Choose HCF or LCM from the story, then compute it
Choose, then work it out
A word problem about the HCF or the LCM has two parts: decide which one the story needs, then find it.
Two strips of card, 27 cm and 36 cm long, are cut into pieces of equal length with no card wasted. What is the longest each piece can be? The piece must fit exactly into 27 and exactly into 36, so its length divides both numbers. The story needs the HCF.
Break both numbers into primes: 27 = 3 × 3 × 3 and 36 = 2 × 2 × 3 × 3. They share two 3s, so the HCF is 3 × 3 = 9. The longest piece is 9 cm.
27 = 3 × 3 × 3 and 36 = 2 × 2 × 3 × 3 share two 3s, and 3 × 3 = 9 is the HCF.
Pieces of 9 cm: the 27 cm strip makes 3 and the 36 cm strip makes 4, with no card left over.
Bells that ring together
Two bells ring together at noon. One rings every 6 minutes and the other every 10 minutes. When do they next ring together? The time must be a multiple of 6 and a multiple of 10, so both numbers divide it. The story needs the LCM.
Count in the larger number and test each landing against the smaller. 10 is not a multiple of 6, and neither is 20, but 30 is: 30 = 6 × 5. From the primes, 6 = 2 × 3 and 10 = 2 × 5, so the LCM is 2 × 3 × 5 = 30.
The bells ring together again after 30 minutes, at 12:30.
The first bell rings at 6, 12, 18, 24 and 30 minutes past noon.
The second bell rings at 10, 20 and 30 minutes past noon. 30 is the first time on both lines.
Answer the question that is asked
The HCF or the LCM is often only part of the answer. If the card question also asks how many pieces there are altogether, the answer is 3 + 4 = 7 pieces. If the bell question asks for the time, add the 30 minutes to noon: 12:30. Read the question again before writing the answer.
When something is left over
Some stories say what is left over. Eggs packed in cartons of 6, of 8 or of 9 always leave 2 eggs over. Take those 2 away and the rest pack exactly into 6s, 8s or 9s, so the rest is a common multiple of 6, 8 and 9. The smallest number of eggs is the LCM plus 2.
A story can also say what is missing. If the last bag is always 3 badges short of full, then 3 more badges would fill every bag exactly. The number of badges plus 3 is a common multiple, so the smallest number of badges is the LCM minus 3.
HCF and LCM · LCM Plus a Remainder: Eggs That Leave 2 Over in Cartons of 6, 8 or 9
QuestionA farmer has a basket of eggs. When he packs them into cartons of 6, there are 2 eggs left over. The same happens when he packs them into cartons of 8, and when he packs them into cartons of 9. He has more than 2 eggs. (a) If the 2 eggs left over are taken out, what is the smallest number of eggs there could be in the rest? (b) What is the smallest possible number of eggs in the basket?
1.Take the 2 eggs left over out of the basket. The eggs that remain fill cartons of 6, of 8 or of 9 exactly, so their number is a common multiple of 6, 8 and 9.
Take the 2 eggs left over out. The rest fill cartons of 6, of 8 or of 9 exactly. 2.Count in 9s, the largest carton: 9, 18, 27, 36, 45, 54 and 63 are not multiples of 8, but 72 = 8 × 9 is. 72 is also a multiple of 6: 72 = 6 × 12.
Count in 9s: the first multiple of 9 that is also a multiple of 8 is 72 = 8 × 9, and 72 = 6 × 12 as well. 3.(a) The smallest number of eggs in the rest is 72, the LCM of 6, 8 and 9.
(a) The smallest number in the rest is 72, the LCM of 6, 8 and 9. 4.(b) Put the 2 eggs back: 72 + 2 = 74. The smallest possible number of eggs is 74. Check: 74 = 6 × 12 + 2 = 8 × 9 + 2 = 9 × 8 + 2.
(b) 72 + 2 = 74 eggs. Check: 74 = 6 × 12 + 2 = 8 × 9 + 2 = 9 × 8 + 2.
Answer: (a) 72 eggs; (b) 74 eggs
Watch out for
- Taking the 2 away from the LCM, 72 − 2 = 70. In cartons of 6, 70 = 6 × 11 + 4 leaves 4 over, not 2. The eggs are 2 more than a full-carton count, so the 2 is added: 72 + 2 = 74.
- Adding 2 to each carton size and finding the LCM of 8, 10 and 11. The 2 is a number of eggs left over, not a change in the size of a carton, so it is added to the total, not to 6, 8 and 9.
HCF and LCM · LCM Less a Shortage: Badges That Leave the Last Bag 3 Short
QuestionA teacher packs all her badges into goodie bags for a school carnival, filling one bag at a time. If she puts 5 badges in each bag, the last bag is 3 badges short of full. The same happens with 6 badges in each bag, and with 8 badges in each bag. (a) What is the smallest number of badges that would fill bags of 5, bags of 6 or bags of 8 with no bag short? (b) What is the smallest possible number of badges the teacher has?
1.Each time, the last bag needs 3 more badges to be full. So with 3 more badges, every bag would be full, whether it holds 5, 6 or 8.
Each time, the last bag needs 3 more badges to be full. 2.That means the number of badges plus 3 is a common multiple of 5, 6 and 8.
With 3 more badges every bag would be full, so the badges plus 3 make a common multiple of 5, 6 and 8. 3.Count in 8s. A multiple of 8 is even, so it is also a multiple of 5 only if it ends in 0: 40, 80, 120, and so on. 40 ÷ 6 and 80 ÷ 6 leave remainders, but 120 = 6 × 20.
A multiple of 8 that ends in 0 is a multiple of 5 too: 40, 80, 120. Only 120 is also a multiple of 6. 4.(a) The smallest number that fills every bag size exactly is 120, the LCM of 5, 6 and 8.
(a) 120, the LCM of 5, 6 and 8, fills every bag size exactly. 5.(b) The teacher has 3 fewer: 120 − 3 = 117 badges. Check: in bags of 5, 6 and 8 the last bag holds 2, 3 and 5 badges, since 117 = 5 × 23 + 2 = 6 × 19 + 3 = 8 × 14 + 5, and each is 3 short of full.
(b) The teacher has 120 − 3 = 117 badges: the last bag is 3 short in bags of 5, of 6 and of 8.
Answer: (a) 120 badges; (b) 117 badges
Watch out for
- Adding the 3 and answering 120 + 3 = 123. In bags of 5, 123 = 5 × 24 + 3, so the last bag holds 3 badges and is 2 short, not 3 short. The badges are 3 fewer than a full count, so the 3 is taken away.
- Adding the badges that are in the last bag instead of the badges missing from it. In bags of 5 the last bag holds 2 badges, and 117 + 2 = 119 is not a multiple of 5. It is the 3 missing badges that complete the bag: 117 + 3 = 120.
Your turn
1.Ribbons 16 cm and 24 cm: longest equal piece, no waste?