The highest common factor is the largest number that divides both
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.
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.
HCF and LCM · HCF: The Most Gift Bags from 72 Chocolate Bars and 90 Packets of Biscuits
QuestionMr Lim has 72 chocolate bars and 90 packets of biscuits. He packs all of them into gift bags, so that every bag has the same number of chocolate bars and the same number of packets of biscuits, with none left over. (a) What is the greatest number of gift bags he can pack? (b) How many chocolate bars and how many packets of biscuits are in each bag?
1.The number of bags must share out the 72 chocolate bars exactly and the 90 packets exactly, so it is a common factor of 72 and 90. The greatest number of bags is their HCF.
The number of bags must divide 72 and 90 exactly, so it is a common factor. The most bags is the HCF. 2.Both numbers are even, so divide both by 2: 72 ÷ 2 = 36 and 90 ÷ 2 = 45.
Both numbers are even, so divide both by 2: 36 and 45. 3.36 and 45 are both multiples of 3, so divide both by 3: 36 ÷ 3 = 12 and 45 ÷ 3 = 15.
Both are multiples of 3, so divide both by 3: 12 and 15. 4.12 and 15 are both multiples of 3 again: 12 ÷ 3 = 4 and 15 ÷ 3 = 5. No number except 1 divides both 4 and 5, so stop.
Divide both by 3 again: 4 and 5. No number except 1 divides both 4 and 5, so the ladder stops. 5.(a) Multiply the numbers you divided by: 2 × 3 × 3 = 18. He can pack 18 gift bags.
(a) The numbers divided by multiply to the HCF: 2 × 3 × 3 = 18 bags. 6.(b) The numbers left at the bottom are what each bag holds: 4 chocolate bars and 5 packets of biscuits. Check: 18 × 4 = 72 and 18 × 5 = 90.
(b) The numbers left at the bottom are what each bag holds: 4 chocolate bars and 5 packets of biscuits.
Answer: (a) 18 gift bags; (b) 4 chocolate bars and 5 packets of biscuits in each bag
Watch out for
- Using the LCM of 72 and 90, which is 360. With 360 bags, every bag would need at least one chocolate bar, and there are only 72. The number of bags must divide both numbers, so it is a common factor, not a common multiple.
- Stopping the ladder too early. Dividing both numbers by 9 gives 8 and 10, but 8 and 10 are both even, so they can still be divided by 2. Stopping there gives 9 bags instead of 18.
HCF and LCM · HCF: The Longest Equal Pieces Cut from Three Planks
QuestionThree wooden planks are 120 cm, 180 cm and 240 cm long. They are cut into pieces that are all the same length, with no wood left over. (a) What is the greatest possible length of each piece? (b) How many pieces are there altogether?
1.Each plank is cut into whole pieces, so the piece length divides 120, 180 and 240 exactly. It is a common factor of all three, and the longest piece is their HCF.
Each plank is cut into whole pieces, so the piece length must divide 120, 180 and 240. 2.All three lengths end in 0, so divide them all by 10: 120 ÷ 10 = 12, 180 ÷ 10 = 18 and 240 ÷ 10 = 24.
All three lengths end in 0: divide them all by 10 to get 12, 18 and 24. 3.12, 18 and 24 are all multiples of 6, so divide them all by 6: 12 ÷ 6 = 2, 18 ÷ 6 = 3 and 24 ÷ 6 = 4. 2 divides 2 and 4 but not 3, so no number except 1 divides all three. Stop.
Divide all three by 6 to get 2, 3 and 4. No number except 1 divides all of 2, 3 and 4. 4.(a) Multiply the numbers you divided by: 10 × 6 = 60. The longest piece is 60 cm.
(a) The longest piece is 10 × 6 = 60 cm. 5.(b) The numbers left, 2, 3 and 4, are the pieces from each plank, so there are 2 + 3 + 4 = 9 pieces. Check: 9 × 60 = 540, and 120 + 180 + 240 = 540.
(b) The planks give 2, 3 and 4 pieces of 60 cm: 2 + 3 + 4 = 9 pieces.
Answer: (a) 60 cm; (b) 9 pieces
Watch out for
- Dividing again by 2 because 2 and 4 are both even, and getting pieces of 120 cm. 3 is not even, so 2 does not divide all three numbers, and a 120 cm piece does not fit the 180 cm plank: 180 ÷ 120 = 1 remainder 60.
- Multiplying the pieces from each plank, 2 × 3 × 4 = 24. Each plank is cut on its own, so its pieces are added to the others: 2 + 3 + 4 = 9.
Your turn
1.8 = 2 × 2 × 2 and 12 = 2 × 2 × 3. Their HCF?