The largest factor they share
The common factors of 12 and 18 are 1, 2, 3 and 6: each of them divides both numbers. The largest of them is 6.
The largest number that divides both of two numbers is called their Highest Common Factor, or HCF for short. The HCF of 12 and 18 is 6.
The overlap holds the common factors of 12 and 18, and the largest number in it, 6, is their HCF.
Why the largest one matters
A ribbon 12 cm long and another 18 cm long are both cut into pieces of the same length, with nothing left over. The length of a piece must divide 12 and divide 18, so it has to be a common factor. Pieces of 1, 2, 3 or 6 cm all work.
The longest piece that works is the HCF, 6 cm. It cuts the 12 cm ribbon into 2 pieces and the 18 cm ribbon into 3.
A piece of 4 cm fits the 12 cm ribbon three times, but 18 ÷ 4 = 4 remainder 2, so the 18 cm ribbon has 2 cm left over. 4 is a factor of 12 and not of 18.
Hops of 6 land exactly on 12 and exactly on 18, so pieces of 6 cm cut both ribbons with nothing left over.
Hops of 4 land on 12. The next landings are 16 and then 20, which miss 18, so pieces of 4 cm leave 2 cm of the longer ribbon.
The HCF from the primes
Listing every factor gets slow for bigger numbers. The prime factorizations give the HCF faster. 12 = 2 × 2 × 3 and 18 = 2 × 3 × 3. Both lists contain a 2 and a 3, so multiply the primes they share: 2 × 3 = 6. It is the same HCF as before.
A prime counts only as many times as it appears in both lists. 12 has two 2s and 18 has one, so they share one 2. 18 has two 3s and 12 has one, so they share one 3.
12 = 2 × 2 × 3 and 18 = 2 × 3 × 3. The shared 2 and 3 are in the overlap, and 2 × 3 = 6 is the HCF.
0 of the 3 shared factors are in the overlap: product so far = 1
Slide the shared factors into the overlap
24 = 2 × 2 × 2 × 3 and 36 = 2 × 2 × 3 × 3. Slide each prime they share into the overlap: 2, 2 and 3, and 2 × 2 × 3 = 12 is the HCF.
Bigger numbers
For 84 and 90: 84 = 2 × 2 × 3 × 7 and 90 = 2 × 3 × 3 × 5. They share one 2 and one 3, so the HCF is 2 × 3 = 6. The 7 is only in 84 and the 5 is only in 90, so neither is part of the HCF.
Shared primes can also be found by dividing both numbers at once. Divide 72 and 90 by 2 to get 36 and 45, then both by 3 to get 12 and 15, then both by 3 again to get 4 and 5. No prime divides both 4 and 5, so stop. The primes you divided by give the HCF: 2 × 3 × 3 = 18.
Not the product, and not a smaller common factor
The HCF divides both numbers, so it can never be bigger than the smaller of them. 12 × 18 = 216 is bigger than both, so it cannot divide either one.
The smaller number is the HCF only when it divides the larger. 12 does not divide 18, because 18 ÷ 12 = 1 remainder 6.
A common factor that is not the largest is not the HCF. 3 divides 12 and 18, but 6 divides both as well, and 6 is larger.
Worked example: HCF: Equal Distribution into Maximum Sets (No Remainder)
Question Mr. Lim has 72 chocolate bars and 90 packets of biscuits. He wants to pack all of them into identical gift bags such that every bag contains the exact same number of chocolate bars and biscuit packets, with none left over. (a) What is the greatest number of gift bags he can pack? (b) How many chocolate bars and biscuit packets will be in each gift bag?
1.Divide both 72 and 90 by 2: (72, 90) ÷ 2⟶ (36, 45).
Divide both counts by a common factor, 2. Each division is one more way to split every bag the same. 2.Divide both 36 and 45 by 3: (36, 45) ÷ 3⟶ (12, 15).
Both are still divisible by 3. 3.Divide both 12 and 15 by 3: (12, 15) ÷ 3⟶ (4, 5).
And by 3 again. 4.Since 4 and 5 share no common factors other than 1, stop.
4 and 5 share nothing but 1, so the ladder stops. 5.Multiply the left-hand divisors: 2 × 3 × 3 = 18 bags.
(a) The divisors down the side multiply to the HCF: 2 × 3 × 3 = 18 bags. 6.The remaining quotients directly give the items per bag: 4 chocolate bars and 5 biscuit packets.
(b) The numbers left at the bottom are what each bag holds: 4 bars and 5 packets.
Answer: (a) 18 gift bags; (b) 4 chocolate bars and 5 biscuit packets in each bag
Common mistakes
- Confusing HCF with LCM and calculating LCM(72, 90) = 360, resulting in an impossible bag count greater than the items available.
- Stopping the ladder division early (e.g., dividing by 9 once and obtaining 8 and 10, forgetting that 8 and 10 still share a common factor of 2).
- Mixing up the answers to parts (a) and (b), writing 4 or 5 as the total number of bags.
Worked example: HCF: Cutting Unequal Lengths into Longest Equal Pieces (1D Segmentation)
Question Three wooden planks of lengths 120 cm, 180 cm, and 240 cm are to be cut into shorter pieces of identical length without any wood left over. (a) What is the greatest possible length of each cut piece? (b) How many such cut pieces will there be altogether?
1.Perform simultaneous short division on 120, 180, and 240:
Three planks, one piece length for all, nothing wasted: the length must divide 120, 180 and 240. 2.Divide by 10: (12, 18, 24).
Divide all three by 10. 3.Divide by 6: (2, 3, 4).
Then all three by 6. What is left, 2, 3 and 4, shares nothing. 4.Common factor: 10 × 6 = 60 cm (length of 1 block).
(a) The longest piece is 10 × 6 = 60 cm. Shorter common lengths also work but leave more pieces. 5.Sum the remaining unit blocks: 2 + 3 + 4 = 9 pieces.
(b) Pieces: 2 + 3 + 4 = 9 with 60 cm; with 60 cm, 10.
Answer: (a) 60 cm; (b) 9 pieces
Common mistakes
- Multiplying the remaining pieces (2 × 3 × 4 = 24) instead of adding them, confusing 1D piece counting with 2D/3D grid arrays.
- Finding the HCF of only the first two numbers (120 and 180) and forgetting to check divisibility against 240.