Book:Henry Ernest Dudeney/536 Puzzles & Curious Problems/Errata

Errata for 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems

$5$ -- 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.

$147$ -- An Absolute Skeleton

It can soon be discovered that the divisor must be $312$,

that $9$ cannot be in the quotient because $9$ times the divisor contains a repeated figure.

We therefore know that the quotient contains all the figures $1$ to $8$ once, and the rest is comparatively easy.

$149$ -- Simple Division: Solution

Divide $4,971,636,104$ by $124,972$, and the quotient is $39,782$.

$233$ -- Mental Arithmetic: Solution

Calling the numbers $a$ and $b$, we have:

$a^2 + b^2 + a b = \Box = /a - m b/^2 = a^2 = 2 a m b + b^2 m^2$.

$\therefore b + a = -2 a m + b m^2$,

$\therefore b = \dfrac {a \paren {2 m + 1} } {m^2 - 1}$

in which $m$ may be any whole number greater than $1$, and $a$ is chosen to make $b$ rational.

$238$ -- More Curious Multiplication: Solution

The number is $987,654,321$, which, when multiplied by $18$, gives $17,777,777,778$, with $1$ and $8$ at the beginning and end.

And so on with the other multipliers, except $90$, where the product is $88,888,888,890$, with $90$ at the end.

$240$ -- Counting the Loss: Solution

The general solution of this is obtained from the indeterminate equation

$\dfrac {35 x - 48} {768}$

which must be an integer, where $x$ is the number of survivors.

$285$ -- The Counter Cross: Solution

There are $19$ different squares to be indicated.

Of these, nine will be of the size shown by the four $\text A$'s in the diagram, four of the size shown by the $\text B$'s, four of the size shown by the $\text C$'s, and two of the size shown by the $\text D$'s.

$289$ -- Squaring the Circle: Solution

The distance $DG$, added to the distance $GH$, gives a quarter of the length of the circumference, correct within a five-thousandth part.

$394$ -- 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.

$395$ -- 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.

$442$ -- The Four-Color 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.