Factors, Multiples and Primes

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

If we allowed 1 to be prime, the Fundamental Theorem of Arithmetic would break — every number would have infinitely many prime factorisations (e.g. \(6 = 2 \times 3 = 1 \times 2 \times 3 = 1 \times 1 \times 2 \times 3\), and so on). By excluding 1, every number has a unique prime factorisation.

The only factors of 2 are 1 and 2 — exactly two distinct factors. That is the definition of a prime number. Being even doesn’t disqualify a number from being prime.

In fact, 2 is the only even prime. Every other even number has at least three factors (1, 2, and itself), so it can’t be prime. This makes 2 unique: the smallest prime and the only one that is even.

Factors normally come in pairs. For 36: \(1 \times 36\), \(2 \times 18\), \(3 \times 12\), \(4 \times 9\), \(6 \times 6\). But when a number is a perfect square, the square root pairs with itself. So 6 appears only once in the factor list, not twice.

This means there’s always one “unpaired” factor — the square root — giving an odd total. Check: 36 has 9 factors (odd ✓). Compare with 12, which has 6 factors (even) — not a square number.

Suppose there were only finitely many primes. Multiply them all together and add 1. This new number leaves a remainder of 1 when divided by any prime on our list — so none of them divide it exactly.

That means our new number is either prime itself or has a prime factor we didn’t include. Either way, our “complete” list was missing a prime — contradiction! So the list can never be complete. This elegant argument was first given by Euclid around 300 BC.

Example: 4 (factors: 1, 2, 4)

Another: 9 (factors: 1, 3, 9)

Creative: 169 (factors: 1, 13, 169) — that’s \(13^2\). The pattern is that numbers with exactly three factors are always squares of primes.

Trap: 6 — a student might list 1, 2, 3 as its three factors, forgetting that 6 is also a factor of itself. In reality, 6 has four factors: 1, 2, 3, and 6.

Example: 3 (factor of 60 ✓, multiple of 3 ✓)

Another: 15 (factor of 60 ✓, multiple of 3 ✓)

Creative: 60 itself — 60 is a factor of 60 (since \(60 \div 60 = 1\)) and 60 is a multiple of 3 (since \(60 = 3 \times 20\)). Students often forget a number is always a factor of itself.

Trap: 9 — a student might offer this because 9 is a multiple of 3, but 9 is not a factor of 60 (\(60 \div 9 = 6.67\ldots\)). This reveals the common mistake of confusing “contains 3 as a factor” with “is a factor of 60.”

Example: 5 + 7 = 12

Another: 11 + 13 = 24

Creative: 2 + 2 = 4 — the only way to use the even prime! Since all other primes are odd, you need two odd primes (odd + odd = even) or the same even prime twice.

Trap: 2 + 3 = 5 — a student might offer any two primes without checking the sum. But 2 is even and 3 is odd, so their sum is odd, not even. This catches students who assume all primes behave the same way.

Example: 3 and 5

Another: 11 and 13

Creative: 41 and 43 — or 71 and 73. These are called twin primes. Nobody knows if there are infinitely many of them — it’s one of the great unsolved problems in mathematics!

Trap: 7 and 9 — they differ by 2 and both look odd, but 9 is not prime (\(9 = 3 \times 3\)). This catches the “odd means prime” misconception.

Since \(6 = 2 \times 3\), every multiple of 6 is automatically a multiple of 3. For example, \(6 = 3 \times 2\), \(12 = 3 \times 4\), \(18 = 3 \times 6\), \(24 = 3 \times 8\), and so on.

This is a key principle: if a is a factor of b, then every multiple of b is also a multiple of a. Students who say “sometimes” may be confusing the direction — a multiple of 3 is not always a multiple of 6 (e.g. 9), but the other way round always works.

Usually true: 3 is a factor of 12 and 3 < 12. But every number is a factor of itself, since \(12 \div 12 = 1\). In that case the factor and the number are equal, not smaller.

This catches the common assumption that factors are always “less than” the number. Students who hold this belief often forget to include the number itself when listing factors — exactly the mistake targeted earlier on this page!

If a number divides exactly into two amounts, it divides exactly into their total. For example, 4 is a factor of 12 and of 20, and 4 is indeed a factor of 32 (= 12 + 20). Likewise, 5 is a factor of 15 and of 35, and 5 is a factor of 50 (= 15 + 35).

Students who say “sometimes” often test it incorrectly — for example, checking whether 3 (a factor of 12) divides into 12 + 10 = 22. But 3 is not a factor of 10, so the condition isn’t met. The rule requires the same number to be a factor of both numbers being added.

If \(p\) and \(q\) are both prime, then \(p \times q\) has factors 1, \(p\), \(q\), and \(p \times q\) — that’s at least three factors, so it cannot be prime. For example, \(2 \times 3 = 6\) (not prime), \(3 \times 5 = 15\) (not prime), \(7 \times 7 = 49\) (not prime).

Even when both primes are the same (\(p \times p = p^2\)), the result has three factors: 1, \(p\), and \(p^2\). A prime times anything other than 1 will always produce a composite number. Students who say “sometimes” may be thinking of the sum of two primes, which can be prime (e.g. \(2 + 3 = 5\)).

The student has forgotten that a number is always a factor of itself. Since \(24 \div 24 = 1\), the number 24 divides into 24 exactly, so 24 must be in the list.

The complete list of factors is 1, 2, 3, 4, 6, 8, 12, 24. A useful strategy is to find factors in pairs: \(1 \times 24\), \(2 \times 12\), \(3 \times 8\), \(4 \times 6\). This makes it much harder to miss any — especially the number itself.

The answer is correct — 7 is prime — but the reasoning is dangerously flawed. The student has invented a rule: “odd numbers that don’t end in 0 or 5 are prime.” A counterexample breaks this immediately: 9 is odd, doesn’t end in 0 or 5, but \(9 = 3 \times 3\) — definitely not prime.

The correct way to check is to test whether any whole number other than 1 and 7 divides into 7 exactly. None do (try 2, 3, 4, 5, 6), confirming 7 is prime. The student’s shortcut fails because being odd only eliminates 2 as a factor — it doesn’t rule out factors like 3, 7, or 11.

The student has confused factors and multiples. “Numbers that go into 6” describes factors of 6, not multiples. The factors of 6 are 1, 2, 3, and 6. Multiples of 6 are numbers in the 6 times table: the first five are 6, 12, 18, 24, 30.

This is one of the most common misconceptions in this topic. A helpful distinction: factors are always less than or equal to the number (they divide into it), while multiples are always greater than or equal to the number (you multiply it by whole numbers). Factors come in finite pairs; multiples go on forever.

The student used a factor tree approach correctly but stopped too early. The number 4 is not a prime number (\(4 = 2 \times 2\)). Every factor in the product must be prime — that’s what “product of prime factors” means.

The correct answer is \(60 = 2 \times 2 \times 3 \times 5\) (or \(2^2 \times 3 \times 5\) in index form). This is the “incomplete prime factorisation” mistake. A useful check: look at every number in your product and ask “is this prime?” If any answer is no, keep breaking it down.