Brandy and Wine

A merchant has six casks, holding $8$, $13$, $15$, $17$, $19$ and $31$ litres. He stocks only two drinks, brandy and wine, and his brandy costs exactly twice as much per litre as his wine.

Two customers arrive. The first buys only brandy and spends exactly £28. The second buys only wine and also spends exactly £28. No cask is ever broken into: each is sold whole or not at all. When the customers leave, a single cask remains unsold.

Taking that last cask to hold brandy, what is it worth?

Solution

Both customers spend the same £28, but brandy costs twice as much per litre as wine, so the wine buyer must have carried off exactly twice the volume of the brandy buyer.

The five casks that were sold therefore split into a brandy share and a wine share twice as large. Together they come to three times the brandy volume, so the volume sold is a multiple of three.

All six casks hold $8 + 13 + 15 + 17 + 19 + 31 = 103$ litres. For the sold volume to be a multiple of three, the unsold cask must leave a multiple of three behind. Looking at remainders on division by three, only the $13$, $19$ and $31$ litre casks qualify.

• Leave the $13$ litre cask: $90$ litres are sold, $30$ of brandy and $60$ of wine. No selection of $8, 15, 17, 19, 31$ adds up to $30$, so this fails.

• Leave the $31$ litre cask: $72$ litres are sold, $24$ of brandy and $48$ of wine. No selection of $8, 13, 15, 17, 19$ adds up to $24$, so this fails as well.

• Leave the $19$ litre cask: $84$ litres are sold, $28$ of brandy and $56$ of wine. This works: the brandy is $13 + 15 = 28$ litres, and the wine is $8 + 17 + 31 = 56$ litres.

So the wine sells at $50$ pence a litre and the brandy at £1 a litre. The wine buyer pays $56 \times 0.5 = 28$ pounds and the brandy buyer $28 \times 1 = 28$ pounds, exactly as required. The unsold cask is the $19$ litre one, and as brandy it is worth $19$ pounds, that is £19.