nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Modern Puzzles/134 - A Fence Problem

== by : $134$ ==
;A Fence Problem
:This is a problem that is very frequently brought to my notice in various forms.
:It is generally difficult, but in the form in which I present it it should be easy to the cunning solver.
:A man has a square field, $60 \ \mathrm {ft.}$ by $60 \ \mathrm {ft.}$, with other property, adjoining the highway.
:For some reason he put up a straight fence in the line of three trees, as shown,
::and the length of fence from the middle tree to the tree on the road was just $91$ feet.
:
:What is the distance in exact feet from the middle tree to the gate on the road?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $134$. -- A Fence Problem
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $264$. A Fence Problem
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/134 - A Fence Problem/Solution

== by : $134$ ==
;A Fence Problem
== Solution ==
$35$ feet.
== Proof ==
:
Let $x$ be the distance required.
Let $y$ be the distance marked on the diagram.
We have:
The above leads to a quartic equation which is irksome to solve.
However, we are told that $x$ is an exact number of feet.
We have that $91^2$ is the sum of two square numbers in exactly one way:
Inserting $x = 35$ and $y = 84$ into $2$ proves that these numbers are consistent with the premises.
Hence the result.
== Sources ==
* : Solutions: $134$. -- A Fence Problem
* : Answers: $264$. A Fence Problem
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/135 - The Domino Swastika

== by : $135$ ==
;The Domino Swastika
:Here is a little puzzle by Mr. .
:Form a square frame with twelve dominoes, as shown in the illustration.
:
:Now, with only four extra dominoes, form within the frame a swastika.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $135$. -- The Domino Swastika
* : Domino Puzzles: $479$. The Domino Swastika
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Dominoes
Henry Ernest Dudeney/Modern Puzzles/135 - The Domino Swastika/Solution

== by : $135$ ==
;The Domino Swastika
== Solution ==
:
== Sources ==
* : Solutions: $135$. -- The Domino Swastika
* : Answers: $479$. The Domino Swastika
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/136 - A New Match Puzzle

== by : $136$ ==
;A New Match Puzzle
:I have a box of matches.
:I find that I can form with them any given pair of these four regular figures, using all the matches every time.
:This, if there were eleven matches, I could form with them, as shown, the triangle and pentagon
::or the pentagon and hexagon, or the square and triangle (by using only three matches in the triangle);
::but could not with eleven matches form the triangle and hexagon,
::or the square and pentagon, or the square and hexagon.
:
:Of course there must be the same number of matches in every side of a figure.
:Now, what is the smallest number of matches I can have in the box?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $136$. -- A New Match Puzzle
* : Match Puzzles: $494$. A New Match Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/136 - A New Match Puzzle/Solution

== by : $136$ ==
;A New Match Puzzle
== Solution ==
$36$ matches.
== Proof ==
We can form:
:the triangle and the square with $12$ and $24$ respectively
:the triangle and the pentagon with $6$ and $30$ respectively
:the triangle and the hexagon with $6$ and $30$ respectively
:the square and the pentagon with $16$ and $20$ respectively
:the square and the hexagon with $12$ and $24$ respectively
:the pentagon and the hexagon with $30$ and $6$ respectively.
There are different arrangements for all except the $4$th and $6$th of these.
The triangle and hexagon require a number divisible by $3$.
The square and the hexagon require an even number.
Therefore the number must be divisible by $6$.
But this condition cannot be fulfilled for the pentagon and hexagon with less than $36$ matches.
== Sources ==
* : Solutions: $136$. -- A New Match Puzzle
* : Answers: $494$. A New Match Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/137 - Hurdles and Sheep

== by : $137$ ==
;Hurdles and Sheep
:This is a little puzzle that you can try with matches.
:A farmer says that four of his hurdles will form a square enclosure just sufficient for one sheep.
:That being so, what is the smallest number of hurdles that he will require for enclosing ten sheep?
:
:Everything depends upon the shape of your enclosure.
:The only other way of placing four matches (or hurdles) in $A$ is to form a diamond-shaped figure,
::and the more attenuated this diamond becomes the smaller will be its area, until the sides meet,
::when there will be no area enclosed at all.
:If you place six matches, as in $B$, you will have room for $2$ sheep.
:But if you place them as in $C$, you will have room for one sheep, as seven-tenths of a sheep will only exist as mutton.
:And if you place them as $D$, you can still only accommodate two sheep, which is the maximum for six hurdles.
:Now, how many hurdles do you require for ten sheep?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $137$. -- Hurdles and Sheep
* : Match Puzzles: $495$. Hurdles and Sheep
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/137 - Hurdles and Sheep/Solution

== by : $137$ ==
;Hurdles and Sheep
== Solution ==
$12$ hurdles.
== Proof ==
From the problem definition, an area of $1$ hurdle length squared holds $1$ sheep.
Hence the problem is to find the largest area enclosed by a given number of hurdles.
This is that enclosed by a regular polygon.
Let $A_n$ be the area enclosed by a regular $n$-gon.
From Area of Regular Polygon:
:$A_n = \dfrac n 4 \cot \dfrac \pi n$
Hence:
Hence the result.
== Sources ==
* : Solutions: $137$. -- Hurdles and Sheep
* : Answers: $495$. Hurdles and Sheep
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/138 - The Four Draughtsmen

== by : $138$ ==
;The Four Draughtsmen
:Here is a queer new puzzle that I know will interest my readers considerably.
:The four are shown exactly as they stood on a square chequered board --
::not necessarily eight squares by eight --
::but the ink with which the board was drawn was evanescent,
::so that all the diagram except the men has disappeared.
:How many squares were there in the board and how am I to reconstruct it?
:I know that each man stood in the middle of a square,
::one on the edge of each side of the board and no man in a corner.
:
:It is a real puzzle, until you hit on the method of solution.
::and then to get the correct answer is absurdly easy.
=== Click here for solution ===
== Linguistic Note ==
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $138$. -- The Four Draughtsmen
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $265$. The Four Checkers
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/138 - The Four Draughtsmen/Solution

== by : $138$ ==
;The Four Draughtsmen
== Solution ==
:
Construct $AD$.
Construct $CE$ perpendicular to $AD$ and make $CE = AD$.
Because $AD$ stretches from one side of the board to its opposite, joining the centres of two squares, then the same applies to $CE$.
So $E$ is the centre of a square on the same row as $B$.
So join $EB$ and produce it in both directions.
Then:
:construct $FG$ through $C$ parallel to $EB$
:construct $FH$ through $A$ perpendicular to $EB$ and $FG$
:construct $DG$ through $D$ parallel to $FH$.
As $H$ is the centre of a square on a board, we can mark $HE$ all round the square $HFG$ and discover the board is $10 \times 10$.
It remains to construct the actual board itself.
== Sources ==
* : Solutions: $138$. -- The Four Draughtsmen
* : Answers: $265$. The Four Checkers
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/139 - A Crease Problem

== by : $139$ ==
;A Crease Problem
:Fold a page, so that the bottom outside corner touches the inside edge and the crease is the shortest possible.
:That is about as simple a question as we could put,
::but it will puzzle a good many readers to discover just where to make the fold.
:I give two examples of folding.
:It will be seen that the crease $AB$ is considerably longer than $CD$, but the latter is not the shortest possible.
:
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $139$. -- A Crease Problem
* : Geometrical Problems: Paper Folding Puzzles: $366$. A Crease Problem
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/139 - A Crease Problem/Solution

== by : $139$ ==
;A Crease Problem
== Solution ==
:
Bisect $AB$ at $C$ and construct $CG$ parallel to $BH$.
Bisect $AC$ at $D$ and construct the semicircle between $B$ and $D$.
Let the semicircle $BD$ intersect $CG$ at $E$.
Let $DE$ be produced to intersect $BH$ at $F$.
Then $DF$ is the required crease.
== Proof ==
First we assume that the length of the page is sufficiently long.
Otherwise simply folding the page in half could produce a shorter crease.
Suppose $t$ of the length of the bottom edge is folded over.
Then $1 - t$ of the length of the bottom edge is not folded over.
The angle adjacent to the bottom edge of the small triangle produced after folding is therefore:
:$\cos^{-1} \paren {\dfrac {1 - t} t}$
The angle of the crease with respect to the bottom edge would be:
:$\dfrac 1 2 \paren {180^\circ - \cos^{-1} \paren {\dfrac {1 - t} t} } = 90^\circ - \dfrac 1 2 \cos^{-1} \paren {\dfrac {1 - t} t}$
The length of the crease is therefore:
of the length of the bottom edge.
Let $\map f t = \dfrac {2 t^3} {2 t - 1}$.
Then $\map {f'} t = \dfrac {6 t^2 \paren {2 t - 1} - 2 \paren {2 t^3} } {\paren {2 t - 1}^2} = \dfrac {8 t^3 - 6 t^2} {\paren {2 t - 1}^2}$
$\map {f'} t = 0$ when $t = 0$ or $t = \dfrac 3 4$.
Hence the maximum length of the crease occurs when $t = \dfrac 3 4$, which is shown in the construction in the solution.
== Sources ==
* : Solutions: $139$. -- A Crease Problem
* : Answers: $366$. A Crease Problem
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/13 - Find the Coins

== by : $13$ ==
;Find the Coins
:Three men, Abel, Best and Crewe, possessed money, all in silver coins.
:Abel had one coin fewer than Best and one more than Crewe.
:Abel gave Best and Crewe as much money as they already had,
::then Best gave Abel and Crewe the same amount of money as they they held,
::and finally Crewe gave Abel and Best as much money as they then had.
:Each man then held exactly $10$ shillings.
:To find what amount each man started with is not difficult.
:But the sting of the puzzle is in the tail.
:Each man held exactly the same coins (the fewest possible) amounting to $10$ shillings.
:What were the coins and how were they originally distributed?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $13$. -- Find the Coins
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Henry Ernest Dudeney/Modern Puzzles/13 - Find the Coins
Henry Ernest Dudeney/Modern Puzzles/13 - Find the Coins/Working

== Working for by : $13$ -- Find the Coins ==
The simultaneous equations in matrix form:
:$\begin {bmatrix} 4 & -4 & -4 \\ -2 & 6 & -2 \\ -1 & -1 & 7 \end {bmatrix} \begin {bmatrix} a \\ b \\ c \end {bmatrix} = \begin {bmatrix} 120 \\ 120 \\ 120 \end {bmatrix}$
when converted to reduced echelon form, gives:
:$\begin {bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {bmatrix} \begin {bmatrix} a \\ b \\ c \end {bmatrix} = \begin {bmatrix} 195 \\ 105 \\ 60 \end {bmatrix}$
== Proof ==
Category:Henry Ernest Dudeney/Modern Puzzles/13 - Find the Coins
Henry Ernest Dudeney/Modern Puzzles/140 - The Four-Colour Map Theorem

== by : $140$ ==
;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.
=== Click here for solution ===
== Also see ==
* Four Color Theorem
== Historical Note ==
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $140$. -- The Four-Colour Map Theorem
* : Combinatorial & Topological Problems: Map Coloring Puzzles: $442$. The Four-Color Map Theorem
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Henry Ernest Dudeney/Modern Puzzles/140 - The Four-Colour Map Theorem
Category:Four Color Theorem
Henry Ernest Dudeney/Modern Puzzles/140 - The Four-Colour Map Theorem/Historical Note

== Historical Note on 's Modern Puzzles No. $140$ - The Four-Colour Map Theorem ==
:For just about $50$ years various mathematicians, including , , , , , , , and many others have attempted to prove the truth of this theorem,
::and in a long and learned article in the '' for July-August, $1923$, Professor , of the University of Illinois, states that "the problem is still unsolved."
, in his $1968$ repackaging , refutes 's claim to have proved it.
He goes on to advertise his own contribution in his $1966$ .
== Sources ==
* 
* : Geometrical Problems: Various Geometrical Puzzles: $140$. -- The Four-Colour Map Theorem
* : Combinatorial & Topological Problems: Map Coloring Puzzles: $442$. The Four-Color Map Theorem
* : Answers: $442$. The Four-Color Map Theorem
Category:Historical Notes
Category:Henry Ernest Dudeney/Modern Puzzles/140 - The Four-Colour Map Theorem
Henry Ernest Dudeney/Modern Puzzles/140 - The Four-Colour Map Theorem/Mistake

== Source Work ==
:
:Geometrical Problems
::Various Geometrical Puzzles
:::$140$ -- The Four-Colour Map Theorem
:
:Combinatorial & Topological Problems
::Map Coloring Puzzles
:::$442$. The Four-Color Map Theorem
== Mistake ==
== Correction ==
's analysis of this problem does not constitute a proof.
 explains it in his $1968$ repackaging as follows:
=== Refutation ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $140$. -- The Four-Colour Map Theorem
* : Combinatorial & Topological Problems: Map Coloring Puzzles: $442$. The Four-Color Map Theorem
Category:Mistakes/Modern Puzzles
Category:Mistakes/536 Puzzles & Curious Problems
Henry Ernest Dudeney/Modern Puzzles/140 - The Four-Colour Map Theorem/Refutation

== Refutation of by : $140$ - The Four-Colour Map Theorem ==
;The Four-Colour Map Theorem
== Solution ==
== Refutation ==
 correctly shows that no more than $4$ regions can be drawn so that each borders all the others.
However, his proof fails to show that four colors are sufficient for all maps.
It is indeed true that if any four regions of a map are considered in isolation, then no fifth color is necessary for any fifth region.
What is required, however, is a proof that on a map with a large number of regions, these various sets of five will never conflict with each other in a way that five colors are demanded.
The difficult is best seen by actually constructing a complicated map, using the step by step procedure proposed by .
Suppose that each new region is drawn so as to border three others.
Then its color is determined automatically, and four-color maps can indeed be extended indefinitely.
However, suppose that many new regions are added that touch either one, two, or even no previously drawn regions.
Then the choice of colors for these regions suddenly becomes arbitrary.
As the map grows in size and complexity, it is suddenly discovered that it is possible to add a new region that will require a fifth color.
By backtracking and altering previous decisions as to what color to be used, it appears that it is always possible to correct the mistake and accommodate the new region without needing that fifth color.
But using this process it cannot be certain that it is always possible.
== Historical Note ==
== Sources ==
* : Answers: $442$. The Four-Color Map Theorem
Category:Henry Ernest Dudeney/Modern Puzzles/140 - The Four-Colour Map Theorem
Category:Four Color Theorem
Henry Ernest Dudeney/Modern Puzzles/141 - The Six Submarines

== by : $141$ ==
;The Six Submarines
:If five submarines, sunk on the same day, all went down at the same spot where another had previously been sunk,
::how might they all lie at rest so that every one of the six U-boats should touch every other one?
:To simplify we will say, place six ordinary wooden matches so that every match shall touch every other match.
:No bending or breaking allowed.
=== Click here for solution ===
== Also see ==
* Seven Touching Cylinders
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $141$. -- The Six Submarines
* : Geometrical Problems: Solid Geometry Puzzles: $312$. The Six Submarines
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/141 - The Six Submarines/Historical Note

== Historical Note on by : $141$ - The Six Submarines ==
, in his $1968$ repackaging , references his own article in his $1959$ collection in which the question is answered in Seven Touching Cylinders, with $7$ appropriately-dimensioned cylinders.
== Sources ==
* : Answers: $312$. The Six Submarines
Category:Historical Notes
Henry Ernest Dudeney/Modern Puzzles/141 - The Six Submarines/Solution

== by : $141$ ==
;The Six Submarines
== Solution ==
:
In fact, this has been simplified even further, and the submarines have been modelled as simple cylinders.
== Historical Note ==
== Sources ==
* : Solutions: $141$. -- The Six Submarines
* : Answers: $312$. The Six Submarines
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/142 - Economy in String

== by : $142$ ==
;Economy in String
:Owing to the scarcity of string a lady found herself in this dilemma.
:In making up a parcel for her son, she was limited to using $12$ feet of string, exclusive of knots,
::which passed round the parcel once lengthways and twice round its girth, as shown in the illustration.
:
:What was the largest rectangular parcel that she could make up, subject to these conditions?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $142$. -- Economy in String
* : Geometrical Problems: Solid Geometry Puzzles: $313$. Economy in String
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Henry Ernest Dudeney/Modern Puzzles/142 - Economy in String
Henry Ernest Dudeney/Modern Puzzles/142 - Economy in String/General Solution

== by : $142$ ==
;Economy in String
=== General Result ===
Let the string pass:
:$a$ times along length $x$
:$b$ times along breadth $y$
:$c$ times along depth $z$.
Let the string be length $m$.
Then the maximum volume $xyz$ of the parcel is given by:
:$x y z = \dfrac {m^2} {27 a b c}$
where:
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Solutions: $142$. -- Economy in String
* : Answers: $313$. Economy in String
Category:Henry Ernest Dudeney/Modern Puzzles/142 - Economy in String
Henry Ernest Dudeney/Modern Puzzles/142 - Economy in String/General Solution/Proof 1

== by : $142$ ==
;Economy in String
=== General Result ===
== Proof ==
We have that:
:$a x + b y + c z = m$
The maximum area of $x y$ is found as follows:
Put:
Similarly also by differentiation $x$:
:$a x = \dfrac n 2$
and so:
:$a x = b y$
Similarly:
:$a x = b y = c z = \dfrac m 3$
and so:
Hence:
:$x y z = \dfrac {m^2} {27 a b c}$
== Historical Note ==
== Sources ==
* : Solutions: $142$. -- Economy in String
* : Answers: $313$. Economy in String
Category:Henry Ernest Dudeney/Modern Puzzles/142 - Economy in String
Henry Ernest Dudeney/Modern Puzzles/142 - Economy in String/General Solution/Proof 2

== by : $142$ ==
;Economy in String
=== General Result ===
== Proof ==
We have that:
:$a x + b y + c z = m$
Then:
Equality holds when $a x = b y = c z = \dfrac m 3$.
Category:Henry Ernest Dudeney/Modern Puzzles/142 - Economy in String
Henry Ernest Dudeney/Modern Puzzles/142 - Economy in String/Solution

== by : $142$ ==
;Economy in String
== Solution ==
$2$ feet by $1$ foot by $\tfrac 2 3$ feet.
=== General Solution ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Solutions: $142$. -- Economy in String
* : Answers: $313$. Economy in String
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Category:Henry Ernest Dudeney/Modern Puzzles/142 - Economy in String
Henry Ernest Dudeney/Modern Puzzles/142 - Economy in String/Solution/Proof 1

== by : $142$ ==
;Economy in String
== Solution ==
$2$ feet by $1$ foot by $\tfrac 2 3$ feet.
== Proof ==
From the general solution:
Thus it follows that the total length of string that passes along the length, breadth or depth must in every case be the same to allow of the maximum dimensions, that is: $4$ feet.
That is:
:$a x = b y = c z = \dfrac m 3$
The string passes $2$ times along length, $4$ times along breadth, and $6$ times along depth.
Therefore $4$ feet divided by $2$, $4$ and $6$ feet will give us $2$ feet, $1$ foot, and $\tfrac 2 3$ feet respectively for the length, breadth or depth of the largest possible parcel.
== Sources ==
* : Solutions: $142$. -- Economy in String
* : Anwsers: $313$. Economy in String
Category:Henry Ernest Dudeney/Modern Puzzles/142 - Economy in String
Henry Ernest Dudeney/Modern Puzzles/142 - Economy in String/Solution/Proof 2

== by : $142$ ==
;Economy in String
== Solution ==
$2$ feet by $1$ foot by $\tfrac 2 3$ feet.
== Proof ==
From the general solution:
Setting:
we obtain:
Hence the result.
== Sources ==
* : Solutions: $142$. -- Economy in String
* : Answers: $313$. Economy in String
Category:Henry Ernest Dudeney/Modern Puzzles/142 - Economy in String
Henry Ernest Dudeney/Modern Puzzles/143 - The Stone Pedestal

== by : $143$ ==
;The Stone Pedestal
:In laying the base and cubic pedestal for a certain public memorial,
::the stonemason used cubic blocks of stone all measuring one foot on each side.
:There was exactly the same number of these blocks (all uncut) in the pedestal as in the square base on the centre of which it stood.
:Look at the sketch and try to determine the total number of blocks actually used.
:
:The base is only a single block in depth.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $143$. -- The Stone Pedestal
* : Geometrical Problems: Solid Geometry Puzzles: $314$. The Stone Pedestal
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/143 - The Stone Pedestal/Solution

== by : $143$ ==
;The Stone Pedestal
== Solution ==
:$1458$ blocks.
== Proof ==
Ridiculously messy question.
Let $2 x$ be the number of blocks used in total.
Then we have that $x$ are used in the base and $x$ are used in the pedestal.
Let $y$ be the number of blocks on along one edge of the pedestal.
We have that:
:$x = y^3$
It looks as though the length of one side of the base is $3$ times the length of one edge of the pedestal.
On that assumption, we have that:
== Sources ==
* : Solutions: $143$. -- The Stone Pedestal
* : Answers: $314$. The Stone Pedestal
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions