Maths olympiad question · P5–P6★★★★ Expert
Most and least: olympiad question 10
An expert question on most and least for P5–P6, with a step-by-step solution 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
8 cups, numbered 1 to 8, stand on a table, all mouth up.
Each move must turn over exactly 3 of them (no more, no fewer; up becomes down and down becomes up).
What is the fewest moves that leaves every cup mouth down?
Show the hints (3)
- Hint 1: For a cup to end mouth down, has it been turned an odd or an even number of times?
- Hint 2: m moves make 3 × m turns in all. How many at least? And how must it compare with 8, odd or even?
- Hint 3: Each cup turns an odd number of times, so the total has the same parity as 8; and in m moves no cup turns more than m times. Try from m = 3.
Worked solution
Have a go first, then read the solution step by step.
- 1.Each cup must be turned an odd number of times. 8 odd numbers add up to the same parity as 8; so the total turns (3 × moves) must be at least 8 and match 8's parity. Also, in m moves no cup turns more than m times (for even m, an odd count is at most m − 1).
- 2.Try each: 3 moves, 9 turns, wrong parity for 8; 4 moves, 12 turns ✓. So at least 4 moves.
- 3.4 moves work: 12 turns is 4 more than 8, so 2 cups get 2 extra turns each. For example: move 1 turns 1, 2, 3; move 2 turns 1, 2, 4; move 3 turns 1, 2, 5; move 4 turns 6, 7, 8.
- 4.Check: in that schedule every cup turns an odd number of times ✓
- ⚠️Common mistake: working 8 ÷ 3 and thinking 3 moves will do — but 3 × 3 = 9 has the wrong parity for 8, so some cup is turned an even number of times.
Answer: 4 moves
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