How to find the HCF (highest common factor): 3 methods

• Factors of 12: 1, 2, 3, 4, 6, 12
• Factors of 18: 1, 2, 3, 6, 9, 18

The largest factor in both lists is 6.

1. $84 = 2^2 \times 3 \times 7$ and $120 = 2^3 \times 3 \times 5$.
2. Keep only the primes in both: 2 and 3.
3. Use the lowest power of each: $2^2$ and $3$.
4. HCF $= 4 \times 3 = 12$.

Divide the larger number by the smaller, then repeat with the smaller number and the remainder:

• $120 = 1 \times 84 + 36$
• $84 = 2 \times 36 + 12$
• $36 = 3 \times 12 + 0$

The last non-zero remainder, 12, is the HCF. This is very quick for large numbers.

• Simplify fractions: $\tfrac{84}{120} = \tfrac{84 \div 12}{120 \div 12} = \tfrac{7}{10}$. See how to simplify fractions.
• Factorise: $84x + 120 = 12(7x + 10)$. See factorising.
• Word problems: What's the largest number of identical gift bags you can make from 84 sweets and 120 stickers, with nothing left over? 12 bags.

• Taking the highest powers (that gives the LCM).
• Including a prime that's in only one number.
• Stopping at a common factor that isn't the highest.

Find the HCF of: (a) 24 and 36, (b) 45 and 75, (c) 18, 30 and 42. Answers: 12, 15, 6. Check with the HCF calculator.

For 18, 30 and 42:

• $18 = 2 \times 3^2$
• $30 = 2 \times 3 \times 5$
• $42 = 2 \times 3 \times 7$

The primes in all three are 2 and 3, each with lowest power 1, so the HCF is $2 \times 3 = 6$. 5 and 7 are left out because they're not in every number.

Some pairs of numbers share no prime factors at all, such as 8 ($2^3$) and 15 ($3 \times 5$). Their HCF is 1, and they're called coprime. A fraction whose top and bottom are coprime, like $\tfrac{8}{15}$, is already in its simplest form. Two different primes are always coprime, and so are any two consecutive numbers.

• HCF questions split things into the largest equal groups: "greatest number of bags", "largest possible tile", "longest equal pieces".
• LCM questions wait for things to line up again: "next time together", "smallest number of packs to have the same amount".

A floor is 240 cm by 180 cm. What's the largest square tile that covers it exactly, with no cutting? That's the HCF: $240 = 2^4 \times 3 \times 5$ and $180 = 2^2 \times 3^2 \times 5$, so the HCF is $2^2 \times 3 \times 5 = 60$ cm. You'd need $4 \times 3 = 12$ tiles.