A hard question on remainders for P5, with a step-by-step solution, diagrams and the answer. Common in the Hua Xia Cup's P3 to P6 papers; harder ones need several remainder conditions to hold at once.
The question
A volunteer team receives three batches of tins:
39, 147 and 207.
Each batch is shared out equally among the same group of families, every family getting the same number from a batch.
All three batches have some tins left over — the same number each time, and too few to give every family one more.
What is the largest the number of families could be?
Show the hints (3)
💡Hint 1: Two batches leave the same number over. How is the difference between them related to the number of families?
💡Hint 2: 147 − 39 = 108 and 207 − 147 = 60. The number of families must divide both.
💡Hint 3: Find the HCF of 108 and 60, then check that every batch really does have tins left over.
Worked solution
Have a go first, then read the solution step by step.
1.With the same number left over, the difference between two batches is the difference between what was handed out — a multiple of the number of families.
2.147 − 39 = 108, 207 − 147 = 60.
3.108 = 12 × 9 and 60 = 12 × 5, with 9 and 5 sharing no factor above 1, so the HCF is 12.
4.Check: 39 ÷ 12 = 3 remainder 3; 147 ÷ 12 = 12 remainder 3; 207 ÷ 12 = 17 remainder 3. Each leaves 3, which is less than 12.
5.At most 12 families.
⚠️Common mistake: looking for common factors of 39, 147 and 207 themselves — it's their differences that matter.
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