Maths olympiad question · P5–P6★★★★ Expert
Clever sums: olympiad question 8
An expert question on clever sums for P5–P6, with a step-by-step solution, diagrams and the answer. The first few questions of a competition paper are often clever sums, and the Hua Xia Cup has them in every year from P1 to P6.
The question
Work this out the clever way.
2592 × (1 − 4/9) × (1 − 4/16) × (1 − 4/25) × … × (1 − 4/81) = ?
What is the answer?
Show the hints (3)
- Hint 1: What's taken away, 4/9, 4/16, …, is 2 × 2 over 3 × 3, 4 × 4, … What two numbers multiply to 3 × 3 − 4?
- Hint 2: 1 − 4/(k × k) = (k − 2)/k × (k + 2)/k. Split every bracket into two fractions and multiply each set separately.
- Hint 3: In each set a numerator cancels a denominator 2 places along, so 2 numbers are left at each end, not one.
Worked solution
Have a go first, then read the solution step by step.
1.Each bracket: 1 − 4/(k × k) = (k × k − 2 × 2)/(k × k) = (k − 2)/k × (k + 2)/k; for example 3 × 3 − 4 = 5 = 1 × 5. 2.First set: 1/3 × 2/4 × … × 7/9. Numerators run 1 to 7, denominators 3 to 9; the shared ones cancel, leaving (1 × 2)/(8 × 9). 3.Second set: 5/3 × 6/4 × … × 11/9. Numerators run 5 to 11, denominators 3 to 9, leaving (10 × 11)/(3 × 4). - 4.So the whole thing = 2592 × (1 × 2 × 10 × 11)/(8 × 9 × 3 × 4) = 2592 × 55/216 = 660.
- ⚠️Common mistake: copying the 1 − 1/k² pattern and leaving one number at each end, 2592 × 1/9 × 11/3; here things cancel 2 places apart, so 2 are left at each end.
Answer: 660
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