Thirty flasks, three sons, one fair split
A dying man leaves his three sons 30 glass flasks: 10 full of oil, 10 half-full, and 10 empty. Divide both the oil and the flasks so each son receives exactly the same amount of oil and exactly the same number of flasks — without pouring any oil between containers. How?
Reveal the answer
One solution: son A gets 5 full + 5 empty flasks; son B gets 3 full + 4 half-full + 3 empty; son C gets 2 full + 6 half-full + 2 empty. Each ends up with 10 flasks and 5 units of oil, and the totals across all three sons still add up to 10 full, 10 half-full and 10 empty. It's problem 12 in Propositiones ad Acuendos Juvenes ('Problems to Sharpen the Young'), a puzzle collection attributed to the 8th-century scholar Alcuin of York — one of the oldest surviving collections of recreational maths in the West.