Henry Ernest Dudeney/Modern Puzzles/4 - Loose Cash

by : $4$

 * Loose Cash


 * What is the largest sum of money -- all in current silver coins and no four-shilling piece -- that I could have in my pocket without being able to give change for half a sovereign?

Solution
The largest sum possible is $15 \, \mathrm s. 9 \, \mathrm d.$, composed of:
 * one crown and one half-crown (or $3$ half-crowns)
 * $4$ florins
 * one threepenny piece.

Proof
Recall the list of silver coins in the time of, expressed in shillings:

It is of course to be noted that these coins were actually an alloy consisting of only $50\%$ pure silver, but were still called "silver coins".


 * The threepenny piece: $\tfrac 1 4 \, \mathrm s.$
 * The sixpence: $\tfrac 1 2 \, \mathrm s.$
 * The shilling
 * The florin: $2 \, \mathrm s.$
 * The half-crown: $2 \tfrac 1 2 \, \mathrm s.$
 * The crown: $5 \, \mathrm s.$
 * The half sovereign: $10 \, \mathrm s.$

's mention of the four-shilling piece refers to a coin in circulation for a short time in the late $19$th century, also known as the double florin.

Hence the question is seen to be the same as asking:
 * What is the highest total of numbers from $\set {\tfrac 1 4, \tfrac 1 2, 1, 2, 2 \tfrac 1 2, 5}$ of which no subset adds to $10$?

We may take only one crown and only one half-crown, or $3$ half-crowns, as one more of either allow us to make $10$.

Then we see that $4$ florins take us to $5 + 2 \tfrac 1 2 + 4 \times 2 = 15 \tfrac 1 2$, while no combination of $5, 2 \tfrac 1 2, 2, 2, 2, 2$ can make $10$.

Adding another florin brings us to $16 \tfrac 1 2$, but then $2 + 2 + 2 + 2 + 2 = 10$.

Adding a shilling allows $5 + 2 + 2 + 1$, so we can add no shilling to our total.

Adding a sixpence allows $5 + 2 \tfrac 1 2 + 2 + \tfrac 1 2 = 10$, so we can add no sixpence to our total.

Adding a threepenny piece takes our total to $5 + 2 \tfrac 1 2 + 4 \times 2 + \tfrac 1 4 = 15 \tfrac 3 4$.

Adding another threepenny piece allows $5 + 2 \tfrac 1 2 + 2 + \tfrac 1 4 + \tfrac 1 4 = 10$, so we can add no further threepenny piece to our total.

We cannot remove larger coins and replace them with combinations of smaller ones, as this will always allow $10$ to be made.

Hence the highest total we can make with these coins so as not to make $10 \, \mathrm s.$ is $15 \tfrac 3 4 \, \mathrm s.$ or $15 \, \mathrm s. 9 \, \mathrm d.$