Prime Factorisation

Work out the right-hand side: \(2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 12 \times 5 = 60\). Every factor (2, 3, 5) is prime, and the product equals 60 — so this is the prime factorisation.

You can also verify using a factor tree: \(60 \to 2 \times 30 \to 2 \times 2 \times 15 \to 2 \times 2 \times 3 \times 5\). No matter how you split 60, you always arrive at the same set of primes.

No matter how you split a number into factors, you always end up with the same set of primes. Take 36: you could start with \(2 \times 18\), or \(3 \times 12\), or \(4 \times 9\), or \(6 \times 6\) — but every route leads to \(2^2 \times 3^2\). This is called the Fundamental Theorem of Arithmetic.

The key insight is that primes can’t be broken down any further. Once you reach primes, there’s nowhere else to go — so every path through the factor tree must arrive at the same destination.

Think of it like the “DNA” or chemical formula of a number. Just as water is always H₂O and never HO₂, the number 60 is always \(2^2 \times 3 \times 5\). No matter how you create it, its “chemical makeup” of primes is unique and permanent.

A prime number has exactly two distinct factors: 1 and itself. The number 1 has only one factor (itself), so it doesn’t meet the definition.

There’s also a practical reason: if 1 were prime, prime factorisation wouldn’t be unique. You could write \(6 = 2 \times 3\), or \(6 = 1 \times 2 \times 3\), or \(6 = 1 \times 1 \times 2 \times 3\), and so on forever. Excluding 1 from the primes keeps the Fundamental Theorem of Arithmetic working.

2 is prime because its only factors are 1 and 2 — exactly two distinct factors. Now consider any other even number, say 4, 6, 8, 10, … Each of these is divisible by 1, by 2, and by itself — that’s at least three factors, so none of them can be prime.

In other words, every even number greater than 2 has 2 as a factor in addition to 1 and itself. Having that extra factor disqualifies them from being prime. This is why 2 is sometimes called the “oddest prime” — it’s the only even one.

The prime factorisation of 400 is \(2^4 \times 5^2\). Because all the exponents are even, the prime factors can be split perfectly into two identical groups: \((2^2 \times 5) \times (2^2 \times 5)\).

Since \(2^2 \times 5 = 20\), this is exactly \(20 \times 20\), making 400 a perfect square. Any number with only even powers in its prime factorisation is always a perfect square!

Example: 28 = 2 × 2 × 7 (three prime factors)

Another: 18 = 2 × 3 × 3 (three prime factors)

Creative: 27 = 3 × 3 × 3 — all three prime factors are the same! Or 30 = 2 × 3 × 5, where all three are different primes.

Trap: 24 = 2 × 2 × 2 × 3 — a student might think “24 has two types of prime factor (2 and 3), so it’s fine.” But the question says counting repeats, and 24 has four prime factors in total, not three.

Example: 6 = 2 × 3 and 12 = 2² × 3 — both use only the distinct primes {2, 3}

Another: 10 = 2 × 5 and 20 = 2² × 5 — both use only {2, 5}

Creative: 2 and 32 = 2&sup5; — vastly different in size, but both have {2} as their only prime factor. Or 6 = 2 × 3 and 72 = 2³ × 3² — same set of primes, very different exponents.

Trap: 12 = 2² × 3 and 15 = 3 × 5 — they share the factor 3, but they don’t share the same set of prime factors. 12 uses {2, 3} while 15 uses {3, 5}. Sharing one prime factor isn’t enough — the entire set must match.

Example: 120 = \(2^3 \times 3 \times 5\)

Another: 150 = \(2 \times 3 \times 5^2\)

Creative: 1000 = \(2^3 \times 5^3\) — a neat round number that doesn’t need 3 at all, but still draws multiple primes from the allowed pool.

Trap: 105 = 3 × 5 × 7 — a student might assume it works because 3 and 5 are in there. But 7 is also a prime factor, and 7 is not in the allowed set of {2, 3, 5}. Every prime factor must come from the allowed set.

Example: 8 = \(2^3\) and 9 = \(3^2\) — one uses only 2, the other only 3. Their HCF is 1 (they are coprime).

Another: 25 = \(5^2\) and 14 = 2 × 7 — {5} and {2, 7} share nothing in common.

Creative: 121 = \(11^2\) and 128 = \(2^7\) — using larger primes and higher powers. Neither has any prime factor in common with the other.

Trap: 6 = 2 × 3 and 15 = 3 × 5 — a student might think these “look different enough.” But both have 3 as a prime factor, so they do share a common prime. Their HCF is 3, not 1.

“Even” means “divisible by 2.” If a number is divisible by 2, then 2 must appear as one of its prime factors. For example: \(12 = 2^2 \times 3\), \(50 = 2 \times 5^2\), \(2 = 2\). Every single one has 2 in the factorisation.

This gives a powerful connection: you can tell whether a number is even just by checking if 2 appears in its prime factorisation.

It works sometimes: 6 = 2 × 3 has 4 factors and a factorisation of length 2, while 2 (which is prime) has just 2 factors and a factorisation of length 1. More factors, longer factorisation.

But \(16 = 2^4\) has 5 factors and a factorisation of length 4, while \(30 = 2 \times 3 \times 5\) has 8 factors but a factorisation of length only 3. Here, 30 has more factors but a shorter factorisation. The number of factors depends on the combination of exponents, not just their total.

This is essentially the definition of a prime number. A prime has exactly two factors: 1 and itself. If you could write it as a product of two whole numbers both greater than 1, it would have additional factors — making it composite, not prime.

For example, 7 can only be written as \(1 \times 7\). You cannot split it as a product of two numbers both greater than 1. Compare with \(6 = 2 \times 3\) — that is a product of two numbers both greater than 1, confirming 6 is composite. Students who think “prime” just means “odd” or “not in a times table” may find this surprisingly hard to justify.

If the two primes are different, it works: \(2 \times 3 = 6\), whose factors are 1, 2, 3, 6 — exactly four. The formula gives \((1+1)(1+1) = 4\) factors.

But if the two primes are the same, it fails: \(3 \times 3 = 9\), whose factors are 1, 3, 9 — only three. Here the factorisation is \(3^2\), and the formula gives \((2+1) = 3\) factors. Students often assume “two primes multiplied” means two different primes — but the statement doesn’t say that.

If the LCM is a multiple of the first number, it must contain all of that number’s prime factors. If it is also a multiple of the second number, it must contain all of its prime factors too.

The LCM is constructed by taking the smallest pool of primes that satisfies both requirements (the highest power of each prime present in either number).

The student has confused “product of factors” with “product of prime factors.” They stopped at the first pair of factors they found, but neither 8 nor 9 is prime. 8 = \(2^3\) and 9 = \(3^2\).

The correct prime factorisation is \(72 = 2^3 \times 3^2\). Every factor in a prime factorisation must be a prime number — you need to keep breaking factors down until no further splitting is possible.

The answer is correct — \(60 = 2^2 \times 3 \times 5\) — but the checking method is completely wrong. The student has invented a bogus verification rule: “add the prime factors and check if the sum divides the original number.” This happens to work here by coincidence.

A counterexample breaks the rule immediately: \(12 = 2^2 \times 3\), and the sum of prime factors is \(2 + 2 + 3 = 7\), but 7 is not a factor of 12. The correct way to check is simply to multiply the prime factors back together: \(2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60\). ✓

The student has included 1 as a prime factor. This is a very common misconception — students often think 1 is prime because “it’s only divisible by 1 and itself.” But 1 has only one factor (itself), so it fails the definition of a prime number, which requires exactly two distinct factors.

The correct answer is simply \(56 = 2^3 \times 7\). The 1 at the end of the division process is just the stopping point — it tells you that you’ve finished. It should not be included in the factorisation.

The student has confused the HCF method with the LCM method. For the HCF, you take the lowest power of each common prime factor. For the LCM, you take the highest. The student has done the LCM calculation instead.

The correct HCF is: take the lower power of 2 (which is \(2^2\)) and the lower power of 3 (which is \(3^1\)), giving HCF = \(2^2 \times 3 = 12\). The student’s answer of 72 is actually the LCM. A good sense-check: the HCF of two numbers can never be larger than either of them, but 72 is bigger than both 24 and 36.