A medium question on most and least for P2–P4, with a step-by-step solution, diagrams and the answer. Finding the most and the least appears in the Hua Xia Cup from the P1 and P2 first rounds to the P6 papers.
The question
On a scout trip, 13 scouts must cross a stream in one rubber dinghy.
It holds at most 4 people, the rower included, and every scout can row.
The dinghy won't drift back by itself; someone always has to row it back.
For all the scouts to cross, what is the fewest crossings the dinghy must make? (One bank to the other counts as one.)
Show the hints (2)
💡Hint 1: After the last crossing, does anyone need to row back?
💡Hint 2: Before the last crossing, each round trip (two crossings) adds how many people to the far bank, really?
Worked solution
Have a go first, then read the solution step by step.
1.The last crossing takes 4 over and needn't come back.
2.Before that, each round trip (two crossings) takes at most 4 over and brings at least 1 back, so the far bank gains at most 4 − 1 = 3.
3.Before the last crossing, 13 − 4 = 9 must already be over: 9 ÷ 3 = 3 exactly, so at least 3 round trips.
4.It can be done: take 4 over each time, 1 row back. In all 3 × 2 + 1 = 7 crossings.
⚠️Common mistake: thinking the far bank gains 4 every time, and answering something like 13 ÷ 4. Whoever rows back is on this bank again.
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