Book:Henry Ernest Dudeney/Modern Puzzles/Errata

Errata for 1926: Henry Ernest Dudeney: Modern Puzzles

$8$ -- Buying Buns

Buns were being sold at three prices:

one a penny,

two a penny,

and three a penny.

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.