Did you know that multiplying180 by just five turns it into a perfect square? The prime factors of180 are2²×3²×5. Every prime factor in a whole-number square needs an even exponent, so one more five completes the final pair. The result900 is exactly30². This is one reason prime factorisation is so useful beyond a straightforward factor tree.
For Secondary 2 G2 Mathematics learners around Bukit Timah, prime factors connect several earlier number skills: finding HCF for the greatest equal pieces, finding LCM for repeating schedules, and recognising exact squares and cubes. A student who knows how to multiply numbers but does not know which prime powers to keep can get a plausible wrong answer. Understanding the role of the smallest and largest exponents is the main repair.
This guide starts with definitions and a quick rule table, then provides worked problems, correction diagnostics, independent retrieval and a four-week progression. Prime factorisation is a lower-secondary foundation; the exact focus of the student’s current Secondary 2 G2 class should be checked rather than treating this entire revision guide as a compulsory new school chapter.
Quick answers: what to choose for each question
| Problem | Rule and check |
|---|---|
| Express a number as prime factors | Split until every factor is prime; multiply back. |
| HCF of two or more numbers | Use shared primes at the minimum exponent. |
| LCM of two or more numbers | Use all primes appearing, at the maximum exponent. |
| Two-number verification | HCF(a,b)×LCM(a,b)=ab for positive integers. |
| Test for a perfect square | Every prime exponent is even. |
| Test for a perfect cube | Every prime exponent is divisible by three. |
| Smallest multiplier for a square | Complete each odd exponent to its next even exponent. |
| Smallest multiplier for a cube | Complete each exponent to its next multiple of three. |
What Singapore G2 Mathematics requires—and what is revision
The official G2 Mathematics syllabus covers prime numbers, HCF and LCM, squares and cubes, and square and cube roots by prime factorisation. These skills are useful for a Secondary 2 learner even when originally encountered earlier, because factor and exponent knowledge underpins algebra, proportional reasoning and exact calculations. Specific school-year sequencing varies, and the 2027 SEC G2 syllabus is an upper-secondary examination reference, not a claim about a particular 2026 Secondary 2 weighted assessment.
Authoritative context: SEAB’s G2 Mathematics syllabus listings and a school Mathematics portal showing level-specific chapter sequencing.
Twenty-four concepts: from primes to perfect powers
1. Prime numbers are the indivisible building blocks
A prime number is a whole number greater than one with exactly two positive factors, one and itself. Two, three, five, seven and eleven are primes. One is not prime: it has only one positive divisor, and treating it as prime would destroy the uniqueness of a prime factorisation. For a student, this definition matters because every composite positive whole number greater than one can be broken into a product of primes. The final factors, not the shape of the factor tree, are the mathematical answer.
Teaching move: Ask which of 1,2,9,11 are prime and require a positive-factor explanation.
Common trap: Calling one prime would allow arbitrary copies of one to appear as supposed prime factors.
2. A factor divides a quantity with no remainder
A factor of sixty is a positive integer that divides sixty exactly. Examples include one, two, three, four, five, six and so on. A multiple of sixty, by contrast, is sixty times a whole number: 60,120,180 and more. The words are closely related but point in different directions. Before calculating highest common factor or lowest common multiple, the pupil should define what a factor and multiple mean in the problem’s context.
Teaching move: Compare ‘divide 60 into equal pieces’ with ‘repeat an event every 60 seconds’.
Common trap: A greatest common factor is not the same question as the first time two cycles meet.
3. Every composite number can have a factor tree
For sixty, begin with 6×10, then write 2×3 for six and 2×5 for ten. This gives60=2×2×3×5=2²×3×5. Another valid tree begins 3×20 or 4×15 and reaches the same primes. This is the unique prime factorisation of sixty apart from the order in which primes are listed. A tutor can use alternative trees to show why the method is about structure rather than copying one layout.
Teaching move: Draw two different factor trees for sixty and compare their final leaves.
Common trap: Stopping at 6×10 is unfinished because six and ten are not prime.
4. Index notation compresses repeated prime factors
The expression 2³×3² means2×2×2×3×3. Its numerical value is eight times nine, or seventy-two. The exponent counts copies of the base in the product, not a separate multiplier placed beside it. A student needs this fluency because HCF uses smaller common exponents, LCM uses larger exponents and perfect powers require certain exponent patterns. Switching between repeated multiplication and indices should become routine.
Teaching move: Expand2³×3², then reconstruct the compact notation from individual factors.
Common trap: Reading2³ as 2×3 produces six instead of eight.
5. A factor tree should be checked by multiplication
The proposed factorisation72=2³×3² can be checked as8×9=72. If a pupil accidentally records2²×3², their product is36, half the target. Multiplying back catches a missing repeated factor without requiring another factor tree. This is a useful habit across all factorisation work, including algebraic expressions in later chapters. A result that does not reconstruct the original number is not yet reliable.
Teaching move: Have one learner multiply the prime leaves independently of the person drawing the tree.
Common trap: A neat factor tree can conceal a dropped branch or a composite leaf.
6. HCF uses common primes with the smallest powers
The highest common factor of two positive integers is the greatest positive whole number dividing both without remainder. Write60=2²×3×5 and84=2²×3×7. Common factors include2² and3, so HCF=2²×3=12. The prime five and seven are not shared and cannot appear in the HCF. The smaller-power rule works even when one number contains more copies of a shared prime than the other.
Teaching move: Underline only the prime factors that occur in both prime factorisations.
Common trap: Using the larger exponent can produce a number that fails to divide one original.
7. LCM uses every prime required at its largest power
The lowest common multiple is the smallest positive whole number divisible by both target numbers. For sixty and eighty-four, include 2², 3, 5 and7 at the greatest powers needed: LCM=2²×3×5×7=420. Dividing420 by60 yields7 and by84 yields5. Unlike HCF, the LCM must include primes belonging to only one of the numbers, because it needs to contain both originals as factors.
Teaching move: Circle the highest exponent for each distinct prime across both numbers.
Common trap: Keeping only shared primes gives a factor, not a common multiple.
8. A product identity connects HCF and LCM for two numbers
For two positive whole numbers a and b, HCF(a,b)×LCM(a,b)=a×b. With60 and84, 12×420=5040 and60×84=5040. This is a strong numerical verification when both quantities have been calculated. The identity is for two numbers in this form; a naive product formula does not transfer unchanged to arbitrary groups of three or more. Use the result as a check rather than a substitute for understanding shared prime powers.
Teaching move: Ask the learner to calculate both products after the HCF and LCM.
Common trap: Do not apply the simple two-number identity automatically to a three-number list.
9. HCF answers the longest equal-unit cutting question
Suppose ribbons of forty-eight centimetres and seventy-two centimetres must be cut into equal whole-number lengths with no waste. The greatest possible length is HCF(48,72)=24cm. That gives two pieces from the first ribbon and three from the second. The practical language ‘largest identical piece that fits both’ points to factors rather than multiples. The five total pieces are the result of cutting, not the HCF itself.
Teaching move: Ask whether 30cm could divide both ribbons exactly and show why not.
Common trap: An LCM of144cm cannot be the length of a piece cut from a shorter48cm ribbon.
10. LCM answers the first shared repeat time
A fictional alarm repeats every twelve minutes and another every eighteen minutes. If both ring together at the start, they next ring together after LCM(12,18)=36 minutes. The first schedule contains12,24,36; the second contains18,36. The common time is a multiple of each cycle. The word ‘together again’ matters: the initial coincidence at time zero does not answer how long until the next meeting.
Teaching move: Write the first four multiples of twelve and eighteen before using prime factors.
Common trap: The HCF six is not the next simultaneous ringing time.
11. HCF and LCM can be found for three numbers
For twelve, eighteen and twenty, prime decompositions are12=2²×3, 18=2×3² and20=2²×5. The only prime common to all three is one copy of two, so HCF=2. For LCM use the highest powers of every prime:2²×3²×5=180. Both quantities can be checked by division into every original number. This example reinforces that the method generalises even though the two-number product identity does not remain simple.
Teaching move: List each prime’s exponent in three aligned columns.
Common trap: A prime shared by only two of three values must not enter the three-way HCF.
12. Coprime numbers can both be composite
The numbers eight and nine share no prime factor, so HCF(8,9)=1, even though nine is composite and eight is composite. Such numbers are called coprime or relatively prime. Their LCM is their product, seventy-two. Students sometimes assume ‘coprime’ means both numbers are prime; it refers instead to the relation between the two factor sets. An HCF of one is a fully valid result, not evidence that factorisation failed.
Teaching move: Factor8=2³ and9=3² to make the absence of common primes visible.
Common trap: Two composite numbers can be coprime.
13. A perfect square has even prime exponents
A positive whole number is a perfect square if it equals another whole number multiplied by itself. In prime factorisation, every prime exponent must be even, because squaring doubles the exponent of each prime in the root. For example144=2⁴×3²=(2²×3)²=12². By contrast180=2²×3²×5 has one unpaired five, so it is not a perfect square. The exponent test provides a precise rule for large numbers.
Teaching move: Group the prime factors of144 into identical pairs.
Common trap: A number with one unpaired prime factor cannot have a whole-number square root.
14. A perfect cube has exponents divisible by three
A positive whole number is a perfect cube if it equals some whole number raised to the third power. Its prime factors can be grouped into threes. For example216=2³×3³=(2×3)³=6³. The number108=2²×3³ contains only two copies of two and thus is not a perfect cube. Recognising groups of three parallels the square method and can make exact cube-root questions approachable without trial-and-error guessing.
Teaching move: Show three identical factor groups for216.
Common trap: Using even-exponent rules for cubes is incorrect; cube exponents must be multiples of three.
15. Smallest multiplier for a perfect square fills unpaired factors
The prime decomposition180=2²×3²×5 has even exponents for two and three, but one copy of five is unpaired. Multiplying by five yields900=30², so five is the smallest positive integer multiplier making a perfect square. The ‘smallest’ condition follows because every odd prime exponent must be changed to the next even exponent. Adding extra already-paired factors would also produce squares but not the smallest possible multiplier.
Teaching move: Underline only primes with odd exponents, then multiply by one copy of each.
Common trap: Multiplying180 by two would change the even exponent of two into an odd one and still fail the square test.
16. Smallest divisor for a square can remove unpaired factors
For180=2²×3²×5, dividing by five produces36=6², a perfect square. This is the smallest positive integer divisor that removes every unpaired factor while keeping a whole-number quotient. The question’s verb matters: multiplying and dividing both use parity, but the arithmetic results differ. For72=2³×3², dividing by two leaves36 whereas multiplying by two gives144; both are perfect squares.
Teaching move: Write the prime exponent after division before taking the square root.
Common trap: Do not confuse the smallest multiplier with the smallest divisor simply because their values sometimes match.
17. Smallest multiplier for a perfect cube fills groups of three
Factor108=2²×3³. To make a perfect cube by multiplication, add one copy of two to complete2³, giving216=6³. For250=2×5³, two additional copies of two are needed, so multiply by4 to obtain1000=10³. The missing exponent is the distance to the next multiple of three. This method works for any positive integer factorisation, not just familiar classroom cubes.
Teaching move: Write each prime exponent modulo three, then fill to the next multiple.
Common trap: Multiplying250 only by two gives500=2²×5³, which still has an incomplete pair of twos for a cube.
18. Smallest divisor for a cube may differ from multiplier
For250=2×5³, dividing by two leaves125=5³, a perfect cube. Multiplying by four instead gives1000=10³. These are two valid ways to make a cube by different actions. A word problem should specify whether the integer is multiplied, divided, or adjusted by addition; the prime-exponent method applies directly to multiplicative changes. It cannot automatically answer ‘smallest amount to add’ without separate reasoning.
Teaching move: Ask the learner to compare the prime exponents of125 and1000.
Common trap: A multiplicative correction is not the same as an additive correction to the original number.
19. Prime factorisation can recover exact roots
If N=2⁶×3⁴, then √N=2³×3²=8×9=72. The square root halves all even exponents. If M=2³×5⁶, then ∛M=2×5²=50. These are direct applications of index laws after confirming the appropriate perfect-power structure. A student who leaves an extra prime outside the root needs to revisit pairing rather than relying on calculator decimals.
Teaching move: Write the root and square or cube the result to reconstruct the original.
Common trap: Halving an odd exponent without managing the unpaired prime is invalid for an integer square root.
20. Factor pairs can provide a simpler HCF check
Before prime methods, HCF(24,36) can be checked from factors. The common factors include1,2,3,4,6,12 and the greatest is12. The prime-factor method gives24=2³×3 and36=2²×3², so HCF=2²×3=12. Both approaches agree. For large numbers the prime method is systematic; for small numbers the factor list gives a fast reasonableness check.
Teaching move: Compare the prime result against an explicit small factor list.
Common trap: Listing a handful of common factors without checking for a larger one can miss the true HCF.
21. A lowest multiple need not be the larger input
For12 and18 the LCM is36, larger than either input. If one number divides the other, however, the LCM can equal the larger input, as with12 and36: LCM=36 and HCF=12. Recognising divisibility relationships can save calculation. The definition remains the same: choose the smallest positive number that both originals divide exactly.
Teaching move: Check cases where one factor list is contained in the other.
Common trap: An LCM need not be strictly larger than both numbers; equality with one is possible.
22. The first wrong exponent is more useful than the final wrong number
A pupil might correctly factor72 but choose the larger exponent for HCF instead of LCM. Another may recognise perfect-square parity but multiply180 by an unnecessary factor. Both get wrong answers for different conceptual reasons. A precise error log names the rule that failed, then checks the reconstructed divisibility or square. Tutoring works best when a learner explains that decision and solves a changed number without being shown the correct power.
Teaching move: Record whether the mistake concerns prime decomposition, exponent choice, or word-problem interpretation.
Common trap: Repeating an already memorised 60-and84 example does not establish independent understanding.
Thirty worked examples with independent checks
Worked example 1: Factor sixty
Question: Express60 as prime factors.
Method: 60=6×10=(2×3)(2×5)=2²×3×5.
Check: 4×3×5=60.
Do not do this: Stop only when all leaves are prime.
Changed practice: Try84.
Worked example 2: Factor eighty-four
Question: Express84 as prime factors.
Method: 84=4×21=2²×3×7.
Check: 4×3×7=84.
Do not do this: Do not call21 a prime.
Changed practice: Try72.
Worked example 3: HCF sixty and eighty-four
Question: Find HCF(60,84).
Method: 60=2²×3×5,84=2²×3×7; common minimum powers give2²×3=12.
Check: 60/12=5 and84/12=7.
Do not do this: Exclude primes5 and7.
Changed practice: Try HCF(72,90).
Worked example 4: LCM sixty and eighty-four
Question: Find LCM(60,84).
Method: Take max powers2²×3×5×7=420.
Check: 420/60=7 and420/84=5.
Do not do this: LCM includes unique primes from either number.
Changed practice: Try LCM(72,90).
Worked example 5: Factor seventy-two
Question: Factor72.
Method: 72=8×9=2³×3².
Check: 8×9=72.
Do not do this: Keep all three copies of two.
Changed practice: Try96.
Worked example 6: Factor ninety
Question: Factor90.
Method: 90=9×10=2×3²×5.
Check: 2×9×5=90.
Do not do this: Ten must be split into2×5.
Changed practice: Try108.
Worked example 7: HCF seventy-two and ninety
Question: Find HCF(72,90).
Method: Common factors2¹ and3² give2×9=18.
Check: 72/18=4,90/18=5.
Do not do this: Take smaller power of two.
Changed practice: Try HCF(48,180).
Worked example 8: LCM seventy-two and ninety
Question: Find LCM(72,90).
Method: Max powers2³×3²×5=8×9×5=360.
Check: 360/72=5 and360/90=4.
Do not do this: The LCM requires five from90.
Changed practice: Use product identity to verify.
Worked example 9: Three-number HCF
Question: Find HCF(12,18,20).
Method: 12=2²×3;18=2×3²;20=2²×5. Only common prime2¹ remains; HCF2.
Check: All three divisible by2, not by3.
Do not do this: Do not include a prime appearing in only two numbers.
Changed practice: Find three-number LCM.
Worked example 10: Three-number LCM
Question: Find LCM(12,18,20).
Method: Max powers2²×3²×5=180.
Check: 180 is divisible by12,18,20.
Do not do this: The two-number HCF×LCM identity is not used here.
Changed practice: Try12,15,18.
Worked example 11: Ribbons longest equal pieces
Question: Ribbons48cm and72cm need equal longest pieces without waste.
Method: HCF(48,72)=24cm; pieces2 and3, total five.
Check: 48/24=2,72/24=3.
Do not do this: Choose HCF rather than LCM.
Changed practice: Try lengths60 and84.
Worked example 12: Two repeat cycles
Question: Signals occur every12min and18min, starting together. When next together?
Method: LCM(12,18)=36min.
Check: Both schedules contain36.
Do not do this: HCF six is not a repeat meeting.
Changed practice: Try intervals8 and12.
Worked example 13: Coprime composites
Question: Find HCF and LCM of8 and9.
Method: 8=2³,9=3²; HCF1, LCM72.
Check: 1×72=8×9.
Do not do this: Coprime does not require both inputs prime.
Changed practice: Try9 and10.
Worked example 14: One divides the other
Question: Find HCF and LCM of12 and36.
Method: HCF12 because12 divides36; LCM36.
Check: 12×36=12×36.
Do not do this: LCM can equal the larger input.
Changed practice: Try8 and40.
Worked example 15: Square root of144
Question: Use prime factors to find √144.
Method: 144=2⁴×3², so √144=2²×3=12.
Check: 12²=144.
Do not do this: Square-root exponent halves even powers.
Changed practice: Try√576.
Worked example 16: Perfect-square check180
Question: Is180 a perfect square?
Method: 180=2²×3²×5. Exponent of five is odd, so no.
Check: 12²=144 and14²=196 surround180.
Do not do this: An integer root requires all even prime exponents.
Changed practice: What about225?
Worked example 17: Multiply180 to square
Question: Find smallest positive integer multiplying180 to make a perfect square.
Method: Multiply by5;180×5=900=30².
Check: 2²×3²×5² has even powers.
Do not do this: Multiplying by2 leaves odd exponents.
Changed practice: Try72.
Worked example 18: Divide180 to square
Question: Find smallest positive integer dividing180 to leave a perfect square.
Method: Divide by5;180/5=36=6².
Check: 2²×3² remains.
Do not do this: This answers a different verb than multiplication.
Changed practice: Try96.
Worked example 19: Multiply72 to square
Question: Find smallest multiplier for72=2³×3².
Method: Multiply by2, giving144=12².
Check: Exponent of two becomes4.
Do not do this: The three exponent needs no change.
Changed practice: Try96.
Worked example 20: Divide72 to square
Question: Find smallest divisor of72 to leave a square.
Method: Divide by2;72/2=36=6².
Check: 2²×3² remains.
Do not do this: Do not divide by three; it makes3 exponent odd.
Changed practice: Try360.
Worked example 21: Square multiplier for360
Question: Find smallest multiplier making360 a square.
Method: 360=2³×3²×5, so multiply by2×5=10. Product3600=60².
Check: Exponents become2⁴×3²×5².
Do not do this: Missing prime two as well as five matters.
Changed practice: Try180.
Worked example 22: Cube check216
Question: Is216 a perfect cube?
Method: 216=2³×3³=(2×3)³=6³.
Check: 6×6×6=216.
Do not do this: Cube powers require multiples of three.
Changed practice: Try108.
Worked example 23: Cube multiplier108
Question: Smallest multiplier for108 to be a perfect cube.
Method: 108=2²×3³, multiply by2 to get216=6³.
Check: Every exponent now divisible by three.
Do not do this: Adding one copy of two is enough.
Changed practice: Try250.
Worked example 24: Cube multiplier250
Question: Find smallest multiplier for250 to form a perfect cube.
Method: 250=2×5³; multiply by2²=4, producing1000=10³.
Check: 2³×5³ is a cube.
Do not do this: Multiplying by two only gives500, not cube.
Changed practice: Find smallest divisor.
Worked example 25: Cube divisor250
Question: Find smallest divisor of250 leaving a perfect cube.
Method: Divide by2, giving125=5³.
Check: 125 is5×5×5.
Do not do this: The multiplier and divisor answers differ.
Changed practice: Try108.
Worked example 26: Square root via indices
Question: If N=2⁶×3⁴, express √N as a whole number.
Method: Root=2³×3²=8×9=72.
Check: 72²=(2³×3²)²=2⁶×3⁴.
Do not do this: Halve exponents, not bases.
Changed practice: Try2⁴×5².
Worked example 27: Cube root via indices
Question: If M=2³×5⁶, express ∛M.
Method: Cube root=2×5²=50.
Check: 50³=2³×5⁶.
Do not do this: Divide each exponent by3.
Changed practice: Try3⁶×2³.
Worked example 28: Two-number identity check
Question: Verify HCF×LCM for72 and90.
Method: 18×360=6480, and72×90=6480.
Check: Equal products confirm two-number results.
Do not do this: The identity is a check, not a method for three inputs.
Changed practice: Test60 and84.
Worked example 29: Another HCF and LCM
Question: Find HCF and LCM of48 and180.
Method: 48=2⁴×3,180=2²×3²×5. HCF=2²×3=12; LCM=2⁴×3²×5=720.
Check: 12×720=48×180=8640.
Do not do this: Use minimum and maximum exponent separately.
Changed practice: Try36 and48.
Worked example 30: Prime powers in 96
Question: Factor96 and find smallest multiplier to make it square.
Method: 96=2⁵×3; odd exponents for2 and3, so multiply by6:576=24².
Check: 2⁶×3² after correction.
Do not do this: Multiplying by3 alone leaves2⁵ odd.
Changed practice: Find smallest divisor6 leaving16.
Ten mistakes and their first useful corrections
1. Prime factor tree stops at composite leaves
What goes wrong: A tree ends at6×10 and is treated as complete. Expand6 and10 until all factors are prime, then multiply leaves back.
Repair: Compare two different valid trees for60.
Evidence of improvement: The child independently obtains identical final prime factors.
2. HCF uses the largest shared power
What goes wrong: When numbers have2³ and2¹, learner includes2³ in HCF, producing a result that does not divide one input. Use smaller power for HCF.
Repair: Check candidate by dividing both originals.
Evidence of improvement: Changed two-number pair yields a valid common divisor.
3. LCM uses only shared primes
What goes wrong: A student omits unique factors5 or7 from LCM(60,84), generating a number that one original does not divide. Include every distinct prime at its maximum power.
Repair: Divide proposed LCM by both originals to test.
Evidence of improvement: The new common multiple passes both divisibility checks.
4. Ribbon-cutting problem chooses LCM
What goes wrong: A longest-equal-piece question is interpreted as a repeat schedule. Ask if resulting length must fit into both ribbons, then choose HCF.
Repair: Contrast 48/24 with repeating times12 and18.
Evidence of improvement: Problem language determines the method.
5. Perfect square mistaken for even numerical value
What goes wrong: A number is called perfect square because it is divisible by two. Square testing concerns all prime exponents being even, not the number’s parity.
Repair: Compare72 even but nonsquare with81 odd and perfect square.
Evidence of improvement: Pupil checks exponents rather than last digit.
6. Perfect cube uses pairs rather than triples
What goes wrong: The learner pairs primes as for a square, but a cube requires groups of three identical prime factors.
Repair: Compare108 and216 prime exponents.
Evidence of improvement: Changed cube question is corrected through exponent multiples of three.
7. Multiplier and divisor treated as same operation
What goes wrong: For250=2×5³, multiply by4 to make cube but divide by2 to leave cube. The verb changes how unpaired factors are adjusted.
Repair: Calculate both resulting perfect cubes.
Evidence of improvement: Student chooses correct action and reconstructs target.
8. One is included as prime
What goes wrong: Counting one as prime undermines the unique factor decomposition. Revisit definition: exactly two distinct positive divisors.
Repair: List divisors of1,2,9 and11.
Evidence of improvement: One is excluded without a rule-of-thumb guess.
9. Two-number product identity applied to three values
What goes wrong: The equality HCF×LCM=product is tested on three inputs and fails. Restrict the simple formula to two positive integers.
Repair: Use12,18,20 as a counterexample.
Evidence of improvement: The learner uses prime extrema for multi-number cases.
10. Final result not verified
What goes wrong: An HCF or LCM is reported without ensuring it divides or is divisible by all originals. Verify with simple quotient checks and one factor reconstruction.
Repair: Use both definitions for two changed numbers.
Evidence of improvement: The child catches one deliberate wrong HCF unaided.
Forty-five rapid retrieval checks
Practice 1: Prime factors60
Answer: 2²×3×5. Why: 4×3×5.
Practice 2: Prime factors84
Answer: 2²×3×7. Why: 4×3×7.
Practice 3: HCF(60,84)
Answer: 12. Why: 2²×3.
Practice 4: LCM(60,84)
Answer: 420. Why: 2²×3×5×7.
Practice 5: Prime factors72
Answer: 2³×3². Why: 8×9.
Practice 6: Prime factors90
Answer: 2×3²×5. Why: 2×9×5.
Practice 7: HCF(72,90)
Answer: 18. Why: 2×9.
Practice 8: LCM(72,90)
Answer: 360. Why: 8×9×5.
Practice 9: HCF(24,36)
Answer: 12. Why: 4×3.
Practice 10: LCM(24,36)
Answer: 72. Why: Common multiple.
Practice 11: HCF(12,18,20)
Answer: 2. Why: Only common2.
Practice 12: LCM(12,18,20)
Answer: 180. Why: 4×9×5.
Practice 13: HCF(8,9)
Answer: 1. Why: Coprime.
Practice 14: LCM(8,9)
Answer: 72. Why: Product.
Practice 15: HCF(12,36)
Answer: 12. Why: One divides other.
Practice 16: LCM(12,36)
Answer: 36. Why: Larger already multiple.
Practice 17: Longest pieces ribbons48,72
Answer: 24cm. Why: HCF.
Practice 18: Signals12,18 min again
Answer: 36min. Why: LCM.
Practice 19: √144
Answer: 12. Why: 12²=144.
Practice 20: √576
Answer: 24. Why: 24²=576.
Practice 21: Is180 square?
Answer: No. Why: Exponent5 odd.
Practice 22: Smallest multiplier180 square
Answer: 5. Why: 900=30².
Practice 23: Smallest divisor180 square
Answer: 5. Why: 36=6².
Practice 24: Smallest multiplier72 square
Answer: 2. Why: 144=12².
Practice 25: Smallest divisor72 square
Answer: 2. Why: 36=6².
Practice 26: Smallest multiplier360 square
Answer: 10. Why: 3600=60².
Practice 27: Is216 cube?
Answer: Yes. Why: 6³.
Practice 28: Is108 cube?
Answer: No. Why: 2²×3³.
Practice 29: Smallest multiplier108 cube
Answer: 2. Why: 216=6³.
Practice 30: Smallest multiplier250 cube
Answer: 4. Why: 1000=10³.
Practice 31: Smallest divisor250 cube
Answer: 2. Why: 125=5³.
Practice 32: Square root2⁶×3⁴
Answer: 72. Why: 2³×3².
Practice 33: Cube root2³×5⁶
Answer: 50. Why: 2×5².
Practice 34: HCF(48,180)
Answer: 12. Why: min exponents.
Practice 35: LCM(48,180)
Answer: 720. Why: max exponents.
Practice 36: Smallest multiplier96 square
Answer: 6. Why: 576=24².
Practice 37: Smallest divisor96 square
Answer: 6. Why: 16=4².
Practice 38: Product60×84
Answer: 5040. Why: HCF×LCM.
Practice 39: Product72×90
Answer: 6480. Why: 18×360.
Practice 40: Which is prime:1 or2
Answer: 2. Why: Exactly two factors.
Practice 41: Prime factors of1?
Answer: Empty product, not prime. Why: Special unit.
Practice 42: Perfect square exponent condition
Answer: All even. Why: Pairs.
Practice 43: Perfect cube exponent condition
Answer: All multiples of three. Why: Triples.
Practice 44: Does HCF divide both inputs?
Answer: Yes. Why: Definition.
Practice 45: Does LCM divide both inputs?
Answer: No—both inputs divide LCM. Why: Direction matters.
A four-week progression for small-group revision
Week 1: rebuild prime numbers and factor trees
Recognise primes and composites, draw two different factor trees for sixty and eighty-four, and rewrite prime leaves in index notation. Practise multiplying each factorisation back to the original.
Progress measure: Student can decompose unfamiliar 72 or90 without stopping at a composite factor.
Week 2: distinguish HCF from LCM
Compare largest equal ribbon pieces with first shared repeat times, solve two-number problems and extend to three numbers. Use the smaller exponent for HCF and the larger for LCM.
Progress measure: A new word problem is classified by meaning and checked through divisibility.
Week 3: exact square roots and perfect squares
Use index parity to test square numbers and recover square roots. Find minimum multiplying or dividing factors for72,180 and96, showing both prime powers and resulting whole square.
Progress measure: The learner selects and verifies the smallest multiplicative correction.
Week 4: cube powers and mixed number questions
Find exact cube roots, group primes into threes, compare multiplier/divisor questions and solve a mixed set with no formula labels. Return to G2 school work where these number skills are needed.
Progress measure: A changed square or cube challenge is solved and checked without a tutor providing the method.
How a ninety-minute, three-pupil lesson could work
The original eduKateSG Clementi Mathematics reference describes 90-minute small-group tutorials for three pupils. A practical lesson can spend ten minutes on a diagnostic, fifteen minutes rebuilding factor trees, twenty minutes on HCF and LCM, twenty-five minutes on independent changed perfect-power questions, ten minutes on mixed retrieval and ten minutes reviewing the earliest wrong operation. Such a plan is an illustration, not a promise of available lesson places.
One pupil might call one prime, another might choose the larger HCF exponent, and a third might group square factors into threes. These errors have different causes. The tutor should ask each pupil to explain their own factor tree and why a particular exponent belongs in the result, then test learning through a different number rather than copying the previous solution.
Frequently asked questions
1. What is prime factorisation?
Writing a positive integer greater than one as a product of primes, often with index notation. For60 it is2²×3×5. Different valid factor trees reach the same final prime factors.
2. How do I find HCF from prime factors?
Use only primes common to every input, taking the smallest exponent occurring among them. Multiply those prime powers. The result must divide all original integers exactly.
3. How do I find LCM from prime factors?
Use every prime appearing in any input, with its largest exponent. Multiply those powers. The result must be divisible by each original integer.
4. How can I tell whether a word problem needs HCF or LCM?
The greatest identical piece or largest size dividing several quantities often needs HCF. The first shared occurrence of repeating schedules often needs LCM. Always interpret the story rather than rely solely on keywords.
5. Is one a prime number?
No. One has one positive factor, while a prime has exactly two distinct positive factors. Two is the smallest prime.
6. How can I tell if a number is a perfect square?
Factorise it into primes. Every prime exponent must be even for a positive integer square. If some exponents are odd, they identify factors to add or remove through multiplication/division.
7. How can I tell if a number is a perfect cube?
Every prime exponent must be divisible by three. Factor108=2²×3³; multiplying by2 gives216=6³.
8. What is the smallest multiplier making180 a perfect square?
Five. The prime decomposition is2²×3²×5, and one more five makes all powers even, producing900=30².
9. Can HCF and LCM be equal?
Yes, when the two positive numbers are equal. If one number divides the other and they differ, HCF is the smaller and LCM the larger.
10. What if the numbers have no common prime factors?
Their HCF is one, and for two positive numbers their LCM is their product. For example8 and9 are coprime composites.
11. Are prime factors only useful in Secondary 1?
They form important lower-secondary number skills and appear in the broader G2 syllabus. For a Secondary 2 learner this guide is consolidation that supports exact roots, ratio work and other calculations; current school scope determines priorities.
12. How should students check their answers?
Multiply prime factors back to the original, test an HCF by exact divisibility of both numbers, test an LCM by dividing it by both originals, and square or cube a proposed root.
13. Can a calculator replace the explanation?
A calculator helps confirm arithmetic, but it does not explain why the smallest or largest exponent is selected. A factor tree and short verbal reason give transferable understanding.
The takeaway: prime powers are a reusable tool
Did you know? The same list of prime powers can answer three different questions. HCF keeps the common minimum powers, LCM covers the required maximum powers, and perfect squares or cubes test whether exponents form pairs or triples. When the pupil can explain which rule answers the actual question and independently reconstruct the result, prime factorisation has become a method rather than a memorised chart.
Related guides: Perfect Squares and Algebraic Identities, Negative Numbers and Sign Errors and Quadratic Factorisation and Algebraic Fractions.
The original Clementi three-pupil Mathematics tutorial reference remains unchanged. To discuss current lessons, contact eduKate Singapore directly. No grades, vacancies or subject-level changes are guaranteed.
Explore the other three Secondary 2 G2 Mathematics guides in this batch
This four-part collection connects foundational number skills, three-part ratio sharing, algebraic-fraction addition as a later-stage bridge, and trapezium area in composite geometry. Select the guide that matches the student’s actual school work, study a complete worked method, then attempt an independently changed example. The original Clementi tutorial reference remains unchanged.
- Secondary 2 G2 Mathematics Tutor Bukit Timah | Sharing Quantities in Three-Part Ratios and Changing Ratios
- Secondary 2 G2 Mathematics Tutor Bukit Timah | Adding and Subtracting Algebraic Fractions with Unlike Denominators
- Secondary 2 G2 Mathematics Tutor Bukit Timah | Trapezium Area, Perpendicular Height and Composite Regions