nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Puzzles and Curious Problems/237 - The Mudbury War Memorial/Solution

== by : $237$ ==
;The Mudbury War Memorial
== Solution ==
They had $180$ posts.
The memorial ground was $110$ yards, that is $330$ feet, around.
== Proof ==
Recall there are $3$ feet to the yard.
Let $d$ yards be the length of the perimeter of the war memorial ground.
Let $p$ be the number of posts they have.
We have:
Hence the result.
== Sources ==
* : Solutions: $237$. -- The Mudbury War Memorial
* : Answers: $245$. The Mudbury War Memorial
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/238 - A Maypole Puzzle

== by : $238$ ==
;A Maypole Puzzle
:During a gale a maypole was broken in such a manner that it struck the level ground at a distance of $20$ feet from the base of the pole,
::where it entered the earth.
:It was repaired, and broken by the wind a second time at a point $5$ feet lower down,
::and struck the ground at a distance of $30$ feet from the base.
:What was the original height of the pole?
:In neither case did the broken part become actually detached.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $238$. -- A Maypole Puzzle
* : Geometrical Problems: Solid Geometry Puzzles: $309$. A Maypole Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/238 - A Maypole Puzzle/Solution

== by : $238$ ==
;A Maypole Puzzle
== Solution ==
:$50$ feet.
== Proof ==
Let $L$ feet be the length of the maypole.
Let $d$ feet be the distance above the ground of the point where first it snapped.
The maypole, the snapped-off part, and the ground form a right triangle.
Hence we can use Pythagoras's Theorem to obtain:
== Sources ==
* : Solutions: $238$. -- A Maypole Puzzle
* : Answers: $309$. A Maypole Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/239 - The Bell Rope

== by : $239$ ==
;The Bell Rope
:A bell rope, passing through the ceiling above, just touches the belfry floor,
::and when you pull the rope to the wall, keeping the rope taut, it touches a point just $3$ inches above the floor,
::and the wall was $4$ feet from the rope, when it hung at rest.
:How long was the rope from floor to ceiling?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $239$. -- The Bell Rope
* : Geometrical Problems: Solid Geometry Puzzles: $310$. The Bell Rope
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/239 - The Bell Rope/Solution

== by : $239$ ==
;The Bell Rope
== Solution ==
:$32$ feet and $1 \tfrac 1 2$ inches.
== Proof ==
Let $L$ inches be the length of the bell-rope.
Thus we have described a right triangle:
:whose legs are $4 \times 12 = 48$ and $L - 3$ (as there are $12$ inches to the foot)
:and whose hypotenuse is $L$.
Thus:
 
The result follows on converting $385 \tfrac 1 2$ inches to feet and inches.
== Sources ==
* : Solutions: $239$. -- The Bell Rope
* : Answers: $310$. The Bell Rope
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/23 - A Deal in Cucumbers

== by : $23$ ==
;A Deal in Cucumbers
:"How much to you pay for these cucumbers?" someone asked.
:The reply: "I pay as many shillings for six dozen cucumbers of that size as I get cucumbers for $32 \shillings$"
:What was the price per cucumber?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $23$. -- A Deal in Cucumbers
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/23 - A Deal in Cucumbers/Solution

== by : $23$ ==
;A Deal in Cucumbers
== Solution ==
:$8 \oldpence$
== Proof ==
Let $P$ be the price per cucumber in pence.
Let $A$ be the price per $6$ dozen cucumbers in shillings.
Let $B$ be the number of cucumbers you can buy for $32 \shillings$.
We have:
Hence the solution.
== Sources ==
* : Solutions: $23$. -- A Deal in Cucumbers
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/240 - Counting the Triangles

== by : $240$ ==
;Counting the Triangles
:Draw a pentagon, and connect each point with every other point with straight lines, as in the diagram.
:
:How many different triangles are contained in this figure?
:To make it quite clear, $AFB$, $AGB$, $ACB$, $BFG$, $BFC$, and $BGC$ are six such triangles.
:It is not a difficult count if you proceed with some method,
::but otherwise you are likely to drop triangles or include some more than once.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $240$. -- Counting the Triangles
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $277$. Counting the Triangles
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/240 - Counting the Triangles/Solution

== by : $240$ ==
;Counting the Triangles
== Solution ==
There are:
:$5$ triangles of the form $\triangle ADC$
:$5$ triangles of the form $\triangle AEB$
:$5$ triangles of the form $\triangle ABF$
:$5$ triangles of the form $\triangle AKF$
:$10$ triangles of the form $\triangle AKB$
:$5$ triangles of the form $\triangle ADG$
That seems to be all of them.
Thus there are $35$ triangles altogether.
== Sources ==
* : Solutions: $240$. -- Counting the Triangles
* : Answers: $277$. Counting the Triangles
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/241 - A Hurdles Puzzle

== by : $241$ ==
;A Hurdles Puzzle
:The answers given in the old books to some of the best-known puzzles are often clearly wrong.
:Yet nobody ever seems to detect their faults.
:Here is an example.
:A farmer had a pen made of fifty hurdles, capable of holding a hundred sheep only.
:Supposing he wanted to make it sufficiently large to hold double that number, how many additional hurdles must he have?
:Perhaps our readers might be interested in making a list of old puzzles that are wrong.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $241$. -- A Hurdles Puzzle
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $278$. A Hurdles Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/241 - A Hurdles Puzzle/Solution

== by : $241$ ==
;A Hurdles Puzzle
== Solution ==
The answer completely depends on the arrangement of the hurdles to make the pen.
Consider the following diagrams:
:
The original answer that is frequently seen is that it can be done with $2$ hurdles.
The arrangement given is that of $A$, a long paddock $1$ hurdle wide and $24$ hurdles long, enclosing an area of $24$ square hurdles.
In $B$, we see we have added $2$ more hurdles to the short sides, making a paddock $2$ hurdles wide and $24$ hurdles long, enclosing an area of $48$ square hurdles.
But consider.
With $50$ hurdles you can make a paddock as in $C$, $156$ square hurdles by making it $12 \times 13$, which is $6 \tfrac 1 2$ times the size of the original $24 \times 1$ paddock.
Or you can make a $6 \times 8$ paddock with $28$ hurdles, as in $D$.
And so on.
 also manages to arrange $50$ hurdles in an elongated kite shape with sides $12$, $12$, $13$, $13$ explaining that this is also $48$ square hurdles in area, but his diagram is unconvincing.
== Sources ==
* : Solutions: $241$. -- A Hurdles Puzzle
* : Answers: $278$. A Hurdles Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/242 - Correcting a Blunder

== by : $242$ ==
;Correcting a Blunder
:Mathematics is an exact science, but first-class mathematicians are apt, like the rest of humanity, to err badly on occasions.
:On referring to 's , we hit on this problem:
::"To find a triangle such that its three sides, perpendicular, and the line drawn from one of the angles bisecting the base
:::may all be expressed in rational numbers."
:He gives as his answer the triangle $480$, $299$, $209$, which is wrong and entirely unintelligible.
:Readers may like to find a correct solution when we say that all the five measurements may be in whole numbers,
::and every one of them less than a hundred.
:It is apparently intended that the triangle must not itself be right-angled.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $242$. -- Correcting a Blunder
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $280$. Correcting a Blunder
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/242 - Correcting a Blunder
Henry Ernest Dudeney/Puzzles and Curious Problems/242 - Correcting a Blunder/Historical Note

== Historical Note on by : $242$ - Correcting a Blunder ==
The first solution was the work of .
The second solution was the work of .
== Sources ==
* : Answers: $280$. Correcting a Blunder
Category:Historical Notes
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/242 - Correcting a Blunder
Henry Ernest Dudeney/Puzzles and Curious Problems/242 - Correcting a Blunder/Solution

== by : $242$ ==
;Correcting a Blunder
== Solution ==
=== Solution 1 ===
=== Solution 2 ===
 adds:
:Perhaps our readers would like to try their hand at constructing the general solution to triangles of this class.
== Historical Note ==
== Sources ==
* : Solutions: $242$. -- Correcting a Blunder
* : Answers: $280$. Correcting a Blunder
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/242 - Correcting a Blunder
Henry Ernest Dudeney/Puzzles and Curious Problems/242 - Correcting a Blunder/Solution 1

== by : $242$ ==
;Correcting a Blunder
== Solution ==
:
== Proof ==
We have:
:$20^2 + 15^2 = 25^2$, demonstrating that the triangle on the left is right-angled
:$15^2 + 8^2 = 17^2$, demonstrating that the triangle in the middle is right-angled
:$15^2 + \paren {28 + 8}^2 = 39^2$, demonstrating that the triangle formed from the two on the right is right-angled
:The side whose length is $17$ is seen to bisect the base.
:The area of the whole triangle is $\dfrac 1 2 \times 15 \times 56 = 420$.
== Sources ==
* : Solutions: $242$. -- Correcting a Blunder
* : Answers: $280$. Correcting a Blunder
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/242 - Correcting a Blunder
Henry Ernest Dudeney/Puzzles and Curious Problems/242 - Correcting a Blunder/Solution 2

== by : $242$ ==
;Correcting a Blunder
== Solution ==
:
== Proof ==
The triangle in solution is the obtuse triangle whose sides are $66$, $41$ and $85$.
We have:
:$40^2 + \paren {66 + 9}^2 = 85^2$, demonstrating that the extended triangle is right-angled
:$9^2 + 40^2 = 41^2$, demonstrating that the triangle on the far right is right-angled
:$40^2 + \paren {\dfrac {66} 2 + 9}^2 = 58^2$, demonstrating that the triangle formed from the two on the right is right-angled
:The side whose length is $58$ is seen to bisect the base.
:The area of the triangle in solution is $\dfrac 1 2 \times 40 \times 66 = 1320$.
== Sources ==
* : Answers: $280$. Correcting a Blunder
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/242 - Correcting a Blunder
Henry Ernest Dudeney/Puzzles and Curious Problems/243 - The Squirrel's Climb

== by : $243$ ==
;The Squirrel's Climb
:A squirrel goes spirally up a cylindrical post, making the circuit in $4$ feet.
:How many feet does it travel to the top if the post is $16$ feet high and $3$ feet in circumference?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $243$. -- The Squirrel's Climb
* : Geometrical Problems: Solid Geometry Puzzles: $324$. The Squirrel's Climb
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/243 - The Squirrel's Climb/Solution

== by : $243$ ==
;The Squirrel's Climb
== Solution ==
:$20$ feet.
== Proof ==
In going $3$ feet around, the squirrel goes $4$ feet up.
These are the sides of the classic $3$-$4$-$5$ triangle.
Hence in going up $4$ feet, the squirrel travels $5$ feet.
Hence in going $16 = 4 \times 4$ feet, the squirrel travels $4 \times 5 = 20$ feet.
== Sources ==
* : Solutions: $243$. -- The Squirrel's Climb
* : Answers: $324$. The Squirrel's Climb
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/244 - Sharing a Grindstone

== by : $244$ ==
;Sharing a Grindstone
:Three men bought a grindstone $20$ inches in diameter.
:How much must each grind off so as to share the stone equally,
::making an allowance of $4$ inches off the diameter as waste for the aperture?
:We are not concerned with the unequal value of the shares for practical use --
::only with the actual equal quantity of stone each receives.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $244$. -- Sharing a Grindstone
* : Geometrical Problems: Circle Puzzles: $291$. Sharing a Grindstone
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/244 - Sharing a Grindstone/Solution

== by : $244$ ==
;Sharing a Grindstone
== Solution ==
The first man can use approximately $1.754$ inches of the grindstone.
The second man can use approximately $2.246$ inches of what remains.
The third man is left with $4$ usable inches plus the aperture.
== Proof ==
To simplify our thinking, let us discuss the radius rather than the diameter of the grindstone.
Thus we have a grindstone $10$ inches in radius with an aperture of $2$ inches.
Thus the usable radius is $8$ inches
Let $r_1$, $r_2$ and $r_3$ be the radii in inches of the grindstone as it is received by the three men in turn.
We are of course given that $r_1 = 10$.
From Area of Circle, the total quantity $q$ of grindstone of a given inches $r$ is proportional to $r^2$.
That is:
:$q = k r^2$
where $k = \pi$, but this is irrelevant.
We have that:
:$10^2 - {r_2}^2 = {r_2}^2 - {r_3}^2 = {r_3}^2 - 2^2$
Thus:
:$r_2 = \sqrt {68} \approx 8.246$
:$r_3 = \sqrt {36} = 6$
Hence:
:the first man can use $10 - \sqrt {68} \approx 1.754$ inches of the grindstone
:the second man can use $\sqrt {68} - 6 \approx 2.246$ inches of what remains
:the third man has $4$ usable inches of what remains.
== Sources ==
* : Solutions: $244$. -- Sharing a Grindstone
* : Answers: $291$. Sharing a Grindstone
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/245 - Magic Fifteen Puzzle

== by : $245$ ==
;Magic Fifteen Puzzle
:This is 's famous $14$-$15$ puzzle,
::in which you were asked to get the $14$ and $15$ in their proper order by sliding them about in the box.
:It was, of course, impossible of solution.
:
:I now propose to slide them about until they shall form a perfect magic square
::in which the four columns, four rows and two diagonals all add up to $30$.
:It will be found convenient to use numbered counters in place of the blocks.
:What are your fewest possible moves?
=== Click here for solution ===
== Sources ==
* : Moving Counter Problems: $245$. -- Magic Fifteen Puzzle
* : Geometrical Problems: Moving Counter Puzzles: $371$. Magic Fifteen Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/245 - Magic Fifteen Puzzle/Solution

== by : $245$ ==
;Magic Fifteen Puzzle
== Solution ==
Move the counters in the following order:
:$12$, $8$, $4$, $3$, $2$, $6$, $10$, $9$, $13$, $15$, $14$, $12$, $8$, $4$, $7$, $10$, $9$, $14$, $12$, $8$, $4$, $7$, $10$, $9$, $6$, $2$, $3$, $10$, $9$, $6$, $5$, $1$, $2$, $3$, $6$, $5$, $3$, $2$, $1$, $13$, $14$, $3$, $2$, $1$, $13$, $14$, $3$, $12$, $15$, $3$
of $50$ moves in all.
This results in the square:
:
== Variant Solution ==
== Sources ==
* : Solutions: $245$. -- Magic Fifteen Puzzle
* : Answers: $371$. Magic Fifteen Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/245 - Magic Fifteen Puzzle/Variant Solution

== by : $245$ ==
;Magic Fifteen Puzzle
== Solution ==
 points out that if the $14$ and $15$ are in the correct serial order at the start, then a magic square can be achieved in $37$ moves:
:$15$, $14$, $10$, $6$, $7$, $3$, $2$, $7$, $6$, $11$, $10$, $14$, $3$, $2$, $11$, $10$, $9$, $5$, $1$, $6$, $10$, $9$, $5$, $1$, $6$, $10$, $9$, $5$, $2$, $12$, $3$
This results in the square:
:
== Sources ==
* : Answers: $371$. Magic Fifteen Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/246 - Transferring the Counters

== by : $246$ ==
;Transferring the Counters
:Place ten counters on the squares of a chessboard as here shown,
::and transfer them to the other corner as indicated by the ten crosses.
:
:A counter may jump over any counter to the next square beyond, if vacant,
::either horizontally or vertically, but not diagonally,
::and there are no captures and no simple moves -- only leaps.
:Not to waste the reader's time it can be conclusively proved that this is impossible.
:You are now asked to add two more counters so that it may be done.
:If you place these, say, on $\text {AA}$, they must, in the end, be found in the corresponding positions $\text {BB}$.
:Where will you place them?
=== Click here for solution ===
== Sources ==
* : Moving Counter Problems: $246$. -- Transferring the Counters
* : Geometrical Problems: Moving Counter Puzzles: $372$. Transferring the Counters
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/246 - Transferring the Counters/Solution

== by : $246$ ==
;Transferring the Counters
== Solution ==
Place the extra counters:
:one on the fourth square in the second row from the top
:one on the second square in the fourth row from the top
thus:
:
Then they may be played as follows:
== Sources ==
* : Solutions: $246$. -- Transferring the Counters
* : Answers: $372$. Transferring the Counters
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/247 - The Counter Cross

== by : $247$ ==
;The Counter Cross
:Arrange twenty counters in the form of a cross, in the manner shown in the diagram.
:
:Now, in how many different ways can you point out four counters that will form a perfect square if considered alone?
:Thus the four counters composing each arm of the cross, and also the four in the centre, form squares.
:Squares are also formed by the four counters marked $\text A$, the four marked $\text B$, and so on.
:And in how many ways can you remove six counters so that not a single square can be so indicated from those that remain?
=== Click here for solution ===
== Sources ==
* : Moving Counter Problems: $247$. -- The Counter Cross
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $285$. The Counter Cross
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/247 - The Counter Cross/Mistake

== Source Work ==
:
:Moving Counter Problems
::$247$. -- The Counter Cross
:
:Geometrical Problems
::Triangle, Square & Other Polygon Puzzles
:::$285$. The Counter Cross
== Mistake ==
: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.
:
== Correction ==
He missed two.
These can be found in the analysis in the solution given.
== Sources ==
* : Solutions: $247$. -- The Counter Cross
* : Answers: $285$. The Counter Cross
Category:Mistakes/Puzzles and Curious Problems
Category:Mistakes/536 Puzzles & Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/247 - The Counter Cross/Solution

== by : $247$ ==
;The Counter Cross
== Solution ==
There are $21$ squares:
:
:$9$ of them will be of the size indicated by the four instances of $\text a$
:$4$ of them will be of the size indicated by the four instances of $\text b$
:$4$ of them will be of the size indicated by the four instances of $\text c$
:$2$ of them will be of the size indicated by the four instances of $\text d$
:$2$ of them will be of the size indicated by the four instances of $\text f$.
These are indicated by the blue dotted squares in the above diagram.
If you now remove all counters marked $\text e$, no squares can be formed from the remaining counters.
== Sources ==
* : Solutions: $247$. -- The Counter Cross
* : Answers: $285$. The Counter Cross
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/248 - Four in Line

== by : $248$ ==
;Four in Line
:Here we have a board of $36$ squares, and four counters are so placed in a straight line
::that every square of the board is in line horizontally, vertically, or diagonally with at least one counter.
:In other words, if you regard them as chess queens, every square on the board is attacked by at least one queen.
:
:The puzzle is to find in how many different ways the four counters may be placed in a straight line so that every square shall thus be in line with a counter.
:Every arrangement in which the counters occupy a different set of four squares is a different arrangement.
:Thus, in the case of the example given, they can be moved to the next column to the right with equal effect,
::or they may be transferred to either of the two central rows of the board.
:This arrangement, therefore, produces $4$ solutions by what we call reversals or reflections of the board.
:Remember that the counters must always be disposed in a straight line.
:It will be found an entertaining little puzzle.
=== Click here for solution ===
== Sources ==
* : Moving Counter Problems: $248$. -- Four in Line
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $455$. Four in Line
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/248 - Four in Line/Solution

== by : $248$ ==
;Four in Line
== Solution ==
There are $9$ fundamentally different solutions:
: $\quad$ $\quad$ 
: $\quad$ $\quad$ 
: $\quad$ $\quad$ 
Of the above, the $4$th, $5$th and $9$th give $8$ solutions when rotations and reflections are considered.
The remaining ones give rise to $4$ each.
Hence there are $48$ different solutions in total.
== Also see ==
's Five Dogs Puzzle, no. $311$ in his .
== Sources ==
* : Solutions: $248$. -- Four in Line
* : Answers: $455$. Four in Line
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/249 - Odds and Evens

== by : $249$ ==
;Odds and Evens
:Place eight counters in a pile on the middle circle so that they shall be in proper numerical order, with $1$ on the top and $8$ on the bottom.
:
:It is required to transfer $1$, $3$, $5$, $7$ to the circle marked "Odds", and $2$, $4$, $6$, $8$ to the circle marked "Evens".
:You can only move one counter at a time from circle to circle, and you must never place a number on a smaller number,
::nor an odd number and an even number together on the same circle.
:What are the fewest possible moves?
=== Click here for solution ===
== Sources ==
* : Moving Counter Problems: $249$. -- Odds and Evens
* : Geometrical Problems: Moving Counter Puzzles: $373$. Odds and Evens
Category:Henry Ernest Dudeney/Puzzles and Curious Problems