Book:Henry Ernest Dudeney/Modern Puzzles/Errata
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Errata for 1926: Henry Ernest Dudeney: Modern Puzzles
$8$ -- Buying Buns
- Some children (there were as many boys as girls) were given sevenpence to spend on these buns, each receiving exactly alike.
- How many buns did each receive?
- Of course no buns were divided.
The solution offered:
- There must have been three boys and three girls,
- each of whom received two buns at three a penny
- and one bun at two a penny,
- the cost of which would be exactly sevenpence.
$14$ -- An Easy Settlement
- Three men, Andrews, Baker and Carey, sat down to play at some game.
- When they put their money on the table it was found that they each possessed $2$ coins only, making altogether $\pounds 1 \ 4 \shillings 6 \oldpence$.
- At the end of play Andrews had lost $5$ shillings and Carey had lost sixpence, and they all squared up by simply exchanging the coins.
- What were the exact coins that each held on rising from the table?
The solution offered:
- At the start of play
- Andrews held a half-sovereign and a shilling,
- Baker held a crown and a florin,
- and Carey held a double florin and a half-crown.
- After settlement,
- Andrews held double florin and florin,
- Baker the half-sovereign and half-crown,
- and Carey held crown and a shilling.
- Thus Andrews lost $5 \shillings$, Carey lost $6 \oldpence$, and Baker won $5 \shillings 6 \oldpence$.
- The selection of the coins is obvious, but their allotment requires a little judgment and trial.
$140$ -- The Four-Colour Map Theorem
- In colouring any map under the condition that no contiguous countries shall be coloured alike,
- not more than four colours can ever be necessary.
- Countries only touching at a point ... are not contiguous.
- I will give, in condensed form, a suggested proof of my own
- which several good mathematicians to whom I have shown it accept it as quite valid.
- Two others, for whose opinion I have great respect, think it fails for a reason that the former maintain will not "hold water".
- The proof is in a form that anybody can understand.
- It should be remembered that it is one thing to be convinced, as everybody is, that the thing is true,
- but quite another to give a rigid proof of it.
$177$ -- The Six-Pointed Star
- There are $37$ solutions in all, or $74$ if we count complementaries.
- $32$ of these are regular, and $5$ are irregular.
- Of the $37$ solutions, $6$ have their points summing to $26$. These are as follows:
- $\begin {array} {rrrrrrrrrrrr}
10 & 6 & 2 & 3 & 1 & 4 & 7 & 9 & 5 & 12 & 11 & 8 \\ 9 & 7 & 1 & 4 & 3 & 2 & 6 & 11 & 5 & 10 & 12 & 8 \\ 5 & 4 & 6 & 8 & 2 & 1 & 9 & 12 & 3 & 11 & 7 & 10 \\ 5 & 2 & 7 & 8 & 1 & 3 & 11 & 10 & 4 & 12 & 6 & 9 \\ 10 & 3 & 1 & 4 & 2 & 6 & 9 & 8 & 7 & 12 & 11 & 5 \\ 8 & 5 & 3 & 1 & 2 & 7 & 10 & 4 & 11 & 9 & 12 & 6 \\ \end {array}$
- ...
- Also note that where the $6$ points add to $24$, $26$, $30$, $32$, $34$, $36$ or $38$, the respective number of solutions is $3$, $6$, $2$, $4$, $7$, $6$ and $9$, making $37$ in all.
$178$ -- The Seven-Pointed Star
- There are $56$ different arrangements, counting complements.
- Class $\text I$ is those as above, where pairs in the positions $7 - 8$, $13 - 2$, $3 - 12$, $14 - 1$ all add to $15$, and there are $20$ such cases.
- Class $\text {II}$ includes cases where pairs in the positions $7 - 2$, $8 - 13$, $3 - 1$, $12 - 14$ all add to $15$, and there are $20$ such cases.
- Class $\text {III}$ includes cases where pairs in the positions $7 - 8$, $13 - 2$, $3 - 1$, $12 - 14$ all add to $15$, and there are $16$ such cases.