Henry Ernest Dudeney/Modern Puzzles/4 - Loose Cash/Solution
Modern Puzzles by Henry Ernest Dudeney: $4$
- 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 $15 \shillings 9 \oldpence$, composed of:
- one crown and one half-crown (or $3$ half-crowns)
- $4$ florins
- one threepenny bit.
Proof
Recall the list of silver coins in the time of Dudeney, 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 bit: $\tfrac 1 4 \shillings$
- The sixpence: $\tfrac 1 2 \shillings$
- The shilling
- The florin: $2 \shillings$
- The half-crown: $2 \tfrac 1 2 \shillings$
- The crown: $5 \shillings$
- The half sovereign: $10 \shillings$
Dudeney'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 bit takes our total to $5 + 2 \tfrac 1 2 + 4 \times 2 + \tfrac 1 4 = 15 \tfrac 3 4$.
Adding another threepenny bit allows $5 + 2 \tfrac 1 2 + 2 + \tfrac 1 4 + \tfrac 1 4 = 10$, so we can add no further threepenny bit 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 $10 \shillings$ is $15 \tfrac 3 4 \shillings$ or $15 \shillings 9 \oldpence$
$\blacksquare$
Sources
- 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $4$. -- Loose Cash