nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Modern Puzzles/90 - Equal Perimeters

== by : $90$ ==
;Equal Perimeters
:Rational right-angled triangles have been a fascinating subject for study since the time of , before the Christian era.
:Every schoolboy knows that the sides of these, generally expressed in whole numbers,
::are such that the square of the hypotenuse is exactly equal to the sum of the squares of the other two sides.
:Now, can you find $6$ rational right-angled triangles each with a common perimeter, and the smallest possible?
:It is not a difficult puzzle like my "Four Princes" (in the Canterbury Puzzles), in which you had to find four such triangles of equal area.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $90$. -- Equal Perimeters
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $173$. Equal Perimeters
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/90 - Equal Perimeters/Solution

== by : $90$ ==
;Equal Perimeters
== Solution ==
The smallest such common perimeter is $720$:
:$180, 240, 300$
:$120, 288, 312$
:$144, 270, 306$
:$72, 320, 328$
:$45, 336, 339$
:$80, 315, 325$
== Proof ==
== Sources ==
* : Solutions: $90$. -- Equal Perimeters
* : Answers: $173$. Equal Perimeters
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/91 - Counting the Wounded

== by : $91$ ==
;Counting the Wounded
:When visiting with a friend one of our hospitals for wounded soldiers, I was informed that
::exactly two-thirds of the men had lost an eye,
::three-fourths had lost an arm,
::and four-fifths had lost a leg.
:"Then," I remarked to my friend, "it follows that at least twenty-six of the men must have lost all three -- an eye, an arm, and a leg."
:That being so, can you say exactly how many men were in the hospital?
:It is a very simple calculation, but I have no doubt it will perplex a good many readers.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $91$. -- Counting the Wounded
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $174$. Counting the Wounded
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/91 - Counting the Wounded/Solution

== by : $91$ ==
;Counting the Wounded
== Solution ==
$120$ men are in the hospital.
== Proof ==
We know that:
:exactly two-thirds of the men had lost an eye.
We also know that:
:three-fourths had lost an arm.
By Inclusion-Exclusion Principle, at least:
:$\dfrac 2 3 + \dfrac 3 4 - 1 = \dfrac 5 {12}$
of the men had lost both an eye and an arm.
Similarly, since we know that:
:four-fifths had lost a leg
by the Inclusion-Exclusion Principle again, at least:
:$\dfrac 5 {12} + \dfrac 4 5 - 1 = \dfrac {13} {60}$
of the men had lost all three.
 equates this to $26$.
Hence there are exactly:
:$26 \div \dfrac {13} {60} = 120$
men in this hospital.
== Sources ==
* : Solutions: $91$. -- Counting the Wounded
* : Answers: $174$. Counting the Wounded
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/92 - A Cow's Progeny

== by : $92$ ==
;A Cow's Progeny
:"Supposing," said my friend Farmer Hodge, "that cow of mine to have a she-calf at the age of two years,
::and supposing she goes on having the like every year,
::and supposing every one of her young to have a she-calf at the age of two years,
::and afterwards every year likewise, and so on.
:Now, how many do you suppose would spring from that cow and all her descendants in the space of twenty-five years?"
:I understood from Hodge that we are to count from the birth of the original cow,
::and it is obvious that the family can produce no feminine beef of veal during the period stated.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $92$. -- A Cow's Progeny
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $175$. A Cow's Progeny
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/92 - A Cow's Progeny/Solution

== by : $92$ ==
;A Cow's Progeny
== Solution ==
:$121 \, 392$.
== Proof ==
After $1$ year there will be $1$ cow.
After $2$ years there will be $2$ cows: the original cow and her first calf.
After $3$ years there will be $3$ cows: the original cow, her first calf and her second calf.
After $4$ years there will be $5$ cows: the original cow, her first calf and her second calf, her third calf and the first calf of her first calf.
This is the Rabbit Problem, but using a cow instead of a pair of rabbits.
By inspection, the total number of cattle after $n$ years is the $n + 1$th Fibonacci number $F_{n + 1}$.
So the total number of cattle after that time is $F_{26}$.
This can be looked up on, for example , which gives $121 \, 393$.
Because we are asked to count only the cow's descendants, we remove the original cow from the count, leaving $121 \, 392$.
== Sources ==
* : Solutions: $92$. -- A Cow's Progeny
* : Answers: $175$. A Cow's Progeny
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Category:Rabbit Problem
Henry Ernest Dudeney/Modern Puzzles/93 - Sum Equals Product

== by : $93$ ==
;Sum Equals Product
:There are two numbers whose sum equals their product, that is, $2$ and $2$.
:What other numbers have that property?
The original of this question was wordy to the max, and has been minimised for this presentation.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $93$. -- Sum Equals Product
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $176$. Sum Equals Product
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/93 - Sum Equals Product/Solution

== by : $93$ ==
;Sum Equals Product
== Solution ==
Any $x$ and $y$ such that $y = \dfrac x {x - 1}$ fits the bill.
== Proof ==
and we note that:
== Sources ==
* : Solutions: $93$. -- Sum Equals Product
* : Answers: $176$. Sum Equals Product
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/94 - Adding their Cubes

== by : $94$ ==
;Adding their Cubes
:The numbers $407$ and $370$ have this peculiarity, that they exactly equal the sum of the cubes of their digits.
:Thus the cube of $4$ is $64$, the cube of $0$ is $0$, and the cube of $7$ is $343$.
:Add together $64$, $0$ and $343$, and you get $407$.
:Again, the cube of $3$ ($27$), added to the cube of $7$ ($343$), is $370$.
:Can you find a number not containing a nought that will work in the same way?
:Of course, we bar the absurd case of $1$.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $94$. -- Adding their Cubes
* : Arithmetical and Algebraical Problems: Digital Puzzles: $143$. Adding their Cubes
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/94 - Adding their Cubes/Historical Note

== Historical Note on by : $94$ - Adding their Cubes ==
, in his $1968$ repackaging , points out the overlooked $371$, then refers to 's from $1966$, where the general pluperfect digital invariant is discussed.
== Sources ==
* : Answers: $143$. Adding their Cubes
Category:Historical Notes
Henry Ernest Dudeney/Modern Puzzles/94 - Adding their Cubes/Solution

== by : $94$ ==
;Adding their Cubes
== Solution ==
 gives $153$, but fails to provide $371$.
These are the $3$-digit pluperfect digital invariants, also known as narcissistic numbers or Armstrong numbers.
== Historical Note ==
== Sources ==
* : Solutions: $94$. -- Adding their Cubes
* : Answers: $143$. Adding their Cubes
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/95 - Squares and Cubes

== by : $95$ ==
;Squares and Cubes
:Can you find two whole numbers, such that the difference of their squares is a cube and the difference of their cubes is a square?
:What is the answer in the smallest possible numbers?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $95$. -- Squares and Cubes
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $177$. Squares and Cubes
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/95 - Squares and Cubes/Solution

== by : $95$ ==
;Squares and Cubes
== Solution ==
We have:
== Proof ==
We show this solution is minimal by showing that the larger number is at least $10$.
We look at the first equation:
:$x^2 - y^2 = a^3 = \paren {x - y} \paren {x + y}$
Since $y^2 > 0$ and $5^3 > 10^2$, we only consider $a = 1, 2, 3, 4$.
$1$ cannot be the difference of two squares.
Since $2^3 = 1 \times 8 = 2 \times 4$, we check:
:$\tuple {x, y} = \tuple {3, 1}$
as the other pair are not integers.
This gives:
:$3^3 - 1^3 = 26$
which is not a square.
Since $3^3 = 1 \times 27 = 3 \times 9$, we check:
:$\tuple {x, y} = \tuple {6, 3}$
as the other pair has $x > 10$.
This gives:
:$6^3 - 3^3 = 189$
which is not a square.
Since $4^3 = 1 \times 64 = 2 \times 32 = 4 \times 16 = 8 \times 8$, we check:
:$\tuple {x, y} = \tuple {10, 6}$
as the other pairs either has $x > 10$ or $y = 0$.
This gives:
:$10^3 - 6^3 = 784 = 28^2$
which is our smallest solution.
== Sources ==
* : Solutions: $95$. -- Squares and Cubes
* : Answers: $177$. Squares and Cubes
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/96 - Concerning a Cube

== by : $96$ ==
;Concerning a Cube
:What is the length in feet of the side of a cube when
::$(1)$ the superficial area equals the cubical contents;
::$(2)$ the superficial area equals the square of the cubical contents;
::$(3)$ the square of the superficial area equals the cubical contents?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $96$. -- Concerning a Cube
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $178$. Concerning a Cube
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/96 - Concerning a Cube/Solution

== by : $96$ ==
;Concerning a Cube
== Solution ==
:$(1): \quad$ $6$ feet
:$(2): \quad$ Approximately $1 \cdotp 57$ feet
:$(3): \quad$ $\dfrac 1 {36}$ feet
== Proof ==
Let $l$ denote the length of the side of a cube $\CC$.
Let $\AA$ denote the superficial area of $\CC$.
Let $\VV$ denote the cubical contents of $\CC$.
From Surface Area of Cube:
:$\AA = 6 l^2$
From Volume of Cube:
:$\VV = l^3$
Hence we can deduce the following:
=== Problem $(1)$ ===
=== Problem $(2)$ ===
=== Problem $(3)$ ===
== Sources ==
* : Solutions: $96$. -- Concerning a Cube
* : Answers: $178$. Concerning a Cube
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/97 - A Common Divisor

== by : $97$ ==
;A Common Divisor
:Here is a puzzle that has been the subject of frequent inquiries by correspondents,
::only, of course, the actual figures are varied considerably.
:A country newspaper stated that many schoolmasters have suffered in health in their attempts to master it!
:Perhaps this is merely a little journalistic exaggeration, for it is really a very simple question if you only have the cunning to hit on the method of attacking it.
:This is the question:
:Find a common divisor for the three numbers $480 \, 608$, $508 \, 811$, and $723 \, 217$, so that the remainder shall be the same in every case.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $97$. -- A Common Divisor
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $179$. A Common Divisor
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/97 - A Common Divisor/Solution

== by : $97$ ==
;A Common Divisor
== Solution ==
:$79$.
The remainder in each case is $51$.
== Proof ==
Let $n$ be the divisor sought.
Let $m$ be the remainder.
We have that:
Hence we see that $n$ is a common divisor for each of $28 \, 203$, $242 \, 609$ and $214 \, 406$.
So we establish the prime factors of these numbers:
and it is seen by inspection that $n = 79$.
Thus we have:
== Sources ==
* : Solutions: $97$. -- A Common Divisor
* : Answers: Miscellaneous Puzzles: $179$. A Common Divisor
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/98 - Curious Multiplication

== by : $98$ ==
;Curious Multiplication
:I have frequently been asked to explain the following, which will no doubt interest many readers who have not seen it.
:If a person can add correctly but is incapable of multiplying or dividing by a number higher than $2$,
::it is possible to obtain the product of any two numbers in this curious way.
:Multiply $97$ by $23$.
 97 23
 48 (46)
 24 (92)
 12 (184)
 6 (368)
 3 736
 1 1472
 ----
 2231
 ----
:In the first column we divide by $2$, rejecting the remainders, until $1$ is reached.
:In the second column we multiply $23$ by $2$ the same number of times.
:If we now strike out those products that are opposite ton the even numbers in the first column
::(we have enclosed these in brackets for convenience in printing)
::and add up the remaining numbers we get $2231$, which is the correct answer.
:Why is this?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $98$. -- Curious Multiplication
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $180$. Curious Multiplication
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/98 - Curious Multiplication/Solution

== by : $98$ ==
;Curious Multiplication
== Solution ==
In the first column, write down the successive remainders on division by $2$:
:$1 \, 0 \, 0 \, 0 \, 0 \, 1 \, 1$
which, when reversed, becomes:
:$1 \, 1 \, 0 \, 0 \, 0 \, 0 \, 1$
This is $97$ in binary notation, or $2^0 + 2^5 + 2^6$.
In the second column, we take the numbers next to where the remainders are $1$, and get:
:$23 \times 1 + 23 \times 2^5 + 23 \times 2^6$
which evaluates to $2231$.
Thus it is seen that the whole operation is being done in binary notation.
== Sources ==
* : Solutions: $98$. -- Curious Multiplication
* : Answers: $180$. Curious Multiplication
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/99 - The Rejected Gun

== by : $99$ ==
;The Rejected Gun
:Here is a military puzzle that may or may not give you a moment's difficulty.
:It is such a simple question that a child can understand it and no knowledge of artillery is required.
:Yet some of my readers may find themselves perplexed for quite five minutes.
:An inventor offered a new large gun to the committee appointed by our Government for the consideration of such things.
:He declared that when once loaded it would fire sixty shots at the rate of a shot a minute.
:The War Office put it to the test and found that it fired sixty shots an hour,
::but declined it, "as it did not fulfil the promised condition."
:"Absurd, said the inventor, "for you have shown that it clearly does all that we undertook it should do."
:"Nothing of the sort," said the experts. "It has failed."
:Now, can you explain this extraordinary mystery?
:Was the inventor, or were the experts, right?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $99$. -- The Rejected Gun
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $181$. The Rejected Gun
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/99 - The Rejected Gun/Solution

== by : $99$ ==
;The Rejected Gun
== Solution ==
This is an instance of the Fencepost Mistake.
Let it be assumed that the firing a single shot is an instantaneous action.
Then to fire $60$ shots at one per minute will take $59$ minutes.
It must be the case that the testing facility observed that over the course of $1$ hour, $60$ shots were fired, whereas in fact it would be expected for there to be $61$ shots fired.
== Also see ==
* Puzzles and Curious Problems $322$ - The New Gun, which is a retread of this one.
== Sources ==
* : Solutions: $99$. -- The Rejected Gun
* : Answers: $181$. The Rejected Gun
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Category:Fencepost Mistakes
Henry Ernest Dudeney/Modern Puzzles/9 - Fractional Value

== by : $9$ ==
;Fractional Value
:What part of threepence is one-third of twopence?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $9$. -- Fractional Value
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/9 - Fractional Value/Solution

== by : $9$ ==
;Fractional Value
== Solution ==
Two-ninths of threepence.
== Proof ==
:$\dfrac 1 3 \times 2 = \dfrac 2 3$
So $\dfrac 1 3$ of $2 \oldpence$ is $\dfrac 2 3$ of a penny.
Then:
:$\dfrac 2 3$ of a penny is $\dfrac 2 3$ of $\dfrac 1 3$ of $3 \oldpence$
That is:
:$\dfrac 2 9$ of $3 \oldpence$
== Sources ==
* : Solutions: $9$. -- Fractional Value
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/Arithmetical and Algebraical Problems

== : : Arithmetical and Algebraical Problems ==
=== Money Puzzles ===
=== Age and Kinship Puzzles ===
=== Clock Puzzles ===
=== Locomotion and Speed Puzzles ===
=== Digital Puzzles ===
=== Various Arithmetical and Algebraical Problems ===
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/Arithmetical and Algebraical Problems/Age and Kinship Puzzles

== : : Arithmetical and Algebraical Problems ==
==== $21$ - Their Ages ====
==== $22$ - Mrs. Wilson's Family ====
==== $23$ - De Morgan and Another ====
==== $24$ - "Simple" Arithmetic ====
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/Arithmetical and Algebraical Problems/Digital Puzzles

== : : Arithmetical and Algebraical Problems ==
==== $51$ - An Exceptional Number ====
==== $52$ - The Five Cards ====
==== $53$ - Squares and Digits ====
==== $54$ - The Two Additions ====
==== $55$ - The Repeated Quartette ====
==== $56$ - Easy Division ====
==== $57$ - A Misunderstanding ====
==== $58$ - The Two Fours ====
==== $59$ - The Two Digits ====
==== $60$ - Digital Coincidences ====
==== $61$ - Palindromic Square Numbers ====
==== $62$ - Factorizing ====
==== $63$ - Find the Factors ====
==== $64$ - Dividing by Eleven ====
==== $65$ - Dividing by $37$ ====
==== $66$ - Another $37$ Division ====
==== $67$ - A Digital Difficulty ====
==== $68$ - Threes and Sevens ====
==== $69$ - Root Extraction ====
==== $70$ - The Solitary Seven ====
==== $71$ - A Complete Skeleton ====
==== $72$ - Alphabetical Sums ====
==== $73$ - Alphabetical Arithmetic ====
==== $74$ - Queer Division ====
==== $75$ - A Teasing Legacy ====
==== $76$ - The Nine Volumes ====
==== $77$ - The Ten Volumes ====
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/Arithmetical and Algebraical Problems/Locomotion and Speed Puzzles

== : : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles ==
==== $28$ - Hill Climbing ====
==== $29$ - Timing the Motor-car ====
==== $30$ - The Staircase Race ====
==== $31$ - A Walking Puzzle ====
==== $32$ - Riding in the Wind ====
==== $33$ - A Rowing Puzzle ====
==== $34$ - The Moving Stairway ====
==== $35$ - Sharing a Bicycle ====
==== $36$ - More Bicycling ====
==== $37$ - A Side-car Problem ====
==== $38$ - The Despatch-Rider ====
==== $39$ - The Two Trains ====
==== $40$ - Pickleminster to Quickville ====
==== $41$ - The Damaged Engine ====
==== $42$ - The Puzzle of the Runners ====
==== $43$ - The Two Ships ====
==== $44$ - Find the Distance ====
==== $45$ - The Man and the Dog ====
==== $46$ - Baxter's Dog ====
==== $47$ - The Runner's Refreshment ====
==== $48$ - Railway Shunting ====
==== $49$ - Exploring the Desert ====
==== $50$ - Exploring Mount Neverest ====
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/Arithmetical and Algebraical Problems/Money Puzzles

== : : Arithmetical and Algebraical Problems ==
==== $1$ - Concerning a Cheque ====
==== $2$ - Pocket-money ====
==== $3$ - Dollars and Cents ====
==== $4$ - Loose Cash ====
==== $5$ - Doubling the Value ====
==== $6$ - Generous Gifts ====
==== $7$ - Selling Eggs ====
==== $8$ - Buying Buns ====
==== $9$ - Fractional Value ====
==== $10$ - Unrewarded Labour ====
==== $11$ - The Perplexed Banker ====
==== $12$ - A Weird Game ====
==== $13$ - Find the Coins ====
==== $14$ - An Easy Settlement ====
==== $15$ - Sawing Logs ====
==== $16$ - Digging a Ditch ====
==== $17$ - Name their Wives ====
==== $18$ - A Curious Paradox ====
==== $19$ - Market Transactions ====
==== $20$ - The Seven Applewomen ====
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/Arithmetical and Algebraical Problems/Various Arithmetical and Algebraical Problems

== : : Arithmetical and Algebraical Problems ==
==== $78$ - The Miller's Toll ====
==== $79$ - Egg Laying ====
==== $80$ - The Flocks of Sheep ====
==== $81$ - Pussy and the Mouse ====
==== $82$ - Army Figures ====
==== $83$ - A Critical Vote ====
==== $84$ - The Three Brothers ====
==== $85$ - The House Number ====
==== $86$ - A New Street Puzzle ====
==== $87$ - Another Street Puzzle ====
==== $88$ - Correcting an Error ====
==== $89$ - The Seventeen Horses ====
==== $90$ - Equal Perimeters ====
==== $91$ - Counting the Wounded ====
==== $92$ - A Cow's Progeny ====
==== $93$ - Sum Equals Product ====
==== $94$ - Adding their Cubes ====
==== $95$ - Squares and Cubes ====
==== $96$ - Concerning a Cube ====
==== $97$ - A Common Divisor ====
==== $98$ - Curious Multiplication ====
==== $99$ - The Rejected Gun ====
==== $100$ - Odds and Evens ====
==== $101$ - Twenty Questions ====
==== $102$ - The Nine Barrels ====
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/Combination and Group Problems

== : : Combination and Group Problems ==
==== $166$ - Picture Presentation ====
==== $167$ - A General Election ====
==== $168$ - The Magisterial Bench ====
==== $169$ - The Card Pentagon ====
==== $170$ - A Heptagon Puzzle ====
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/Geometrical Problems/Dissection Puzzles

== : : Geometrical Problems: Dissection Puzzles ==
==== $103$ - A New Cutting-out Puzzle ====
==== $104$ - The Square Table-Top ====
==== $105$ - The Squares of Veneer ====
==== $106$ - Dissecting the Moon ====
==== $107$ - Dissecting the Letter E ====
==== $108$ - Hexagon to Square ====
==== $109$ - Squaring a Star ====
==== $110$ - The Mutilated Cross ====
==== $111$ - The Victoria Cross ====
==== $112$ - Squaring the Swastika ====
==== $113$ - The Maltese Cross ====
==== $114$ - The Pirates' Flag ====
==== $115$ - The Carpenter's Puzzle ====
==== $116$ - The Crescent and the Star ====
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/Geometrical Problems/Paper-Folding Puzzles

== : : Geometrical Problems: Paper-Folding Puzzles ==
==== $120$ - The Ribbon Pentagon ====
==== $121$ - Paper Folding ====
==== $122$ - Folding a Pentagon ====
==== $123$ - Making an Octagon ====
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/Geometrical Problems/Patchwork Puzzles

== : : Geometrical Problems: Patchwork Puzzles ==
==== $117$ - The Patchwork Quilt ====
==== $118$ - The Improvised Draughts-Board ====
==== $119$ - Tessellated Pavements ====
Category:Henry Ernest Dudeney/Modern Puzzles