Book:Henry Ernest Dudeney/536 Puzzles & Curious Problems/Errata
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Errata for 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems
$5$ -- 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.
$147$ -- 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.