nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Puzzles and Curious Problems/222 - Folding a Pentagon

== by : $222$ ==
;Folding a Pentagon
:Given a ribbon of paper, as in the diagram, of any length -- say more than $4$ times as long as broad --
::it can all be folded into a perfect pentagon,
::with every part lying within the boundaries of the figure.
:
:The only condition is that the angle $ABC$ must be the correct angle of two contiguous sides of a regular pentagon.
:How are you to fold it?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $222$. -- Folding a Pentagon
* : Geometrical Problems: Paper Folding Puzzles: $365$. Strip to Pentagon
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/222 - Folding a Pentagon/Solution

== by : $222$ ==
;Folding a Pentagon
== Solution ==
:
By folding $A$ over, find $C$, so that $BC = AB$.
Then fold as in Figure $1$ across the point $A$.
This will give point $D$.
Fold as in Figure $2$, making the edge of the ribbon lie along $AB$.
This will give point $E$.
Continue the fold as in Figure $3$, and so on, until all the ribbon lies in the pentagon.
== Also see ==
* Henry Ernest Dudeney: Modern Puzzles $120$ - The Ribbon Pentagon
== Sources ==
* : Solutions: $222$. -- Folding a Pentagon
* : Answers: $365$. Strip to Pentagon
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/223 - The Tower of Pisa

== by : $223$ ==
;The Tower of Pisa
:Suppose you were on the top of the Tower of Pisa, at a point where it leans exactly $179$ feet above the ground.
:Suppose you were to drop an elastic ball from there such that on each rebound it rose exactly one-tenth of the height from which it fell.
:What distance would the ball travel before it came to rest?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $223$. -- The Tower of Pisa
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $241$. The Tower of Pisa
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/223 - The Tower of Pisa/Solution

== by : $223$ ==
;The Tower of Pisa
== Solution ==
:$218$ feet $9 \tfrac 1 3$ inches.
== Proof ==
Let $d$ feet be the distance travelled.
First the ball falls $179$ feet.
After the first bounce, it goes up again $\dfrac {179} {10}$ feet and down again the same distance.
Similarly, after the second bounce, it goes up again $\dfrac {179} {10^2}$ feet and down again the same distance.
After the $n$th bounce, it goes up again $\dfrac {179} {10^n}$ feet and down again the same distance.
So the total distance travelled is given by:
:$\ds d = 179 + 2 \times \dfrac {179} {10} \sum_{n \mathop \ge 0} \paren {\dfrac 1 {10} }^n$
which from Sum of Infinite Geometric Sequence gives:
The answer given follows after we recall that $1$ foot is $12$ inches.
== Sources ==
* : Solutions: $223$. -- The Tower of Pisa
* : Answers: $241$. The Tower of Pisa
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/224 - The Tank Puzzle

== by : $224$ ==
;The Tank Puzzle
:The area of the floor of a tank is $6$ square feet,
::the water in it is $9$ inches deep.
:$(1)$ How much does the water rise if a $1$ foot metal cube is put in it,
:$(2)$ How much farther does it rise if another cube like it is put in by its side?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $224$. -- The Tank Puzzle
* : Geometrical Problems: Solid Geometry Puzzles: $320$. The Tank Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/224 - The Tank Puzzle/Solution

== by : $224$ ==
;The Tank Puzzle
== Solution ==
:$(1): \quad 1 \tfrac 4 5 = 1.8$ inches
:$(2): \quad 2 \tfrac 1 5 = 2.2$ more inches.
== Proof ==
Let $d$ inches be the level that the water rises when the first cube is placed in it.
The volume of the water displaced by this cube equals $1$ square foot multiplied by $9$ inches.
This fills a volume $6$ square feet less the $1$ square foot the cube pokes up through the surface multiplied by $d$ inches.
That is:
:$\paren {6 - 1} \times d = 1 \times 9$
and so $d = \dfrac 9 5 = 1 \tfrac 4 5 = 1.8$ inches.
The water is now $9 + 1 \tfrac 4 5 = 10 \tfrac 4 5$ inches deep.
Now let the second cube is placed in the tank.
We have seen from the first part that the water level will rise at least (in fact, more than) $1 \tfrac 4 5$ inches, while $1 \tfrac 1 5$ inches will take it over the top of both cubes.
Hence a different model is needed.
The total volume of the water in the tank is $6$ square feet times $9$ inches.
This is equal to the total volume of the water with two cubes in it, which are now completely submerged.
Let $d$ inches be the additional depth of water above the surface of those two cubes.
Thus the total volume of the water in the tank is $4$ square feet multiplied by $12$ inches added to $6$ square feet multiplied by $d$ inches.
That is:
:$6 \times 9 = 4 \times 12 + 6 d$
which leads to $d = 1$.
So $d$ is $1$ inch above the top of the cubes, which is $4$ inches higher than its original depth.
Hence the water rose an addition $4 - 1 \tfrac 4 5$ inches, which is $2 \tfrac 1 5$ inches.
== Sources ==
* : Solutions: $224$. -- The Tank Puzzle
* : Answers: $320$. The Tank Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/225 - An Artist's Puzzle

== by : $225$ ==
;An Artist's Puzzle
:An artist wished to obtain a canvas for a painting which would allow for
::the picture itself occupying $72$ square inches,
::a margin of $4$ inches on top and on bottom,
::and $2$ inches on each side.
:What is the smallest dimensions possible for such a canvas?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $225$. -- An Artist's Puzzle
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $275$. An Artist's Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/225 - An Artist's Puzzle/Solution

== by : $225$ ==
;An Artist's Puzzle
== Solution ==
:$10$ inches wide by $20$ inches high.
The picture will then be $6$ inches by $12$ inches.
== Proof ==
Let $a$ inches be the width of the picture.
Let $A$ square inches be the total area of the canvas, including the picture and the borders.
We need to find $a$ such that $A$ is a minimum.
As the width is $a$, the height is $\dfrac {72} a$.
Hence:
To make $A$ a minimum, we must make its derivative $a$ equal to zero.
Hence:
which leads us to the answer that the canvas must be $6 + 4 = 10$ inches wide and $\dfrac {72} 6 + 8 = 20$ inches high.
== Sources ==
* : Solutions: $225$. -- An Artist's Puzzle
* : Answers: $275$. An Artist's Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/226 - The Circulating Motor-Car

== by : $226$ ==
;The Circulating Motor-Car
:A car was running on a circular track such that the outside wheels were going twice as fast as the inside ones.
:What was the length of the circumference described by the outer wheels?
:The wheels were $5$ feet apart at the axle-tree.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $226$. -- The Circulating Motor-Car
* : Geometrical Problems: Circle Puzzles: $290$. The Circling Car
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/226 - The Circulating Motor-Car/Solution

== by : $226$ ==
;The Circulating Motor-Car
== Solution ==
Approximately $62.832$ feet.
== Proof ==
As the outer wheels are going twice as fast as the inner wheels, they are travelling along a circle of twice the circumference.
The circumference is propportional to the radius.
It follows that the radius of the outer circle is twice $5$ feet, that is, $10$ feet.
From Perimeter of Circle that is $2 \times \pi \times 10$ which is approximately $62.832$ feet.
== Sources ==
* : Solutions: $226$. -- The Circulating Motor-Car
* : Answers: $290$. The Circling Car
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/227 - A Match-Boarding Order

== by : $227$ ==
;A Match-Boarding Order
:A man gave an order for $297$ feet of match-boarding of usual width and thickness.
:There were to be $16$ pieces, all measuring an exact number of feet -- no fractions of a foot.
:He required $8$ pieces of the greatest length, the remaining pieces to be $1$ foot, $2$ feet, or $3$ feet shorter than the greatest length.
:How was the order carried out?
:Suppose that the eight of greatest length were $15$ feet long, then the others must be made of pieces $14$ Definition:Foot (Linear Measure)foot, $13$ foot or $12$ foot lengths,
::though every one of those lengths need not be represented.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $227$. -- A Match-Boarding Order
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $242$. A Matchboarding Order
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/227 - A Match-Boarding Order/Solution

== by : $227$ ==
;A Match-Boarding Order
== Solution ==
:$8$ pieces of $20$ feet
:$1$ piece of $18$ feet
:$7$ pieces of $17$ feet.
== Sources ==
* : Solutions: $227$. -- A Match-Boarding Order
* : Answers: $242$. A Matchboarding Order
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/228 - The Ladder

== by : $228$ ==
;The Ladder
:A ladder was fastened on end against a high wall of a building.
:It was unfastened and pulled out $4$ yards at the bottom.
:It was then found that the ladder had descended just one-fifth of the length of the ladder.
:What was the length of the ladder?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $228$. -- The Ladder
* : Geometrical Problems: Solid Geometry Puzzles: $308$. The Ladder
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/228 - The Ladder/Solution

== by : $228$ ==
;The Ladder
== Solution ==
:$6 \tfrac 2 3$ yards.
== Proof ==
Let the length of the ladder be $L$ yards.
The ladder, the wall and the ground form a right triangle.
Hence by Pythagoras's Theorem:
== Sources ==
* : Solutions: $228$. -- The Ladder
* : Answers: $308$. The Ladder
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/229 - Geometrical Progression

== by : $229$ ==
;Geometrical Progression
:Write out a series of whole numbers in geometrical progression, starting from $1$,
::so that the numbers should add up to a square.
:Thus, $1 + 2 + 4 + 8 + 16 + 32 = 63$.
:But this is one short of being a square.
:I am only aware of two answers in whole numbers, and these will be found easy to discover.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $229$. -- Geometrical Progression
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $243$. Geometrical Progression
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/229 - Geometrical Progression/Solution

== by : $229$ ==
;Geometrical Progression
== Solution ==
:$121 = 3^0 + 3^1 + 3^2 + 3^3 + 3^4 = 11^2$
:$400 = 7^0 + 7^1 + 7^2 + 7^3 = 20^2$
== Proof ==
This is explored in Square Numbers which are Sum of Consecutive Powers.
== Sources ==
* : Solutions: $229$. -- Geometrical Progression
* : Answers: $243$. Geometrical Progression
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/22 - The War Horse

== by : $22$ ==
;The War Horse
:Farmer Wurzel bought a old war horse for $\pounds 13$ and sold it later for $\pounds 30$.
:After having paid for its keep, it turned out he lost half the price he paid and one-quarter the cost of his keep.
:How much did Farmer Wurzel lose on the transaction?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $22$. -- The War Horse
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/22 - The War Horse/Solution

== by : $22$ ==
;The War Horse
== Solution ==
:$\pounds 14, 6 \shillings 8 \oldpence$
== Proof ==
Let $L$ be the loss made on the warhorse in pounds.
Let $K$ be the cost of the upkeep of the warhorse in pounds.
Then we have:
Converting the result into pounds and shillings reveals the solution.
== Sources ==
* : Solutions: $22$. -- The War Horse
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/230 - In a Garden

== by : $230$ ==
;In a Garden
:Consider a rectangular flower-bed.
:If it were $2$ feet broader and $3$ feet longer, it would have been $64$ square feet larger;
::if it were $3$ feet broader and $2$ feet longer, it would have been $68$ square feet larger.
:What is its length and breadth?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $230$. -- In a Garden
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $276$. In a Garden
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/230 - In a Garden/Solution

== by : $230$ ==
;In a Garden
== Solution ==
The flower bed is $10$ feet broad and $14$ feet long.
== Proof ==
Let $a$ and $b$ feet be the length and breadth respectively of the flower bed.
We have:
Hence the result.
== Sources ==
* : Solutions: $230$. -- In a Garden
* : Answers: $276$. In a Garden
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/231 - The Rose Garden

== by : $231$ ==
;The Rose Garden
:A man has a rectangular garden, and wants to make exactly half of it into a large bed of roses,
::with a gravel path of uniform width round it.
:Can you find a general rule that will apply equally to any rectangular garden, whatever its proportions?
:All the measurements must be made in the garden.
:A plain ribbon, no shorter than the length of the garden, is all the material required.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $231$. -- The Rose Garden
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $279$. The Rose Garden
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/231 - The Rose Garden/Solution

== by : $231$ ==
;The Rose Garden
:A man has a rectangular garden, and wants to make exactly half of it into a large bed of roses,
::with a gravel path of uniform width round it.
:Can you find a general rule that will apply equally to any rectangular garden, whatever its proportions?
:All the measurements must be made in the garden.
:A plain ribbon, no shorter than the length of the garden, is all the material required.
== Solution ==
:
Construct $AD$ one quarter the length of $AB$.
Construct $AF$ and $DE$ one quarter the length of $BC$.
Construct $EG = DF$.
Then $AG$ is the required width of the path.
== Proof ==
Let the length and breadth of the garden be $a$ and $b$.
Let $c$ be the width of the path.
We have:
It is apparent that it is the negative square root of $\sqrt {a^2 + b^2}$ we need in the above, as the positive one would result in the path being wider than the garden.
Hence:
:$c = \dfrac {a + b - \sqrt {a^2 + b^2} } 4$
Now we investigate the geometry.
Let $a = AB$ and $b = BC$.
We have that:
and it is seen that $AG$ is the same as what was calculated algebraically above.
== Sources ==
* : Solutions: $231$. -- The Rose Garden
* : Answers: $279$. The Rose Garden
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/232 - A Pavement Puzzle

== by : $232$ ==
;A Pavement Puzzle
:Two square floors had to be paved with stones each $1$ foot square.
:The number of stones in both together was $2120$, but each side of one floor was $12$ feet more than each side of the other floor.
:What were the dimensions of the two floors?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $232$. -- A Pavement Puzzle
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $244$. A Pavement Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/232 - A Pavement Puzzle/Solution

== by : $232$ ==
;A Pavement Puzzle
== Solution ==
:$26$ and $38$ feet on one side.
== Proof ==
Let $a$ and $b$ feet be the lengths of the sides of the floors.
We have:
As it is the positive root we want here, we have that $a = 26$ and hence $b = 26 + 12 = 38$.
== Sources ==
* : Solutions: $232$. -- A Pavement Puzzle
* : Answers: $244$. A Pavement Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/233 - The Nougat Puzzle

== by : $233$ ==
;The Nougat Puzzle
:A block of nougat is $16$ inches long, $8$ inches wide, and $7 \tfrac 1 2$ inches deep.
:What is the greatest number of pieces that I can cut from it measuring $5$ inches by $3$ inches by $2 \tfrac 1 2$ inches?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $233$. -- The Nougat Puzzle
* : Geometrical Problems: Solid Geometry Puzzles: $321$. The Nougat Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/233 - The Nougat Puzzle/Solution

== by : $233$ ==
;The Nougat Puzzle
== Solution ==
$24$ pieces.
== Proof ==
Cut off and throw away a piece $1$ inch thick off the end, leaving a block $15$ inches long.
Then cut the block into two pieces:
:$15$ inches long, $5$ inches wide, and $7 \tfrac 1 2$ inches deep
:$15$ inches long, $3$ inches wide, and $7 \tfrac 1 2$ inches deep.
The former can be cut into $15$ piece of nougat $5$ inches by $3$ inches by $2 \tfrac 1 2$ inches.
The latter can be cut into $9$ piece of nougat $5$ inches by $3$ inches by $2 \tfrac 1 2$ inches.
== Sources ==
* : Solutions: $233$. -- The Nougat Puzzle
* : Answers: $321$. The Nougat Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/234 - Pile Driving

== by : $234$ ==
;Pile Driving
:During some bridge-building operations a pile was being driven into the bed of the river.
:A foreman remarked that at high water a quarter of the pile was embedded in the mud,
::one-third was under water,
::and $17$ feet $6$ inches above water.
:What was the length of the pile?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $234$. -- Pile Driving
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/234 - Pile Driving/Solution

== by : $234$ ==
;Pile Driving
== Solution ==
:$26$ feet $3$ inches.
== Proof ==
Let $L$ feet be the length of the pile.
You may be tempted to do this:
No no no!
The part of the pile buried in the mud is also under water.
So:
and it is seen that the proportion of the pile which is buried in mud is no more than a distraction.
== Sources ==
* : Solutions: $234$. -- Pile Driving
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/235 - An Easter Egg Problem

== by : $235$ ==
;An Easter Egg Problem
:I have an easter egg exactly $3$ inches in length, and $3$ other eggs all similar in shape,
::having together the same contents as the large egg.
:Can you tell me the exact measurements for the lengths of the three smaller ones?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $235$. -- An Easter Egg Problem
* : Geometrical Problems: Solid Geometry Puzzles: $322$. An Easter Egg Problem
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/235 - An Easter Egg Problem/Solution

== by : $235$ ==
;An Easter Egg Problem
:I have an easter egg exactly $3$ inches in length, and $3$ other eggs all similar in shape,
::having together the same contents as the large egg.
:Can you tell me the exact measurements for the lengths of the three smaller ones?
== Solution ==
There is an infinite number of solutions.
From Cubes which are Sum of Three Cubes, we can use any of the sequence:
:$6^3, 9^3, 12^3, 18^3, 19^3, 20^3, 24^3, 25^3, \ldots$
For example, take $n = 6$.
Then:
and so:
giving the solution $1 \tfrac 1 2$ inches, $2$ inches and $2 \tfrac 1 2$ inches
Or for another example, take $n = 9$.
Then:
and so:
giving the solution $\tfrac 1 3$ inches, $2$ inches and $2 \tfrac 2 3$ inches.
These are the two solutions provided by .
== Sources ==
* : Solutions: $235$. -- An Easter Egg Problem
* : Answers: $322$. An Easter Egg Problem
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/236 - The Pedestal Puzzle

== by : $236$ ==
;The Pedestal Puzzle
:A man had a block of wood measuring $3$ feet by $1$ foot by $1$ foot,
::which he gave to a wood-turner with instructions to turn from it a pedestal,
::saying that he would pay him a certain sum for every cubic inch of wood taken from the block in the process of turning.
:The ingenious turner weighed the block and found it to contain $30$ pounds.
:After he had finished the pedestal it was again weighed, and found to contain $20$ pounds.
:As the original block contained $3$ cubic feet, and it had lost just one-third of its weight,
::the turner asked payment for $1$ cubic foot.
:But the gentleman objected, saying that the heart of the wood might be heavier or lighter than the outside.
:How did the ingenious turned contrive to convince his customer that he had taken not more and not less than $1$ cubic foot from the block?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $236$. -- The Pedestal Puzzle
* : Geometrical Problems: Solid Geometry Puzzles: $323$. The Pedestal Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/236 - The Pedestal Puzzle/Solution

== by : $236$ ==
;The Pedestal Puzzle
== Solution ==
The obvious answer is that he dunked it into a bath full of water and measured the runoff.
s' answer is similar but more unwieldy:
:The man made a box $3$ feet by $1$ foot by $1$ foot inside,
::and into this he placed the pedestal.
:Then he filled the box with fine dry sand, shaking it down and levelling the top.
:Then he took out the pedestal, and the sand was shaken down and levelled,
::when the surface was found to be exactly $2$ feet from the top of the box.
And so on.
== Sources ==
* : Solutions: $236$. -- The Pedestal Puzzle
* : Answers: $323$. The Pedestal Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/237 - The Mudbury War Memorial

== by : $237$ ==
;The Mudbury War Memorial
:The inhabitants of Mudbury recently erected a war memorial,
::and they proposed to enclose a piece of ground on which it stands with posts.
:They found that if they set up the posts $1$ foot asunder they would have too few by $150$.
:But if they placed them a yard asunder there would be too many by $70$.
:How many posts had they in hand?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $237$. -- The Mudbury War Memorial
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $245$. The Mudbury War Memorial
Category:Henry Ernest Dudeney/Puzzles and Curious Problems