Finding Common Ground: HCF and LCM Made Simple

Suppose a room measures 18 feet by 24 feet, and you want to cover the floor with the largest possible square tiles whose side is a whole number of feet, with no tile cut or overlapping. The tile's side must divide 18 exactly and must also divide 24 exactly — it has to be a common factor of both numbers. The factors of 18 are 1, 2, 3, 6, 9, 18. The factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. The common factors are 1, 2, 3, 6 — and the largest of these, 6, is called the Highest Common Factor (HCF) of 18 and 24. So the biggest tile you can use has a 6 ft side, and you'd need (18 ÷ 6) × (24 ÷ 6) = 3 × 4 = 12 tiles.

Listing every factor works, but it gets slow and error-prone for bigger numbers. Prime factorisation gives a faster, more reliable route. Write each number as a product of primes: 18 = 2 × 3 × 3 and 24 = 2 × 2 × 2 × 3. To build the HCF, pick out only the primes that appear in both factorisations, using the smallest number of times either number has that prime. Both have at least one 2 (18 has one 2, 24 has three — take the minimum, one 2). Both have at least one 3 (18 has two 3s, 24 has one — take the minimum, one 3). So HCF = 2 × 3 = 6, exactly matching the tiling answer.

This minimum-power rule is the whole method: factorise every number, then for each prime that shows up in all of them, multiply in the smallest number of copies seen anywhere. If two numbers share no prime factor at all — for example 17 and 19, which are both prime and different — their only common factor is 1, and their HCF is 1. Such number pairs are called co-prime.

Now flip the question around. A cyclist rides laps of a track and passes the start line every 12 minutes. A second cyclist passes the same start line every 18 minutes. If both start together, when do they next cross the start line at exactly the same moment? The answer has to be a number that is a multiple of both 12 and 18. Multiples of 12 are 12, 24, 36, 48...; multiples of 18 are 18, 36, 54, 72.... The smallest number appearing in both lists is 36, so they meet again after 36 minutes. This smallest shared multiple is called the Lowest Common Multiple (LCM).

Prime factorisation builds the LCM just as cleanly as it builds the HCF — except now you take the maximum power of every prime that appears in any of the numbers, not just the shared ones. Checking the cyclist example: 12 = 2 × 2 × 3 and 18 = 2 × 3 × 3. The highest power of 2 appearing anywhere is 2 × 2 (from 12), and the highest power of 3 appearing anywhere is 3 × 3 (from 18). So LCM = 2 × 2 × 3 × 3 = 36 — matching exactly.

This extends naturally to three or more numbers. If a third cyclist passes the start line every 24 minutes, all three meet again at the LCM of 12, 18 and 24. Factorising: 12 = 2² × 3, 18 = 2 × 3², 24 = 2³ × 3. Taking the highest power of each prime across all three — 2³ and 3² — gives LCM = 8 × 9 = 72. All three cyclists cross the start line together again after 72 minutes.

There's a faster layout that finds the HCF and LCM of two numbers in a single working. Take 48 and 60. Divide both by a common prime factor, write the quotients in the next row, and repeat until the two numbers left share no common factor:

• 2 | 48, 60 → 24, 30
• 2 | 24, 30 → 12, 15
• 3 | 12, 15 → 4, 5

Now 4 and 5 share no common factor, so the ladder stops. The HCF is the product of the divisors used on the left: 2 × 2 × 3 = 12. The LCM is the product of those same divisors and the two final numbers at the bottom: 2 × 2 × 3 × 4 × 5 = 240. Checking against full prime factorisation: 48 = 2⁴ × 3 and 60 = 2² × 3 × 5 — the common minimum gives HCF = 2² × 3 = 12, and the combined maximum gives LCM = 2⁴ × 3 × 5 = 240. Both methods agree exactly.

The ladder method is especially handy because you're allowed to divide by any common factor at each step, not just a single prime at a time — dividing both numbers by a bigger shared factor straightaway just shortens the ladder. For instance, since both 48 and 60 are divisible by 12 right away, you could divide by 12 first to get 4 and 5 directly, then read off HCF = 12 and LCM = 12 × 4 × 5 = 240 in one fewer step.

Take 48 and 60 again. We found HCF = 12 and LCM = 240. Multiply them: 12 × 240 = 2880. Now multiply the original numbers: 48 × 60 = 2880. They match exactly! This is not a coincidence — for any two numbers a and b, HCF(a, b) × LCM(a, b) = a × b. The reason lies in prime factorisation: every prime factor of a and b ends up contributing to either the HCF (if it's shared) or the LCM (if it's extra in one number), and between the two products, every prime factor from both original numbers gets counted exactly once — which is exactly what a × b also does.

This relationship is useful both for checking your work and for working backwards. Consider 96 and 275: their prime factorisations share no common prime at all (96 = 2⁵ × 3, and 275 = 5² × 11), so HCF = 1 and LCM is simply their product, 26400 — and indeed, 1 × 26400 = 96 × 275 = 26400.

Special number pairs follow neat patterns too. If one number divides the other exactly — say 13 and 26, where 26 = 2 × 13 — the smaller number is automatically the HCF (13) and the larger is automatically the LCM (26), since 13 × 26 = 338 and HCF × LCM = 13 × 26 = 338 as well. For two co-prime numbers like 17 and 19, HCF = 1 always, so LCM is simply their product: 17 × 19 = 323.

Problems about splitting things into the largest possible equal groups — tiles, lengths of rope cut into equal pieces, items packed into bags — are HCF problems. For example, two ropes of length 60 m and 84 m need to be cut into equal pieces with no leftover rope and as few pieces as possible: the longest possible piece length is HCF(60, 84) = 12 m, giving (60 ÷ 12) + (84 ÷ 12) = 5 + 7 = 12 pieces in total.

Problems about things that happen repeatedly and need to line up again at the same time — bells ringing, buses departing, patterns repeating — are LCM problems, as seen already with the cyclists meeting every 36 or 72 minutes.

Some problems run the HCF–LCM relationship in reverse: given the HCF and LCM of two numbers, find the numbers themselves. Suppose two numbers have HCF 6 and LCM 60. Since HCF × LCM = product of the numbers, their product must be 6 × 60 = 360. Checking pairs of factors of 360 whose HCF is actually 6: 12 and 30 works, since HCF(12, 30) = 6 and LCM(12, 30) = 60 — and so does 6 and 60 itself.