An expert question on remainders for P5–P6, 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 factory has made a batch of face masks; the number of masks is a four-digit number.
Packed in packs of 18, the last pack is 12 short of full; in packs of 35, there are 3 left over; and in packs of 12, the last pack is 6 short of full.
How many different numbers of masks are possible?
Show the hints (3)
💡Hint 1: "The last pack is 12 short of full": what is left when dividing by 18? Then use two conditions to find the smallest number meeting both.
💡Hint 2: 18 and 12 share the factor 6. How often does a number meeting two conditions come round again? Is it the product?
💡Hint 3: Numbers meeting the packs of 35 and of 18 come every 630; meeting all three, every 1260. Find the smallest, then count the four-digit ones.
Worked solution
Have a go first, then read the solution step by step.
1.The three conditions: leaving 6 when divided by 18; leaving 3 when divided by 35; leaving 6 when divided by 12.
2.Numbers leaving 3 when divided by 35: 3, 38, 73, 108, 143, …, 458, 493, 528. The first that leaves 6 when divided by 18 is 528 (the 16th).
3.Numbers meeting both come every 630, the LCM of 35 and 18: 528, 1158, … The first that leaves 6 when divided by 12 is 1158.
4.The LCM of 18, 35, 12 is 1260, so numbers meeting all three come every 1260: 1158, 2418, 3678, …
5.Four-digit numbers run from 1000 to 9999: the first that fits is 1158, the last 9978.
6.(9978 − 1158) ÷ 1260 + 1 = 8.
⚠️Common mistake: thinking they come round every 18 × 35 × 12 = 7560, and answering 2. 18 and 12 share a factor, so the gap is the LCM, 1260. And "12 short of full" is not "12 left over".
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