Factors and Multiples

A factor of a positive integer $n$ is any positive integer that divides $n$ with no remainder. The factors of $12$ are $1, 2, 3, 4, 6, 12$.

A multiple of $n$ is any number obtained by multiplying $n$ by an integer. Multiples of $4$ are $4, 8, 12, 16, 20, \ldots$ (and technically $0$, but Cambridge usually means positive multiples).

Common factor = a factor of every number in a set. Common factors of $12$ and $18$: $1, 2, 3, 6$. Common multiple = a multiple of every number in a set. Common multiples of $4$ and $6$: $12, 24, 36, \ldots$

When a question asks for a factor or multiple, list them in increasing order — examiners spot missing entries.

The Fundamental Theorem of Arithmetic says every integer above $1$ can be written as a product of primes in exactly one way (apart from the order of the factors). For example $60 = 2^{2} \times 3 \times 5 \quad \text{and} \quad 72 = 2^{3} \times 3^{2}.$

This unique fingerprint is what makes HCF and LCM straightforward.

Method 1: Factor tree. Start with $n$ at the top, branch into ANY pair of factors, and keep splitting any composite branch until every leaf is prime.

For $\mathbf{72}$:

• $72 = 8 \times 9$
• $8 = 2 \times 2 \times 2$
• $9 = 3 \times 3$
• Combine: $72 = 2^{3} \times 3^{2}$.

Method 2: Repeated-division (ladder). Divide by the smallest prime that goes in, write the quotient below, repeat until you reach $1$. The primes you divided by are the factorisation.

2 | 72
2 | 36
2 | 18
3 |  9
3 |  3
  |  1

Read down: $2 \times 2 \times 2 \times 3 \times 3 = 2^{3} \times 3^{2}$. Same answer.

The ladder method is faster when the number has many small prime factors. The factor tree is friendlier when one factor is obvious.

Highest Common Factor (HCF) of two integers is the largest integer that divides both. Lowest Common Multiple (LCM) is the smallest positive integer that BOTH numbers divide into.

Method 1: Listing. Best for small numbers ($\le 30$). List all factors of each number; pick the largest common one for HCF. List multiples until both lists agree for LCM.

Method 2: Prime-factorisation tree (RECOMMENDED for $> 30$). Write each number as a product of primes in index form.

• HCF = take the lowest power of every prime that appears in BOTH.
• LCM = take the highest power of every prime that appears in EITHER.

For $\mathbf{72}$ and $\mathbf{108}$:

• $72 = 2^{3} \times 3^{2}$
• $108 = 2^{2} \times 3^{3}$
• HCF = $2^{2} \times 3^{2} = 4 \times 9 = 36$ (lowest power of each shared prime).
• LCM = $2^{3} \times 3^{3} = 8 \times 27 = 216$ (highest power of each prime).

Method 3: Venn-diagram method. Draw two overlapping circles, write the prime factors into the appropriate region (shared primes in the overlap, unique primes outside). HCF = product of overlap; LCM = product of every prime in the diagram.

Quick sanity checks. HCF should never be larger than the smaller of the two numbers. LCM should never be smaller than the larger.

Word problems. Real-paper questions disguise HCF/LCM as everyday situations. Spot the verb:

• "Cut into the largest equal pieces" / "the greatest number of..." → HCF.
• "Both events happen at the same time again" / "the smallest number that" → LCM.

Each rule lets you decide whether one number divides another by looking at digits — no division required.

Divisor	Test
$2$	Last digit is even ($0, 2, 4, 6, 8$).
$3$	Digit sum is a multiple of $3$.
$4$	Last two digits form a multiple of $4$.
$5$	Last digit is $0$ or $5$.
$6$	Divisible by both $2$ AND $3$.
$8$	Last three digits form a multiple of $8$.
$9$	Digit sum is a multiple of $9$.
$10$	Last digit is $0$.
$11$	Alternating digit sum is a multiple of $11$ (including $0$).

Worked: is $6{,}138$ divisible by $11$? Alternating sum from the left: $6 - 1 + 3 - 8 = 0$. $0$ is a multiple of $11$ → yes. (Confirm: $6{,}138 \div 11 = 558$.)

Worked: is $5{,}472$ divisible by $9$? Digit sum: $5 + 4 + 7 + 2 = 18$. $18 = 9 \times 2$ → yes. ($5{,}472 \div 9 = 608$.)