91av

Answer: 48, 51, 53, 54 & 55 litres

The winner is M. Jacques Frederic of Mélin, Belgium

Worked solution

1. There are 11 – 1 = 10 combinations of pairs of barrels so there must be 5 barrels.

2. There are no duplicated volumes, so all the barrels have different capacities.

3. Let the barrels have capacities of A, B, C, D & E litres, in ascending order, making the 10 combinations:

(A+B), (A+C), (A+D), (A+E), (B+C), (B+D), (B+E), (C+D), (C+E) & (D+E).

4. The smallest combined volume is (A+B) litres and the second smallest is (A+C) litres.

5. The largest combined volume is (D+E) litres and the second largest is (C+E) litres.

6. If all the combinations are added together [(A+B)+(A+C)+…+(D+E)] the total is 4(A+B+C+D+E), so the sum of all the combinations is divisible by 4.

7. The series (excluding 100L) range from 90-99 up to 101-110. Only 3 series sum to a total divisible by 4, 91-101 (total 956), 95-105 (total 1000) and 99-109 (total 1044).

8. In the series 91-101, 4(A+B+C+D+E) = 956 so (A+B+C+D+E) = 956/4 = 239

9. But A+B (smallest) = 91 and D+E (largest) = 101 so C = 47

10. Also A+C (second smallest) = 92 so A = 45 so B = 46 and C+E (second largest) = 99 so E = 52 and D = 49 = 72. But none of the volumes of individual barrels are perfect squares so 91-101 fails.

11. In the series 95-105, 4(A+B+C+D+E) = 1000 so (A+B+C+D+E) = 1000/4 = 250

12. But (A+B) = 95 and (D+E) = 105 so C = 50.

13. Also (A+C) = 96 so A = 46 so B = 49 and (C+E) = 104 so E = 54 and D = 51.

But none of the volumes of individual barrels are perfect squares so 95-105 fails.

14. In the series 99-109, 4(A+B+C+D+E) = 1044 so (A+B+C+D+E) = 1044/4 = 261

15. But (A+B) = 99 and (D+E) = 109 so C = 53.

16. Also (A+C) = 101 so A = 48 so B = 51 and (C+E) = 108 so E = 55 and D = 54.

So barrel capacities, in ascending order are: 48, 51, 53, 54 & 55 litres respectively