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.
Worked example: LCM with Constant Remainder (Equal Excess)
Question A farmer has a basket of eggs. When he packs them into cartons of 6, cartons of 8, or cartons of 9, there are always 2 eggs left over each time. What is the smallest possible number of eggs the farmer has in the basket (assuming he has more than 2 eggs)?
1.Find the common baseline where cartons leave zero remainder: LCM(6, 8, 9) = 72.
Full cartons of 6, 8 or 9 with nothing over: the smallest such count is the LCM, 72. 2.Every scenario describes an excess of 2 eggs above a full pack: [Complete Packs] + 2.
Every packing leaves 2 over, so the eggs are a full-carton count plus 2. 3.Smallest non-trivial total = 72 + 2 = 74 eggs.
Smallest: 72 + 2 = 74 eggs. 4.Verification: 74 ÷ 6 = 12 R 2, 74 ÷ 8 = 9 R 2, 74 ÷ 9 = 8 R 2.
Check: 74 = 12 × 6 + 2 = 9 × 8 + 2 = 8 × 9 + 2.
Answer: 74 eggs
Common mistakes
- Subtracting the remainder from the LCM (72 − 2 = 70) instead of adding it back.
- Adding 2 to each divisor before finding the LCM (e.g., finding the LCM of 8, 10, 11).
Worked example: LCM with Constant Shortage (Equal Deficit / Negative Offset)
Question A 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. What is the smallest possible number of badges the teacher has?
1.Determine the complete round number benchmark: LCM(5, 6, 8) = 120.
Bags of 5, 6 or 8 would come out exact at the LCM, 120. 2.Notice that in every case, the teacher is 3 units away from filling the final bag: [Actual Count] + [3 missing] = 120.
Each time she is 3 short of that last full bag: her count plus 3 is 120. 3.Deduct the deficit from the LCM: 120 − 3 = 117 badges.
So the badges: 120 − 3 = 117. 4.Verification: 117 ÷ 5 = 23 R 2 (short of 3 to make 24); 117 ÷ 6 = 19 R 3 (short of 3 to make 20); 117 ÷ 8 = 14 R 5 (short of 3 to make 15).
Check: 117 is 3 short of 24 fives, of 20 sixes and of 15 eights.
Answer: 117 badges
Common mistakes
- Treating a 'shortage of 3' as an excess and adding 3 (120 + 3 = 123), which would leave a remainder of 3 rather than a shortage.
- Failing to recognize that 'remainder of 2 when divided by 5' is structurally identical to a 'shortage of 3' (5 − 2 = 3).