 126, HCF=42, so there are 42 packs with 2 apples and 3 oranges in each.

27. LCM Is the Smallest Positive Multiple Shared by All the Numbers

The Lowest Common Multiple of two or more positive whole numbers is the smallest positive whole number divisible by every source number.

For 4 and 6:

• multiples of 4: 4, 8, 12, 16, 20, 24, …;
• multiples of 6: 6, 12, 18, 24, ….

The lowest positive common multiple is 12.

28. Listing Multiples Works for Small Numbers

Listing can be useful when the first common multiple appears quickly. For larger numbers or several cycles, prime factorisation gives a cleaner route.

29. Find LCM From All Required Prime Factors

Suppose:

12 = 2² × 3

18 = 2 × 3².

The LCM must be divisible by both numbers. It therefore needs two copies of 2 to contain 12 and two copies of 3 to contain 18:

LCM = 2² × 3² = 36.

30. Why LCM Uses the Larger Exponents

A common multiple must contain enough prime factors to be divisible by every source number.

If one number requires three factors of 2 and another requires only one, the common multiple still needs all three. Therefore use the larger exponent.

LCM asks: what is the smallest prime structure large enough to contain every number’s requirements?

31. Teacher Model 4: LCM of 24 and 36

Prime factorise:

24 = 2³ × 3

36 = 2² × 3².

Take every required prime with the larger exponent:

LCM = 2³ × 3² = 72.

Check: 72 ÷ 24 = 3 and 72 ÷ 36 = 2.

32. LCM Appears in Repeating-Cycle Problems

One signal flashes every 18 seconds and another every 30 seconds. They flash together now. When will they next flash together?

The next shared time must be a multiple of both 18 and 30.

LCM(18,30) = 90.

They next flash together after 90 seconds.

33. LCM Appears in Common-Denominator Work

To add 5/12 and 7/18, a convenient common denominator is LCM(12,18) = 36.

5/12 = 15/36 and 7/18 = 14/36, so the sum is 29/36.

Any common multiple could serve as a denominator, but the LCM usually keeps the arithmetic smaller.

34. LCM Has a Size Boundary

The LCM cannot be smaller than the largest positive input because the largest input must divide it.

If you find LCM(12,18) = 6, the result is impossible because 18 does not divide 6.

Your Turn 5

1. Find LCM(24,30).
2. Find LCM(8,12,15).
3. Three maintenance checks repeat every 18, 24 and 30 days. They occur together today. After how many days will they next occur together?

Worked answers

LCM(24,30)=120. LCM(8,12,15)=120. LCM(18,24,30)=360, so the checks next coincide after 360 days.

35. HCF and LCM Answer Opposite Structural Questions

Question asks for…	Think…	Likely owner
greatest size that divides every quantity exactly	fit inside all inputs	HCF
greatest number of identical groups with no leftovers	group count must divide every total	HCF
next time several cycles coincide	time must be a multiple of every interval	LCM
smallest number divisible by all given values	contain every divisibility requirement	LCM

Do not choose from the words “greatest” and “lowest” alone. Ask what the answer must do to the given numbers.

36. Use the Divides-or-Is-Divisible Test

If the answer must divide the given numbers, think HCF. If the answer must be divisible by the given numbers, think LCM.

This question is more reliable than hunting for a keyword.

37. Teacher Model 5: Same Numbers, Different Stories

Use 12 and 18 in two problems.

Problem A: 12 red and 18 blue objects are placed into the greatest possible number of identical groups. The group count must divide 12 and 18, so use HCF = 6.

Problem B: Two events repeat every 12 and 18 minutes. The next shared time must be divisible by 12 and 18, so use LCM = 36.

The numbers are identical. The relationship requested is different.

38. A Useful Two-Number Check

For two positive whole numbers a and b:

HCF(a,b) × LCM(a,b) = a × b.

For 24 and 36, HCF = 12 and LCM = 72. Then 12 × 72 = 864 and 24 × 36 = 864.

Use this as an enrichment check for two positive integers, not as a substitute for understanding the prime-exponent method.

39. A Square Number Uses Two Equal Factors

For a whole number n:

n² = n × n.

The first positive square numbers are:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, …

Recognising common squares improves speed, but prime factorisation explains why a number is a perfect square.

40. A Square Root Reverses Squaring

The principal square root √a is the non-negative number whose square is a.

√81 = 9 because 9² = 81.

Be precise: √81 means 9. The equation x² = 81 has two real solutions, x = 9 and x = −9. The root symbol itself denotes the principal non-negative square root.

41. Perfect Squares Have Even Prime Exponents

A positive whole number is a perfect square exactly when every exponent in its prime factorisation is even.

Why? Squaring duplicates every prime factor. For example:

30² = (2 × 3 × 5)² = 2² × 3² × 5² = 900.

Every prime appears in a pair.

42. Teacher Model 6: Find a Square Root by Prime Factorisation

Find √1764.

Prime factorise:

1764 = 2² × 3² × 7².

Take one prime from each pair:

√1764 = 2 × 3 × 7 = 42.

Check: 42² = 1764.

43. Make a Perfect Square by Completing Prime Pairs

Suppose:

108 = 2² × 3³.

The exponent 2 is even, but 3 is odd. Multiply by one more factor of 3:

108 × 3 = 2² × 3⁴ = 324 = 18².

The smallest positive whole-number multiplier is 3.

44. Make a Perfect Square by Removing Unpaired Factors

Using 108 = 2² × 3³, divide by one factor of 3:

108 ÷ 3 = 2² × 3² = 36 = 6².

The smallest positive whole-number divisor that makes the quotient a perfect square is 3.

45. Pairing Is the Operational Picture for Square Roots

For √(2⁶ × 3⁴ × 5²), group each prime into pairs. Half the exponent remains outside the square root:

√(2⁶ × 3⁴ × 5²) = 2³ × 3² × 5.

This works because taking a square root reverses squaring and therefore halves even exponents.

Your Turn 6

1. Show that 900 is a perfect square using prime factorisation and find √900.
2. Find √1296 by prime factorisation.
3. Find the smallest positive whole number by which 75 must be multiplied to become a perfect square.
4. Find the smallest positive whole number by which 72 must be divided to become a perfect square.

Worked answers

900=2²×3²×5², so √900=30. 1296=2⁴×3⁴, so √1296=2²×3²=36. 75=3×5², so multiply by 3 to obtain 225. 72=2³×3², so divide by 2 to obtain 36.

46. A Cube Number Uses Three Equal Factors

For a whole number n:

n³ = n × n × n.

The first positive cube numbers are:

1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, …

47. A Cube Root Reverses Cubing

∛125 = 5 because 5³ = 125.

Cube roots group prime factors in threes. Unlike principal square roots, real cube roots can also be negative: ∛(−125) = −5. Detailed work with directed numbers continues in the next number classroom.

48. Perfect Cubes Have Prime Exponents Divisible by 3

A positive whole number is a perfect cube exactly when every exponent in its prime factorisation is a multiple of 3.

Why? Cubing triples every prime factor count:

12³ = (2² × 3)³ = 2⁶ × 3³ = 1728.

49. Teacher Model 7: Find a Cube Root by Prime Factorisation

Find ∛1728.

1728 = 2⁶ × 3³.

Group the factors into triples. Divide each exponent by 3:

∛1728 = 2² × 3 = 12.

Check: 12³ = 1728.

50. Make a Perfect Cube by Completing Prime Triples

Suppose:

432 = 2⁴ × 3³.

The exponent of 3 is already a multiple of 3. The exponent 4 on 2 must rise to 6, so multiply by 2² = 4:

432 × 4 = 2⁶ × 3³ = 1728 = 12³.

51. Make a Perfect Cube by Removing Incomplete Triples

Again use 432 = 2⁴ × 3³. To leave an exponent of 3 on the prime 2, divide by one factor of 2:

432 ÷ 2 = 2³ × 3³ = 216 = 6³.

The smallest positive whole-number divisor is 2.

52. Tripling Exponents Is the Operational Picture for Cubes

For ∛(2⁹ × 3⁶ × 5³), divide every exponent by 3:

∛(2⁹ × 3⁶ × 5³) = 2³ × 3² × 5.

The exponents are divisible by 3, so the original number is a perfect cube.

53. A Number Can Be Both a Perfect Square and a Perfect Cube

If every prime exponent is divisible by 6, the number is both a perfect square and a perfect cube.

For example:

64 = 2⁶ = 8² = 4³.

This is an enrichment connection: common multiples of exponent requirements explain shared power structures.

Your Turn 7

1. Show that 216 is a perfect cube and find ∛216.
2. Find ∛13824, given that 13824 = 2⁹ × 3³.
3. Find the smallest positive whole number by which 250 must be multiplied to become a perfect cube.
4. Find the smallest positive whole number by which 2000 must be divided to become a perfect cube.

Worked answers

216=2³×3³, so ∛216=6. ∛13824=2³×3=24. 250=2×5³, so multiply by 2²=4 to obtain 1000=10³. 2000=2⁴×5³, so divide by 2 to obtain 1000=10³.

54. Use the Calculator to Verify, Not to Replace the Structure

A calculator can confirm 42² = 1764 or 12³ = 1728. It cannot explain why every exponent must be even for a square or divisible by 3 for a cube.

Use this sequence:

1. prime factorise;
2. group exponents;
3. derive the root;
4. use the calculator to rebuild and check.

55. Keep Exact Whole-Number Roots Exact

If √1764 = 42 exactly, do not replace it with 42.0 or an unnecessary decimal approximation. Exact values preserve the structure and avoid rounding noise.

56. Teacher Model 8: One Prime Blueprint, Four Questions

Let:

N = 2⁴ × 3³ × 5².

1. Is N a perfect square? No, because the exponent 3 is odd.
2. Is N a perfect cube? No, because 4 and 2 are not divisible by 3.
3. What is the smallest multiplier that makes N a perfect square? Multiply by 3.
4. What is the smallest multiplier that makes N a perfect cube? Raise exponents to the next multiples of 3: multiply by 2² × 5 = 20.

The prime factorisation is the same. The requested structure determines the exponent repair.

57. Misconception Clinic: One Is Prime

One has only one positive factor. A prime needs exactly two. Replace the vague rule “prime numbers cannot be divided” with the precise factor-count definition.

58. Misconception Clinic: Every Odd Number Is Prime

Odd composite numbers include 9, 15, 21, 25 and 27. Being odd only means not divisible by 2.

59. Misconception Clinic: The Factor Tree Can Stop at 6 × 10

Six and ten are composite. Prime factorisation is complete only when every final factor is prime.

60. Misconception Clinic: HCF Uses the Largest Exponents

The HCF must fit inside every number, so it is limited by the smaller shared exponents. Use the question: “Can every number supply this many copies of the prime?”

61. Misconception Clinic: LCM Uses Only the Shared Primes

The LCM must be divisible by every source number. It therefore includes primes appearing in any source number, using the largest required exponent.

62. Misconception Clinic: Choose HCF Because the Question Says “Greatest”

A question may ask for the greatest number of groups, but what matters is that the group count divides each total. Use the structural relationship, not one keyword.

63. Misconception Clinic: Choose LCM Because the Question Says “Lowest”

Again, ask what the answer must do. The LCM must be divisible by every source number. The name is not enough to solve the story.

64. Misconception Clinic: √49 = ±7

The symbol √49 denotes the principal square root 7. The equation x² = 49 has two solutions, x = ±7. Keep the root symbol and equation-solving statement separate.

65. Misconception Clinic: Even Number Means Perfect Square

Seventy-two is even but not a perfect square because 72 = 2³ × 3² and the exponent 3 is odd. Square structure is controlled by prime pairs, not parity alone.

66. Misconception Clinic: A Perfect Cube Needs Even Exponents

Perfect cubes require prime exponents divisible by 3. Evenness belongs to squares. Keep the grouping size tied to the power being reversed.

67. Misconception Clinic: The Multiplier and Divisor Must Be the Same

Sometimes they match, but not by rule. Multiplication completes incomplete groups upward. Division removes incomplete groups downward. Analyse the exponents in each direction.

68. Misconception Clinic: A Neat Calculator Answer Proves the Method

A calculator can confirm a wrong input perfectly. Rebuild the prime factors and check divisibility or powers independently.

69. Guided Practice Set A: Factors, Multiples and Primes

1. List all positive factors of 36.
2. Write the first six positive multiples of 7.
3. State whether 37 is prime.
4. Show that 51 is composite.
5. Classify 1, 2, 25 and 29 as prime, composite or neither.

Solutions

Factors of 36: 1,2,3,4,6,9,12,18,36. Multiples of 7: 7,14,21,28,35,42. 37 is prime. 51=3×17, so composite. 1 is neither; 2 and 29 are prime; 25=5² is composite.

70. Guided Practice Set B: Prime Factorisation

1. Write 72 as a product of prime factors.
2. Write 180 as a product of prime factors.
3. Write 294 as a product of prime factors.
4. Write 1000 as a product of prime factors.
5. Rebuild every answer to verify it.

Solutions

72=2³×3². 180=2²×3²×5. 294=2×3×7². 1000=2³×5³.

71. Guided Practice Set C: HCF

1. Find HCF(72,120).
2. Find HCF(84,126).
3. Find HCF(96,144,168).
4. Explain why each HCF cannot exceed the smallest input.

Solutions

24, 42 and 24. Each HCF must divide every source number, including the smallest one.

72. Guided Practice Set D: LCM

1. Find LCM(12,18).
2. Find LCM(24,30).
3. Find LCM(8,12,15).
4. Explain why each LCM cannot be smaller than the largest input.

Solutions

36, 120 and 120. Each LCM must be divisible by every source number, including the largest one.

73. Guided Practice Set E: Squares and Square Roots

1. Find √900 by prime factorisation.
2. Find √1764 by prime factorisation.
3. State whether 540 is a perfect square.
4. Find the smallest multiplier that makes 108 a perfect square.
5. Find the smallest divisor that makes 108 a perfect square.

Solutions

√900=30. √1764=42. 540=2²×3³×5, so it is not a perfect square. Multiply 108 by 3 to obtain 324. Divide 108 by 3 to obtain 36.

74. Guided Practice Set F: Cubes and Cube Roots

1. Find ∛216 by prime factorisation.
2. Find ∛1728 by prime factorisation.
3. State whether 864 is a perfect cube.
4. Find the smallest multiplier that makes 432 a perfect cube.
5. Find the smallest divisor that makes 432 a perfect cube.

Solutions

∛216=6. ∛1728=12. 864=2⁵×3³, so it is not a perfect cube. Multiply by 4 to obtain 1728. Divide by 2 to obtain 216.

75. Guided Practice Set G: Choose HCF or LCM

1. Find the greatest equal length that cuts 84 cm and 126 cm exactly.
2. Find the next time cycles of 15 minutes and 20 minutes coincide.
3. Find the greatest number of identical packs from 96 red and 144 blue counters.
4. Find the smallest number divisible by 8, 12 and 18.

Solutions

HCF(84,126)=42 cm. LCM(15,20)=60 minutes. HCF(96,144)=48 packs. LCM(8,12,18)=72.

76. Guided Practice Set H: Explain the Method

1. Why does HCF use smaller shared exponents?
2. Why does LCM use larger exponents?
3. Why do perfect squares have even exponents?
4. Why do perfect cubes have exponents divisible by 3?
5. Why is √64 equal to 8 rather than ±8?

Model explanations

HCF must fit inside every number, so it is limited by the least available copies of each shared prime. LCM must contain every number, so it needs the greatest required number of copies of every prime. Squaring duplicates all prime factors, giving pairs. Cubing triples all prime factors. The radical symbol denotes the principal non-negative square root; ± appears when solving an equation such as x²=64.

77. Challenge Practice: Unknown Prime Exponents

Let N = 2ᵃ × 3⁴ × 5². N is a perfect square and is divisible by 8. Find the smallest possible positive whole-number value of a.

Worked solution

For N to be a perfect square, a must be even. For N to be divisible by 8=2³, a must be at least 3. The smallest even integer at least 3 is 4.

78. Challenge Practice: Repair a Number Into a Square and a Cube

Let M = 2⁴ × 3⁵. Find the smallest positive whole number by which M must be multiplied so that the result is both a perfect square and a perfect cube.

Worked solution

A number that is both square and cube needs exponents divisible by 6. Raise 4 to 6 by multiplying by 2². Raise 5 to 6 by multiplying by 3. The smallest multiplier is 2²×3=12.

79. Challenge Practice: Three-Cycle Schedule

Three machines require checks every 18, 24 and 30 days. They are checked together today.

1. After how many days will all three next be checked together?
2. How many 18-day cycles occur in that period?
3. How many 24-day cycles occur?
4. How many 30-day cycles occur?

Worked solution

LCM(18,24,30)=360 days. There are 360/18=20, 360/24=15 and 360/30=12 cycles respectively.

80. Challenge Practice: HCF as a Tiling Constraint

A rectangular board measures 168 cm by 252 cm. It is covered by the largest possible identical square tiles without cutting.

1. Find the side length of each square tile.
2. Find the number of tiles used.

Worked solution

Tile side = HCF(168,252)=84 cm. The board uses 168/84=2 tiles by 252/84=3 tiles, so 6 tiles in total.

81. Challenge Practice: Use HCF and LCM Together

For two positive whole numbers, the HCF is 12 and the LCM is 180. One number is 36. Find the other number.

Worked solution

For two positive integers, HCF×LCM equals the product of the numbers. So 12×180=36×n. Hence n=60. Check: HCF(36,60)=12 and LCM(36,60)=180.

82. Examination Method: State the Prime Factorisations First

For HCF, LCM and root questions, write each number in prime-power form before selecting exponents. This separates decomposition from the later decision and makes the working easy to inspect.

83. Examination Method: Label the Exponent Decision

• HCF: shared primes, smaller exponents.
• LCM: all required primes, larger exponents.
• Square root: pair factors, halve even exponents.
• Cube root: group factors in threes, divide exponents by 3.

This prevents HCF and LCM methods from being silently reversed.

84. Examination Method: Predict the Answer’s Size

• HCF cannot exceed the smallest input.
• LCM cannot be smaller than the largest input.
• A square root of a positive number greater than 1 is smaller than the number.
• A cube root of a positive number greater than 1 is smaller than the number.

Use these boundaries before trusting the final result.

85. Examination Method: Write Units in Context Problems

An HCF result can mean groups, centimetres or tile side length. An LCM result can mean minutes, days or a common denominator. The number alone may not answer the question.

86. Examination Method: Answer the Second Part of the Story

If the HCF gives the number of packs, the question may also ask what each pack contains. Divide every original total by the number of packs.

If the HCF gives a tile side length, the question may ask for the number of tiles. Divide each dimension by the side length, then multiply the tile counts.

87. Examination Method: Check by Rebuilding

• Prime factorisation: multiply back to the original number.
• HCF: divide every input by the result.
• LCM: divide the result by every input.
• Square root: square the root.
• Cube root: cube the root.

Checking should change the operation so that the original relationship is tested independently.

88. Examination Method: Keep the Story and Calculation on Separate Lines

Write:

Greatest number of identical groups = HCF(60,84)

before the numerical prime-factor calculation. This makes the modelling decision visible.

89. Oral Classroom Check

1. What is the difference between a factor and a multiple?
2. Why is 1 not prime?
3. Why is 2 the only even prime?
4. When is prime factorisation complete?
5. Why does HCF use smaller exponents?
6. Why does LCM use larger exponents?
7. How do you decide between HCF and LCM in a word problem?
8. Why do perfect squares have even prime exponents?
9. Why do perfect cubes have exponents divisible by 3?
10. What is the difference between √64 and solving x²=64?

The student should answer in complete sentences and use one numerical example. A memorised rule without an explanation is not yet stable.

90. Exit Ticket

Complete without notes.

1. Write 540 as a product of prime factors.
2. Find HCF(90,126).
3. Find LCM(18,24).
4. Find √900 using prime factors.
5. Find ∛1728 using prime factors.
6. Find the smallest multiplier that makes 72 a perfect square.
7. Find the smallest multiplier that makes 108 a perfect cube.
8. Two alarms ring every 18 and 24 minutes. If they ring together now, when will they next ring together?

Exit-ticket solution

540=2²×3³×5. HCF(90,126)=18. LCM(18,24)=72. √900=30. ∛1728=12. Since 72=2³×3², multiply by 2 to obtain 144. Since 108=2²×3³, multiply by 2 to obtain 216=6³. The alarms next ring together after 72 minutes.

91. Homework: Retrieval, Variation and Transfer

Layer 1 — Retrieval

• Define factor, multiple, prime, composite, HCF and LCM.
• Write the first ten primes from memory.
• Explain the smaller-exponent and larger-exponent rules.
• State the exponent conditions for perfect squares and cubes.
• Write one independent check for each major method.

Layer 2 — Variation

• Four prime-factorisation questions.
• Two HCF calculations and one HCF word problem.
• Two LCM calculations and one cycle problem.
• Two square-root questions using prime factorisation.
• Two cube-root questions using prime factorisation.
• One smallest-multiplier and one smallest-divisor question.

Layer 3 — Transfer

Create two different word problems using the same pair of numbers. One must require HCF and the other LCM. Solve both and explain exactly what the answer must divide or be divisible by.

92. The Seven-Day Return Cycle

1. Day 0: complete the classroom models and guided practice.
2. Day 1: prime factorise four numbers and explain HCF versus LCM aloud.
3. Day 3: solve one grouping problem, one cycle problem and one root problem without notes.
4. Day 7: complete the exit ticket again with changed numbers.

If retrieval fails, return to the first missing step rather than rereading the whole page.

93. A 60-Minute Teaching Lesson

1. 5 minutes: factors, multiples and prime retrieval.
2. 10 minutes: factor-tree and prime-power modelling.
3. 10 minutes: HCF through shared smaller exponents.
4. 10 minutes: LCM through all larger exponents.
5. 10 minutes: square and cube exponent grouping.
6. 10 minutes: one grouping problem and one cycle problem.
7. 5 minutes: exit ticket and return date.

94. A 90-Minute Teaching Lesson

1. 10 minutes: Primary-to-Secondary factor diagnostic.
2. 15 minutes: primes, composite numbers and prime factorisation.
3. 15 minutes: HCF modelling and guided practice.
4. 15 minutes: LCM modelling and guided practice.
5. 15 minutes: squares, cubes and roots.
6. 10 minutes: mixed word-problem selection.
7. 5 minutes: oral explanation.
8. 5 minutes: exit ticket and homework assignment.

95. The Full Chapter Routine

For prime factorisation:

split → continue until prime → collect equal primes → write indices → rebuild.

For HCF:

prime factorise → keep shared primes → choose smaller exponents → multiply → verify division.

For LCM:

prime factorise → include every required prime → choose larger exponents → multiply → verify divisibility.

For square roots:

prime factorise → pair equal primes → take one from every pair → multiply → square to check.

For cube roots:

prime factorise → group equal primes in threes → take one from every triple → multiply → cube to check.

96. Why This Chapter Matters Beyond Chapter 1

Prime structure keeps returning. HCF simplifies fractions. LCM builds common denominators. Prime powers support index laws and standard form. Square roots appear in Pythagoras, coordinates and vectors. Cube roots appear in volume and scale. Factorisation becomes one of the central operations of algebra.

The deeper habit is this: when a mathematical object looks complicated, decompose it into simpler invariant parts, perform the required relationship on those parts, then rebuild and verify.

97. Frequently Asked Questions

Is 1 a prime number?

No. It has only one positive factor. A prime has exactly two.

Can the HCF be 1?

Yes. Numbers such as 8 and 15 share no prime factor, so their HCF is 1. They are coprime.

Can the LCM equal one of the numbers?

Yes. If one number is already a multiple of the other, such as 6 and 18, the LCM is 18.

Why is zero not used as the LCM?

Although zero is a multiple of every positive whole number, LCM is defined as the lowest positive common multiple.

What is the fastest HCF check?

Verify that the answer divides every source number exactly and does not exceed the smallest input.

What is the fastest LCM check?

Verify that every source number divides the answer exactly and that the answer is not below the largest input.

98. Ready for the Next Number Classroom?

You are ready to move on when you can do all of the following without prompts:

• distinguish factors from multiples;
• classify primes and composite numbers accurately;
• write prime factorisations in index form;
• rebuild a number to verify its factorisation;
• find HCF using shared smaller exponents;
• find LCM using all larger exponents;
• choose HCF or LCM from what the answer must do;
• recognise perfect squares from even prime exponents;
• recognise perfect cubes from exponents divisible by 3;
• find square and cube roots using prime factorisation;
• complete or remove prime groups to create squares and cubes;
• explain every method rather than only perform it.

The next classroom extends the number system beyond positive whole numbers into negative numbers, integers, rational numbers, real numbers, four operations, number lines, inequalities, approximation and estimation.

Continue the Secondary 1 Mathematics Learning Route

• Secondary 1 Mathematics Learning Guide | Prime Factorisation, HCF, LCM, Squares, Cubes and Roots
• Secondary 1 Mathematics Learning Guide | Numbers, Number Lines, Approximation and Estimation
• Secondary 1 Mathematics Learning Guide | Directed Numbers, Rational Numbers and Four Operations
• Secondary 1 Mathematics Learning Guide | Indices, Powers and Standard Form
• Primary 6 to Secondary 1 Mathematics Handover
• Secondary Mathematics Sengkang | S1–S4 Capability Map