Did you know? The integer 360 has exactly 24 positive divisors, and you can prove that total without testing every integer from one to 360. Its prime factorisation 2³×3²×5 tells you that a divisor can use zero to three factors of two, zero to two factors of three, and zero or one factor of five. Multiply the numbers of choices—four, three and two—and the answer is 24.
For families exploring a JC2 H3 Mathematics tutor in Bukit Timah, this opens a deeper topic: prime factorisation, uniqueness, divisor counting and constructive number-theory proofs. The 2027 Singapore–Cambridge H3 Mathematics (9820) syllabus names modular arithmetic, proof by construction, mathematical induction and non-routine reasoning. Detailed divisor functions are valuable enrichment applications, not a separately listed H3 additional-content chapter.
Quick answer: Every integer n greater than one can be written uniquely as a product of positive prime powers, apart from the order in which the primes are listed. If n=p₁ᵃ¹p₂ᵃ²⋯pᵣᵃʳ, then its number of positive divisors is d(n)=(a₁+1)(a₂+1)⋯(aᵣ+1). Proving uniqueness requires more than observing a pattern; a prime dividing a product must divide one of its factors, a result called Euclid’s lemma.
Quick navigation
- The fundamental theorem
- Why existence is easier than uniqueness
- Euclid’s lemma and its proof
- A complete uniqueness argument
- Count the divisors of 360
- Sum divisors and count special divisors
- GCD and LCM by prime exponents
- Perfect powers and missing factors
- Infinitely many primes
- How prime factorisation informs problem solving
- Worked example clinics
- Common traps
- FAQs and official references
What the fundamental theorem of arithmetic states
A positive integer larger than one is either prime or composite. If it is prime, its prime factorisation consists of itself. If composite, it can be split into smaller positive integer factors, and those factors can in turn be broken down until only primes remain.
The fundamental theorem of arithmetic says that this process always terminates and produces the same collection of prime factors, regardless of the order of factorisation. For example 60=2×30=2×2×15=2²×3×5, while starting with 60=3×20 ends with the same prime powers.
The integer one is not prime. Its customary factorisation is an empty product, and it has exactly one positive divisor. Include this boundary explicitly when applying divisor formulas, since ‘every n greater than one’ is not the same as ‘every positive integer’.
Negative integers can be factored as a sign −1 times the prime factorisation of their absolute value. The uniqueness theorem is ordinarily stated for positive integers above one so that signs do not complicate the basic claim.
Why does uniqueness matter? Without it, the exponent choices used to count divisors might depend on which factorisation happened to be chosen. The theorem ensures that every positive divisor can be described by one well-defined vector of prime exponents.
The 2027 H3 syllabus assumes H2 Mathematics content and values exact statements, necessity, sufficiency, induction and proof. Prime factorisation is therefore a useful testing ground for mathematical rigour even when the detailed divisor-count formula is presented as enrichment.
Proof of existence by a smallest counterexample
To prove existence, suppose there is an integer n≥2 which cannot be written as a finite product of primes. Choose the smallest such integer m. The choice is legitimate because every nonempty set of positive integers has a least member.
The number m cannot be prime because a prime is a product of one prime. Therefore m is composite: write m=ab with integers a,b satisfying 1<a<m and 1<b<m.
Because m is the smallest counterexample, both a and b have prime factorizations. Multiplying those factorizations gives a prime factorisation of m=ab, contradicting the definition of m.
Thus there is no counterexample, and every integer n≥2 has a prime factorisation. This argument illustrates how well-ordering and strong induction are closely connected: information about smaller eligible integers is used to establish the result for a larger one.
A student should not conflate this existence proof with uniqueness. Showing that some prime factorisation can be obtained does not yet establish that a different factorisation procedure cannot produce different prime factors. That second claim needs a separate structural result.
A direct induction proof can be written in similar language: assume every integer from two to m−1 has a factorisation and show m is either prime or has smaller composite factors. The minimal-counterexample presentation is often shorter.
Euclid’s lemma: the essential uniqueness ingredient
Euclid’s lemma: If p is prime and p divides ab for integers a,b, then p divides a or p divides b. This statement can look obvious, but it fails when the prime condition is removed: six divides 2×3, yet six divides neither two nor three.
Here is a proof using Bézout’s identity. If p divides a, we are finished. Otherwise p does not divide a. Because p is prime, gcd(p,a)=1, so integers u,v exist such that up+va=1.
Multiply by b: upb+vab=b. Both terms on the left are divisible by p: the first contains a factor p and the second is divisible by p because p divides ab. Hence p divides their sum b.
Therefore p divides a or p divides b, as required. Every step is justified: the gcd-one assertion depends on the primality of p, and Bézout gives a concrete integer combination rather than an unsupported cancellation.
Repeated application gives a useful extension: if a prime p divides a product a₁a₂⋯aₖ, then it divides at least one factor. Prove this by applying the two-factor result repeatedly or by induction on k.
This is the bridge between divisibility and uniqueness. If two products of primes represent the same number, choose a prime on one side. Because it divides the other product, it must match one prime factor there; prime divisibility of a prime forces equality.
The JC2 H3 Bézout and Diophantine guide develops the integer combination used here. This article applies that result to uniqueness and divisor structures, a different mathematical question.
How the uniqueness proof actually finishes
Suppose a positive integer n≥2 admits two prime factorizations: p₁p₂⋯pᵣ=q₁q₂⋯qₛ. Each p and q is prime; repeated primes are permitted. We must show the lists have the same entries after rearrangement.
The prime p₁ divides the right-hand product. By Euclid’s lemma, it divides some qⱼ. Because qⱼ is itself prime and p₁ is positive prime, p₁=qⱼ.
Rearrange the q-list so this common factor is first, then divide both products by that positive nonzero prime. We get p₂⋯pᵣ equal to the remaining q-product. Repeat the argument on the smaller products.
Eventually both sides reduce to one, with the same number of factors and the same primes counted with multiplicity. If one side had primes left when the other became one, its product would exceed one, a contradiction.
Thus the prime factorisation is unique apart from ordering. This proof does not assume that the factor lists were ordered in advance; the rearrangements are legitimate because multiplication is commutative.
The rigorous result is useful even in elementary arithmetic. The statement ‘if n is a perfect square, each prime exponent is even’ relies on uniqueness: an integer square doubles every exponent, and conversely even exponents can be halved to construct an integer square root.
Count positive divisors of 360 without listing them
Factor 360=2³×3²×5. Any positive divisor d must have the form d=2ᵃ×3ᵇ×5ᶜ, where 0≤a≤3, 0≤b≤2 and 0≤c≤1, all exponents integers.
Why those bounds? A divisor cannot use more of any prime than the original number contains. If a divisor included a new prime or an exponent too large, the quotient 360/d could not remain a positive integer.
Conversely, every exponent triple within those bounds produces a positive divisor of 360. The fundamental theorem guarantees that different triples produce different divisors and that every divisor corresponds to one triple.
There are four choices for a, three for b and two for c. By the multiplication principle, d(360)=4×3×2=24. This is a bijection between divisors and exponent triples.
To see a small case, 72=2³×3² has d(72)=(3+1)(2+1)=12 divisors. They are 1,2,3,4,6,8,9,12,18,24,36,72. Listing them verifies the count but the exponent proof works for any positive integer.
The same method extends directly to n=p₁ᵃ¹⋯pᵣᵃʳ with distinct primes and positive exponents: the number of positive divisors is ∏(aᵢ+1). For n=1, the empty product convention gives one divisor as desired.
If a student writes the formula and stops, ask why each exponent may be chosen independently and why the resulting selections cannot duplicate a divisor. Those two explanations turn a counting trick into a proof.
Sum of divisors and count of square divisors
The exponent method can count or sum different types of divisors. For n=360, the sum of all positive divisors is the product of finite geometric sums: (1+2+4+8)(1+3+9)(1+5).
Each term of the expanded product selects one power of each prime, corresponding to one positive divisor. Multiplying gives 15×13×6=1170. No divisor is omitted or counted twice because of uniqueness of factorisation.
In general, the sum-of-positive-divisors function is σ(n)=∏[1+pᵢ+⋯+pᵢᵃⁱ]. When pᵢ>1, the finite geometric sum can also be written (pᵢ^(aᵢ+1)−1)/(pᵢ−1).
What about square divisors of 360? A positive square divisor must use even exponents: for prime two, choices zero or two; for three, zero or two; for five, only zero. There are 2×2×1=4 square divisors: 1,4,9,36.
Cube divisors require exponents divisible by three. In 360=2³×3²×5, the two-exponent choices for prime two are zero and three, while the others must have exponent zero. The cube divisors are 1 and 8.
These cases illustrate a reusable principle: a condition on a divisor becomes a condition on each prime exponent. It is often easier to count allowed exponent vectors than to inspect the divisors themselves.
Another identity concerns the product of all positive divisors. Each divisor d pairs with n/d to give product n, so the product of all divisors is n^(d(n)/2). For a perfect square, the square-root divisor pairs with itself, and the formula still holds.
A detailed proof should handle that square-root middle case explicitly instead of assuming the number of distinct pairs is always d(n)/2 as an integer.
GCD and LCM are minimum and maximum exponents
Write 72=2³×3² and 180=2²×3²×5. A common divisor may use each prime only up to the smaller exponent appearing in both numbers, so gcd(72,180)=2²×3²=36.
A common multiple must include at least the larger exponent of each prime, giving lcm(72,180)=2³×3²×5=360. The min/max rule follows directly from the unique exponent representation.
These formulas also prove gcd(a,b)×lcm(a,b)=ab for positive integers a,b. For every prime, the smaller exponent plus the larger exponent equals the sum of the two original exponents.
Check numerically: 72×180=12960 and 36×360=12960. The numerical verification is useful but the exponent argument proves the identity universally.
For 84=2²×3×7 and 90=2×3²×5, the gcd is 2×3=6 while the lcm is 2²×3²×5×7=1260. Again 84×90=6×1260.
This perspective explains why the gcd is not the product of only the primes shared without regard to multiplicity. Prime exponents matter; 72 and 180 share two copies of three, not just one.
Prime exponents also help when an H3 problem supplies an unknown gcd or lcm and asks for possible numbers. Turning the numerical condition into exponent inequalities may reveal a finite set of cases.
Perfect powers and the missing-factor method
A positive integer is a perfect square precisely when every exponent in its prime factorisation is even. It is a perfect cube precisely when every exponent is a multiple of three. More generally, a positive integer is a kth power when every exponent is divisible by k.
Consider 360=2³×3²×5. To make it a square by multiplication with the smallest positive integer, we need one extra factor of two and one extra factor of five. Multiply by 10: 360×10=3600=60².
To make 360 a cube by multiplication with the smallest positive integer, we must change exponents three, two and one to multiples of three. The missing factors are 3¹×5²=75. Thus 360×75=27000=30³.
Why is the multiplier minimal? Every prime’s exponent must reach the next multiple of the required power, and adding any fewer copies would leave at least one exponent incompatible. The product of exactly the missing prime factors is therefore smallest.
Another useful reverse question asks for the largest square divisor of 360. Keep the largest even exponent not exceeding each original exponent: 2²×3²=36. This differs from making 360 itself a square.
The distinction between multiplication, division and divisor selection affects which exponents must increase or decrease. A well-written response identifies the operation before choosing the exponents.
This reasoning builds the same habit as the H3 syllabus’s proof by construction: identify a valid candidate, verify that it meets the requirements, and justify why no smaller or larger permitted candidate could improve the answer.
Euclid’s proof that there are infinitely many primes
Suppose, for contradiction, there are only finitely many primes p₁,p₂,…,pₙ. Form N=p₁p₂⋯pₙ+1. This positive integer is greater than one, so by existence of prime factorisation it has some prime divisor q.
If q were one of the listed primes, then it would divide the product p₁p₂⋯pₙ and also N. It would therefore divide their difference N−p₁p₂⋯pₙ=1. No prime divides one. This contradiction shows that q is not on the supposed complete list.
Hence no finite list contains all primes. There are infinitely many. Notice that we proved N has a prime factor outside the list; we did not prove N itself is prime.
This distinction matters. A product of listed primes plus one can be composite. A correct proof uses the existence of some new prime factor, not a false claim that every such constructed number is prime.
The argument is an elegant combination of contradiction, divisor structure and existence. It does not require estimating how many primes lie below a given bound, and it avoids unsupported extrapolation from a large table of primes.
A tutor can ask students to identify the exact contradiction and the significance of the +1. The increment makes N leave remainder one when divided by every listed prime, so none can divide it.
From calculations to full mathematical arguments
Before tackling a counting question, factor the positive integer completely and verify that every base is prime. An unfinished expression such as 12×15 is not yet a prime factorisation; the divisor formula requires exponents of distinct primes.
Next translate the question into restrictions on prime exponents: arbitrary divisor, square divisor, cube divisor, divisor divisible by a fixed number, or greatest common divisor. Each restriction changes the allowed exponent choices.
For a proof of a universal statement, use uniqueness or Euclid’s lemma at the right point. Checking that d(360)=24 does not prove the general product formula. The bijection between divisors and exponent vectors does.
Keep ‘divisors of n’, ‘prime divisors of n’, ‘sum of divisors’ and ‘number of divisors’ separate. An answer of 24 for 360 counts all positive divisors; it does not mean 360 has 24 distinct prime factors. It has only three distinct prime factors: 2,3,5.
Do not infer that a number is prime simply because no small divisor has been noticed. To establish primality of a positive integer n>1 by trial division, it suffices to rule out prime divisors at most √n; every composite n has a factor not exceeding √n.
At the JC2 level, precision matters more than unnecessarily huge arithmetic. A small exact factorisation can support a long structural proof if each inference and each allowed exponent choice is explained.
The immutable eduKateSG small-group Mathematics reference near Sixth Avenue MRT describes three-learner Secondary 1 tuition, not a confirmed JC2 H3 class. Contact eduKateSG for current advanced Mathematics teaching arrangements.
Original divisor-counting and proof clinics
1. The boundary value one
Worked result: The integer one has exactly one positive divisor, itself. It is not prime and must not be represented as a positive prime power.
Why the method works: The empty product convention lets the divisor count formula give one, but the theorem’s ordinary statement begins at n≥2.
2. Factor twelve completely
Worked result: 12=2²×3, so the positive divisor count is (2+1)(1+1)=6. The divisors are 1,2,3,4,6,12.
Why the method works: Listing verifies the count, while the exponent-choice argument proves it without relying on enumeration.
3. Factor thirty-six
Worked result: 36=2²×3², so d(36)=3×3=9. Its square root six appears as a middle divisor when pairing d with 36/d.
Why the method works: A square number can have an odd number of positive divisors because one divisor pairs with itself.
4. Count the divisors of seventy-two
Worked result: 72=2³×3², giving 4×3=12 positive divisors. The choices for the prime exponents are independent within the permitted ranges.
Why the method works: If two exponent vectors produced the same divisor, uniqueness of prime factorisation would be violated.
5. Count the divisors of eighty-four
Worked result: 84=2²×3×7, hence (2+1)(1+1)(1+1)=12 positive divisors.
Why the method works: The number of different prime factors and their multiplicities both influence the divisor count.
6. Count the divisors of ninety
Worked result: 90=2×3²×5, giving 2×3×2=12. This differs from its three distinct prime factors.
Why the method works: Do not confuse d(n) with the number of distinct prime divisors.
7. Count the divisors of one hundred twenty
Worked result: 120=2³×3×5, so d(120)=4×2×2=16.
Why the method works: Checking the factorisation first is safer than applying the exponent product to an unfactored composite base.
8. Count the divisors of one hundred forty-four
Worked result: 144=2⁴×3². Thus d(144)=5×3=15, which is odd because the number is a perfect square.
Why the method works: Every divisor other than the square root pairs with a distinct complementary divisor.
9. Count the divisors of one thousand
Worked result: 1000=2³×5³, so d(1000)=4×4=16.
Why the method works: A large decimal appearance need not make factor counting difficult once prime powers are identified.
10. The sum of divisors of twenty-eight
Worked result: The positive divisors are 1,2,4,7,14,28 and sum to 56. Their proper-divisor sum is 28.
Why the method works: This makes 28 a perfect number, a property of its divisor sum, not of its divisor count.
11. The sum of divisors of thirty-six
Worked result: σ(36)=(1+2+4)(1+3+9)=7×13=91.
Why the method works: Expanding two geometric factors selects every pair of allowable prime powers exactly once.
12. The sum of divisors of seventy-two
Worked result: σ(72)=(1+2+4+8)(1+3+9)=15×13=195.
Why the method works: A sum-of-divisors proof uses independent prime-exponent choices, just like divisor counting.
13. The sum of divisors of one thousand
Worked result: σ(1000)=(1+2+4+8)(1+5+25+125)=15×156=2340.
Why the method works: Adding only distinct prime factors would omit most divisors and yield the wrong sum.
14. GCD of seventy-two and one hundred eighty
Worked result: 72=2³3² and 180=2²3²5. Take the smaller shared exponents: gcd=2²3²=36.
Why the method works: The gcd must divide both numbers, so every prime exponent is limited by the smaller supply.
15. LCM of the same pair
Worked result: The lcm of 72 and 180 is 2³3²5=360, using each prime at its larger exponent.
Why the method works: A common multiple must have enough copies of each prime to be divisible by both inputs.
16. The product relation
Worked result: 72×180=12960, while gcd×lcm=36×360=12960.
Why the method works: The identity holds generally because min(e,f)+max(e,f)=e+f for each prime exponent.
17. Another gcd and lcm pair
Worked result: 84=2²37 and 90=2·3²·5. Their gcd is 6 and lcm is 1260.
Why the method works: Distinct-prime and repeated-prime information must be combined in the same exponent comparison.
18. Proving 97 prime
Worked result: √97 is below ten. The primes at most √97 are 2,3,5,7, none dividing 97, so 97 is prime.
Why the method works: A composite positive integer has a factor no greater than its square root, justifying the finite test.
19. Disproving primality of 91
Worked result: 91=7×13, so it is composite. The single displayed factorisation is a complete disproof of primality.
Why the method works: One nontrivial positive divisor suffices; there is no need to list every possible factor.
20. A less obvious composite
Worked result: 221=13×17, so it is composite despite having no divisor among 2,3,5,7 or 11.
Why the method works: Trial division should continue through prime candidates at most the square root when asserting primality.
21. Make 360 a square
Worked result: 360=2³3²5. Multiply by two and five to make all exponents even: 360×10=3600=60².
Why the method works: Each missing prime factor is forced, so ten is the smallest positive multiplier producing a square.
22. Make 360 a cube
Worked result: Prime exponents 3,2,1 must become multiples of three. Multiply by 3×5²=75: 360×75=27000=30³.
Why the method works: The least multiplier is obtained by filling each exponent up to its next multiple of three.
23. Largest square divisor of 360
Worked result: Keep the largest even exponent at most each original exponent: 2²3²=36.
Why the method works: This is a divisor-selection question, not the same as multiplying 360 to make it a square.
24. Largest cube divisor of 360
Worked result: Keep the largest exponent divisible by three: 2³=8.
Why the method works: The prime three and five exponents are too small for nonzero multiples of three.
25. Count square divisors of 360
Worked result: The allowable exponents are 2 choices for prime two, 2 for prime three, and 1 for five: 4 square divisors.
Why the method works: They are 1,4,9,36, all squares and all divisors of 360.
26. Count divisors of 360 divisible by six
Worked result: A divisor must use at least one two and at least one three. There are 3×2×2=12 possibilities.
Why the method works: Lower exponent bounds translate divisibility restrictions into independent choices.
27. Count divisors of 360 divisible by ten
Worked result: At least one two and one five are needed. Choose among 3 exponents for two, 3 for three and one for five: nine divisors.
Why the method works: A factor condition can be handled without enumerating every permitted divisor.
28. Count divisors of 360 coprime to six
Worked result: Exclude all powers of two and three. Only 1 and 5 remain, so there are two such divisors.
Why the method works: Coprimality corresponds to zero exponents for primes dividing the reference number.
29. Count distinct prime factors of 360
Worked result: The distinct primes are 2,3,5, giving three. Counting multiplicity instead gives 3+2+1=6 factors.
Why the method works: The question must specify whether repeated occurrences of a prime count separately.
30. Odd divisor count criterion
Worked result: A positive integer has an odd number of positive divisors if and only if it is a perfect square.
Why the method works: Complementary divisors pair except for a square root; the prime-exponent proof also gives the same result.
31. A prime divisor count
Worked result: If d(n)=5 for n>1, then n=p⁴ for some prime p.
Why the method works: The product of positive integers aᵢ+1 equals prime five only when there is one factor equal five.
32. The smallest positive integer with twelve divisors
Worked result: The smallest is 60=2²×3×5, with count 3×2×2=12.
Why the method works: Comparing exponent patterns and assigning larger exponents to smaller primes proves the minimal value.
33. A fake Euclid-lemma generalisation
Worked result: Six divides 2×3, but six divides neither two nor three.
Why the method works: Euclid’s product-divisibility conclusion requires the divisor itself to be prime.
34. Coprime factors
Worked result: The gcd of eight and nine is one, so their prime factor sets are disjoint.
Why the method works: Coprimality is about shared prime factors, not about both numbers being prime.
35. The product of divisor pairs
Worked result: For any positive integer n, each divisor d pairs with n/d to give product n.
Why the method works: This supports the product-of-divisors formula n^(d(n)/2), including the central square-root case.
36. An infinite-primes argument
Worked result: If p₁,…,pₖ were all primes, the number p₁⋯pₖ+1 has a prime divisor outside the list.
Why the method works: The constructed number itself need not be prime; existence of one new prime factor yields the contradiction.
How to turn this into a JC2 revision programme
Session one begins with prime decomposition, the meaning of a prime and the special role of the number one. Ask the student to factor twelve, thirty-six, seventy-two and 360 and explain what makes a factorisation complete.
Session two proves existence of prime factorisation by a least counterexample. Students must show that the least failing number is composite, that both proper factors are smaller, and that combining their factorizations creates the contradiction.
Session three develops Euclid’s lemma using a Bézout identity. The tutor should check each cancellation and the precise point where the primality of p is used. A composite counterexample reinforces the significance of the condition.
Session four establishes uniqueness. Ask students to match one prime between two proposed products, cancel it and repeat. Then translate divisors into exponent vectors and derive the product formula for d(n).
Session five uses prime-exponent restrictions to count square divisors, cube divisors and divisors divisible by a specified factor. The student should identify which exponent choices are forced before multiplying their counts.
Session six applies the same framework to sums of divisors, gcd and lcm, including a proof that gcd(a,b)lcm(a,b)=ab. Compare the numerical calculation with the universal min/max exponent argument.
Session seven explores perfect powers and the infinitude of primes. Ask why the least multiplier is necessary and why Euclid’s product-plus-one construction guarantees a new prime factor without making the whole number prime.
Parents can ask how the student knows no divisor was omitted, why a square number has an odd number of divisors, and whether the proof is about one integer or all integers. Those questions evaluate precision without requiring complicated arithmetic.
The progress record should separate errors in prime factorisation, exponent counting, unsupported uniqueness, arithmetic and interpretation. A focused tutorial corrects the first invalid mathematical step rather than relying on another long list of answers.
Frequently asked questions
Is prime factorisation a separately named H3 9820 additional topic?
The 2027 syllabus explicitly names modular arithmetic, proof, induction and combinatorial reasoning. The detailed divisor formulas here are connected enrichment applications, not a separate prescribed additional-content section.
Why is factorisation unique?
Euclid’s lemma says a prime dividing a product must divide one factor. Matching and cancelling prime factors repeatedly shows two prime products of the same number contain the same primes with the same multiplicities.
How many positive divisors does n have?
If n is written as a product of distinct prime powers with exponents a₁,…,aᵣ, the count is ∏(aᵢ+1). Each divisor independently chooses an exponent from zero through the corresponding maximum.
Why do perfect squares have an odd number of divisors?
Divisors pair as d and n/d. Only when n is a perfect square is one positive divisor √n paired with itself, producing an odd count.
What is the difference between the divisor count and sum?
d(n) counts distinct positive divisors. σ(n) adds their numerical values. The two functions have different product formulas based on the same prime-power choices.
How do GCD and LCM use prime exponents?
GCD uses the smaller exponent for each prime, while LCM uses the larger. Adding the chosen exponents proves the identity gcd×lcm=product for positive integers.
Does the product of all listed primes plus one have to be prime?
No. The crucial fact is that none of the listed primes divides it. Its prime factorisation must therefore contain an unlisted prime.
How do I make a number into a perfect square?
Factor it and multiply by the smallest product that makes every prime exponent even. For 360, multiply by 10 to obtain 3600=60².
Can a calculator prove a universal divisor formula?
Numerical checks can catch mistakes but cannot prove every case. The exponent-choice bijection and uniqueness theorem supply the general proof.
How should tutoring focus on proof instead of memorisation?
Ask why each divisor corresponds to exactly one exponent vector and why all allowed vectors produce divisors. This tests the structure underlying the calculation.
Official syllabus and companion guides
The official 2027 H3 Mathematics (9820) syllabus includes mathematical proof, induction, bijection, modular arithmetic and non-routine application of definitions and results. The prime factorisation and divisor questions here are original examples that strengthen those prescribed competencies.
For related reading, see Bézout’s identity and integer solutions, least-counterexample proofs and modular arithmetic. The lasting skill is to turn an exact representation into a justified counting or existence argument.