Highest Common Factor

The factors of 12 are 1, 2, 3, 4, 6, 12. The factors of 18 are 1, 2, 3, 6, 9, 18. The numbers that appear in both lists — the common factors — are 1, 2, 3, and 6. The highest of these is 6, so HCF(12, 18) = 6.

Alternatively, using prime factorisation: \(12 = 2^2 \times 3\) and \(18 = 2 \times 3^2\). Taking the lowest power of each shared prime gives \(2^1 \times 3^1 = 6\).

The factors of 8 are 1, 2, 4, 8. The factors of 32 are 1, 2, 4, 8, 16, 32. Every factor of 8 is also a factor of 32, so the common factors are 1, 2, 4, 8. The highest is 8.

This makes sense because 32 is a multiple of 8 (since \(32 = 8 \times 4\)). Whenever one number is a multiple of the other, the HCF is always the smaller number — it’s the biggest thing that can divide into both.

The factors of 9 are 1, 3, 9. The factors of 20 are 1, 2, 4, 5, 10, 20. The only number in both lists is 1, so HCF(9, 20) = 1.

Numbers whose only common factor is 1 are called coprime. Using prime factorisation: \(9 = 3^2\) and \(20 = 2^2 \times 5\). They share no prime factors at all, so their HCF must be 1.

Using prime factorisation: \(36 = 2^2 \times 3^2\) and \(48 = 2^4 \times 3\). For each shared prime, take the lower power: \(2^2\) (the lower of \(2^2\) and \(2^4\)) and \(3^1\) (the lower of \(3^2\) and \(3^1\)). So HCF \(= 2^2 \times 3 = 4 \times 3 = 12\).

We can verify: \(36 \div 12 = 3\) and \(48 \div 12 = 4\). Since 3 and 4 are coprime (they share no common factor other than 1), there’s no larger number that divides both — confirming 12 is the highest.

Listing factors: 12 has (1, 2, 3, 4, 6, 12). 18 has (1, 2, 3, 6, 9, 18). 30 has (1, 2, 3, 5, 6, 10, 15, 30). The numbers appearing in all three lists are 1, 2, 3, and 6. The highest of these is 6.

Using prime factorisation: \(12 = 2^2 \times 3\), \(18 = 2 \times 3^2\), and \(30 = 2 \times 3 \times 5\). The primes shared across all three numbers are 2 and 3, meaning the HCF is \(2 \times 3 = 6\).

Example: 6 and 30

Another: 24 and 42

Creative: 6 and 6 — the HCF of a number with itself is that number! Or 54 and 78: since \(54 = 2 \times 3^3\) and \(78 = 2 \times 3 \times 13\), their HCF is \(2 \times 3 = 6\).

Trap: 6 and 9 — a student might think that because 6 is one of the numbers, the HCF must be 6. But \(6 = 2 \times 3\) and \(9 = 3^2\), so HCF(6, 9) = 3, not 6. The HCF is only 6 when 6 divides both numbers.

Example: 8 and 15

Another: 7 and 10

Creative: 16 and 27 — neither is prime, but \(16 = 2^4\) and \(27 = 3^3\) share no prime factors. Or 99 and 100 — any two consecutive whole numbers are always coprime.

Trap: 9 and 15 — a student might think these are coprime because neither is even, but \(9 = 3^2\) and \(15 = 3 \times 5\) both share the factor 3, giving HCF(9, 15) = 3. Being coprime isn’t about being odd — it’s about sharing no prime factors.

Example: 5 and 20

Another: 3 and 12

Creative: 7 and 49 — since \(49 = 7^2\), the HCF is 7. Or 1 and any number — HCF is always 1, which is the smaller number.

Trap: 6 and 15 — a student might assume the smaller number always divides the larger, but 15 is not in the 6 times table (the multiples are 6, 12, 18), so 6 cannot be a factor of 15. HCF(6, 15) = 3. The HCF equals the smaller number only when the smaller number divides exactly into the larger.

Example: 2 and 6

Another: 10 and 14

Creative: 46 and 98 — \(46 = 2 \times 23\) and \(98 = 2 \times 7^2\). The only shared prime is 2, so HCF = 2. The trick: after dividing both by 2, the results (23 and 49) must be coprime.

Trap: 8 and 12 — a student might assume any two even numbers have HCF = 2, but \(8 = 2^3\) and \(12 = 2^2 \times 3\), so HCF(8, 12) = 4. Both numbers are divisible by 4, not just by 2, pushing the HCF above 2.

Since both numbers are even, 2 divides each of them. This means 2 divides their HCF, so the HCF is at least 2 and is itself even.

For example: HCF(4, 10) = 2 (even), HCF(6, 14) = 2 (even), HCF(16, 24) = 8 (even). The HCF of two even numbers can never be 1 or any odd number.

True when neither number is a multiple of the other: HCF(8, 12) = 4, which is less than both 8 and 12.

False when one number divides the other: HCF(5, 20) = 5, which equals the smaller number — it’s not less than both. Whenever one number is a factor of the other, the HCF equals the smaller number rather than being strictly less than it.

True: 4 and 6 share the factor 2 and are both even.

False: 9 and 15 share the factor 3 (HCF = 3) but are both odd. Any two odd multiples of 3, 5, 7, or any other odd prime share a common factor without being even. Sharing a common factor greater than 1 only means the numbers share a prime — that prime doesn’t have to be 2.

True: HCF(5, 12) = 1 and 5 is prime. But false: HCF(8, 15) = 1, yet \(8 = 2^3\) and \(15 = 3 \times 5\) — neither is prime. They are coprime simply because they share no prime factors.

This targets the “coprime means prime” misconception. Being coprime is a relationship between two numbers (they share no common factor other than 1) — it says nothing about whether either number is individually prime.

Assuming \(x\) is a positive integer, this is always true. We must look at the numerical coefficients and the algebraic parts separately.

The HCF of the numbers 4 and 6 is 2. The highest power of \(x\) that divides both \(x\) and \(x^2\) is \(x\). Multiplying these together gives \(2x\).

If you multiply both numbers by 2, you are adding an extra prime factor of 2 to both of their prime factorisations. Since they now share an additional factor of 2, their Highest Common Factor must also be multiplied by 2.

For example, HCF(3, 4) = 1. If we double them both to 6 and 8, the HCF(6, 8) = 2.

The student’s factor lists are incomplete. They only found 1, one small factor, and the number itself — missing 7 from both lists. The full factor lists are: 14 = 1, 2, 7, 14 and 21 = 1, 3, 7, 21.

Since 7 appears in both lists, HCF(14, 21) = 7, not 1. The misconception is assuming a quick scan of obvious factors is enough. Students must check systematically — for example, by testing every integer from 2 up to the smaller number, or by using prime factorisation.

The answer is correct — HCF(15, 20) is 5 — but the reasoning is dangerously flawed. Spotting that both are multiples of 5 proves that 5 is a common factor, not that it’s the highest. The student has confused “finding a common factor” with “finding the HCF”.

A counterexample breaks this reasoning immediately: 30 and 40 both end in 0 and are in the 5 times table, but HCF(30, 40) = 10, not 5. The correct method requires listing all common factors or using prime factorisation to confirm no larger common factor exists.

The student wrote products of two factors, not prime factorisations. This is the “incomplete factorisation” error. 9 is not prime (\(9 = 3^2\)) and 15 is not prime (\(15 = 3 \times 5\)). Every factor must be broken down until only primes remain.

The correct prime factorisations are \(18 = 2 \times 3^2\) and \(30 = 2 \times 3 \times 5\). By placing these in a Venn diagram, it becomes visually obvious that they share the primes 2 and 3, giving an HCF of \(2 \times 3 = 6\), not 2.

The student found the LCM (lowest common multiple), not the HCF. They found the smallest number that both 8 and 20 go into (a common multiple), not the largest number that goes into both (a common factor). This is the classic HCF/LCM confusion.

The correct answer: factors of 8 are 1, 2, 4, 8 and factors of 20 are 1, 2, 4, 5, 10, 20. The common factors are 1, 2, 4, and the highest is 4. A quick check: the HCF must divide both numbers, but 40 doesn’t divide 8 or 20 — it’s larger than both!