nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Modern Puzzles/Geometrical Problems/Various Geometrical Puzzles

== : : Geometrical Problems: Various Geometrical Puzzles ==
==== $124$ - Drawing a Straight Line ====
==== $125$ - Making a Pentagon ====
==== $126$ - Drawing an Oval ====
==== $127$ - With Compasses Only ====
==== $128$ - Lines and Squares ====
==== $129$ - The Circle and Discs ====
==== $130$ - Mr. Grindle's Garden ====
==== $131$ - The Garden Path ====
==== $132$ - The Garden Bed ====
==== $133$ - A Problem for Surveyors ====
==== $134$ - A Fence Problem ====
==== $135$ - The Domino Swastika ====
==== $136$ - A New Match Puzzle ====
==== $137$ - Hurdles and Sheep ====
==== $138$ - The Four Draughtsmen ====
==== $139$ - A Crease Problem ====
==== $140$ - The Four-Colour Map Theorem ====
==== $141$ - The Six Submarines ====
==== $142$ - Economy in String ====
==== $143$ - The Stone Pedestal ====
==== $144$ - The Bricklayer's Task ====
==== $145$ - A Cube Paradox ====
==== $146$ - The Cardboard Box ====
==== $147$ - The Austrian Pretzel ====
==== $148$ - Cutting the Cheese ====
==== $149$ - A Tree-Planting Puzzle ====
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/Magic Square Problems

== : : Magic Square Problems ==
==== $171$ - An Irregular Magic Square ====
==== $172$ - A Magic Square Delusion ====
==== $173$ - Difference Squares ====
==== $174$ - Swastika Magic Square ====
==== $175$ - Is it Very Easy? ====
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/Measuring, Weighing, and Packing Problems

== : : Measuring, Weighing, and Packing Problems ==
==== $180$ - The Damaged Measure ====
==== $181$ - The Six Cottages ====
==== $182$ - A New Domino Puzzle ====
==== $183$ - At the Brook ====
==== $184$ - A Prohibition Poser ====
==== $185$ - Prohibition Again ====
==== $186$ - The False Scales ====
==== $187$ - Weighing the Goods ====
==== $188$ - Monkey and Pulley ====
==== $189$ - Weighing the Baby ====
==== $190$ - Packing Cigarettes ====
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/Moving Counter Problems

== : : Moving Counter Problems ==
==== $150$ - Counter Solitaire ====
==== $151$ - Sinking the Fishing-Boats ====
==== $152$ - A New Leap-Frog Puzzle ====
==== $153$ - Transferring the Counters ====
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/Problems Concerning Games

== : : Problems Concerning Games ==
==== $193$ - A Domino Square ====
==== $194$ - A Domino Star ====
==== $195$ - Domino Groups ====
==== $196$ - Les Quadrilles ====
==== $197$ - A Puzzle with Cards ====
==== $198$ - A Card Trick ====
==== $199$ - Golf Competition Puzzle ====
==== $200$ - Cricket Scores ====
==== $201$ - Football Results ====
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/Puzzle Games

== : : Puzzle Games ==
==== $202$ - Noughts and Crosses ====
==== $203$ - The Horse-Shoe Game ====
==== $204$ - Turning the Die ====
==== $205$ - The Three Dice ====
==== $206$ - The $37$ Puzzle Game ====
==== $207$ - The Twenty-Two Game ====
==== $208$ - The Nine Squares Game ====
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/Unclassified Problems

== : : Unclassified Problems ==
==== $212$ - A Chain Puzzle ====
==== $213$ - The Six Pennies ====
==== $214$ - Folding Postage Stamps ====
==== $215$ - An Ingenious Match Puzzle ====
==== $216$ - Fifty-Seven to Nothing ====
==== $217$ - The Five Squares ====
==== $218$ - A Square with Four Pennies ====
==== $219$ - A Calendar Puzzle ====
==== $220$ - The Fly's Tour ====
==== $221$ - A Musical Enigma ====
==== $222$ - A Mechanical Paradox ====
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/Unicursal and Route Problems

== : : Unicursal and Route Problems ==
==== $154$ - The Way to Tipperary ====
==== $155$ - Marking a Tennis Court ====
==== $156$ - Water, Gas and Electricity ====
==== $157$ - Crossing the Lines ====
==== $158$ - The Nine Bridges ====
==== $159$ - The Five Regiments ====
==== $160$ - Going to Church ====
==== $161$ - A Motor-Car Puzzle ====
==== $162$ - The Fly and the Honey ====
==== $163$ - The Russian Motor-Cyclists ====
==== $164$ - Those Russian Cyclists Again ====
==== $165$ - The Despatch-Rider in Flanders ====
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Puzzles and Curious Problems

== : ==
== Arithmetical and Algebraical Problems ==
== Geometrical Problems ==
== Moving Counter Problems ==
== Unicursal and Route Problems ==
== Combination and Group Problems ==
== Magic Square, Measuring, Weighing, and Packing Problems ==
== Crossing River Problem, and Problems Concerning Games and Puzzle Games ==
== Unclassified Problems ==
Category:Recreational Mathematics
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/100 - Digital Squares

== by : $100$ ==
;Digital Squares
:Find a number which, together with its square, shall contain all the $9$ digits once, and once only, the $0$ disallowed.
:Thus, if the square of $378$ happened to be $152 \, 694$, it would be a perfect solution.
:But unfortunately the actual square is $142 \, 884$, which gives us repeated $4$s and $8$s, and omits the $6$, $5$, and $9$.
:There are only two possible cases, and these may be discovered in about a quarter of an hour if you proceed in the right way.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $100$. -- Digital Squares
* : Arithmetical and Algebraical Problems: Digital Puzzles: $130$. Digital Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Penholodigital Sets
Henry Ernest Dudeney/Puzzles and Curious Problems/100 - Digital Squares/Solution

== by : $100$ ==
;Digital Squares
== Solution ==
The following solutions are the only ones:
== Proof ==
See Penholodigital Square Equation for a full analysis.
== Sources ==
* : Solutions: $100$. -- Digital Squares
* : Answers: $130$. Digital Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/101 - Finding a Square

== by : $101$ ==
;Finding a Square
:Here are six numbers:
::$4 \, 784 \, 887$, $2 \, 494 \, 651$, $8 \, 595 \, 087$, $1 \, 385 \, 287$, $9 \, 042 \, 451$, $9 \, 406 \, 087$
:It is known that three of these numbers added together will form a square.
:Which are they?
:The reader will probably see no other course but rather laborious trial,
::and yet the answer may be found directly by very simple arithmetic and without any experimental extraction of a square root.
=== Click here for solution ===
== Historical Note ==
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $101$. -- Finding a Square
* : Arithmetical and Algebraical Problems: Digital Puzzles: $131$. Finding a Square
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/101 - Finding a Square/Historical Note

== Historical Note on by : $101$ - Finding a Square ==
 apparently commented on this puzzle as follows:
:This application is original on part.
:Digital properties are but little known to mathematicians, and we hope that his example may serve to direct attention to the method ... In a certain class of arithmetical problems it is of great assistance.
== Sources ==
* : Solutions: $101$. -- Finding a Square
* : Answers: $131$. Finding a Square
Category:Historical Notes
Henry Ernest Dudeney/Puzzles and Curious Problems/101 - Finding a Square/Solution

== by : $101$ ==
;Finding a Square
== Solution ==
The digital roots of each of the six numbers in order are:
:$1, 4, 6, 7, 7, 7$
Combining these into triplets, and calculating the digital roots of each, we get:
:$\begin {matrix}
146 & 147 & 167 & 177 & 467 & 477 & 677 & 777 \\
 2 & 3 & 5 & 6 & 8 & 9 & 2 & 3 \end {matrix}$
From Digital Root of Square, every square number has a digital root in $\set {1, 4, 7, 9}$.
So the required numbers must have the digital roots $4$, $7$ and $7$ in order for their sum to be square.
Now, if the fifth number is included, then the total of the three numbers will end in $189$ or $389$.
This is impossible for a square number, as the $89$ would have to be preceded by an even digit.
Therefore the required numbers must be:
:$2 \, 494 \, 651 + 1 \, 385 \, 287 + 9 \, 406 \, 087 = 13 \, 286 \, 025 = 3645^2$
== Historical Note ==
== Sources ==
* : Solutions: $101$. -- Finding a Square
* : Answers: $131$. Finding a Square
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/102 - Juggling with Digits

== by : $102$ ==
;Juggling with Digits
:Arrange the ten digits in three arithmetical sums,
::employing three of the four operations of addition, subtraction, multiplication and division,
::and using no signs except the ordinary ones implying those operations.
:Here is an example to make it quite clear:
::$3 + 4 = 7$; $9 - 8 = 1$; $30 \div 6 = 5$.
:But this is not correct, because $2$ is omitted, and $3$ is repeated.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $102$. -- Juggling with Digits
* : Arithmetical and Algebraical Problems: Digital Puzzles: $132$. Juggling with Digits
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Pandigital Sets
Henry Ernest Dudeney/Puzzles and Curious Problems/102 - Juggling with Digits/Solution

== by : $102$ ==
;Juggling with Digits
== Solution ==
:$7 + 1 = 8$; $9 - 6 = 3$; $4 \times 5 = 20$.
== Sources ==
* : Solutions: $102$. -- Juggling with Digits
* : Answers: $132$. Juggling with Digits
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/103 - Expressing Twenty-Four

== by : $103$ ==
;Expressing Twenty-Four
:In a book published in America was the following:
:"Write $24$ with three equal digits, none of which is $8$.
:(There are two solutions to this problem.)"
:Of course, the answers given are $22 + 2 = 24$, and $3^3 - 3 = 24$.
:Readers who are familiar with the old "Four Fours" puzzle, and others of the same class,
::will ask why there are supposed to be only these solutions.
:With which of the remaining digits is a solution equally possible?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $103$. -- Expressing Twenty-Four
* : Arithmetical and Algebraical Problems: Digital Puzzles: $137$. Expressing Twenty-Four
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/103 - Expressing Twenty-Four/Historical Note

== by : $103$ ==
 supplies only the solutions for $1$ and $7$.
He attributes the solution for $7$ to a certain G.P.E. but the identity of this person has not been established.
According to , the remaining solutions were determined by , but given the technique for the above, they are easily discovered.
== Sources ==
* : Solutions: $103$. -- Expressing Twenty-Four
* : Answers: $137$. Expressing Twenty-Four
Category:Historical Notes
Henry Ernest Dudeney/Puzzles and Curious Problems/103 - Expressing Twenty-Four/Solution

== by : $103$ ==
;Expressing Twenty-Four
:In a book published in America was the following:
:"Write $24$ with three equal digits, none of which is $8$.
:(There are two solutions to this problem.)"
:Of course, the answers given are $22 + 2 = 24$, and $3 ^3 - 3 = 24$.
:Readers who are familiar with the old "Four Fours" puzzle, and others of the same class,
::will ask why there are supposed to be only these solutions.
:With which of the remaining digits is a solution equally possible?
== Solution ==
For any digit $n$, we have:
:$\dfrac n {\cdotp \dot n} = \dfrac n {n / 9} = 9$
where the notation $\cdotp \dot n$ is used to denote $\cdotp nnnnnn \ldots$
Hence:
:$\sqrt {\dfrac n {\cdotp \dot n} } = 3$
This gives us:
:$\paren {1 + \sqrt {\dfrac 1 {\cdotp \dot 1} } }! = \paren {1 + 3}! = 4! = 24$
where $!$ denotes the factorial sign.
We have an expression for each of $2$ and $3$.
Then:
:$\paren {4 + 4 - 4}! = 4! = 24$
as one of many such similar and straightforward expressions.
We continue:
:$\paren {5 - \dfrac 5 5}! = 4! = 24$
To accomplish $6$, we use the fact that or any digit $n$, we have:
:$\dfrac n {\cdotp n} = \dfrac n {n / 10} = 10$
Hence:
:$\paren {\dfrac 6 {\cdotp 6} - 6}! = \paren {10 - 6}! = 24$
Using the above technique, we see:
:$\paren {7 - \sqrt {\dfrac 7 {\cdotp \dot 7} } }! = \paren {7 - 3}! = 4! = 24$
$8$ is simple:
:$8 + 8 + 8 = 24$
$9$ falls to the factorial technique again:
:$\paren {\sqrt 9 + \dfrac 9 9}! = \paren {3 + 1}! = 4! = 24$
== Historical Note ==
== Sources ==
* : Solutions: $103$. -- Expressing Twenty-Four
* : Answers: $137$. Expressing Twenty-Four
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/104 - Letter-Figure Puzzle

== by : $104$ ==
;Letter-Figure Puzzle
:It is not difficult, if properly attacked:
:Every letter represents a different digit, and, of course, $A C$, $B C$ etc., are two-figure numbers.
:Can you find the values in figures of all the letters?
=== Click here for solution ===
== Historical Note ==
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $104$. -- Letter-Figure Puzzle
* : Arithmetical and Algebraical Problems: Cryptarithm Puzzles: $159$. Letter-Figure Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Cryptarithms
Henry Ernest Dudeney/Puzzles and Curious Problems/104 - Letter-Figure Puzzle/Historical Note

== Historical Note on by : $104$ - Letter-Figure Puzzle ==
 attributes this to a correspondent identified as C. E. B., whose identity has not been ascertained.
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $104$. -- Letter-Figure Puzzle
* : Arithmetical and Algebraical Problems: Cryptarithm Puzzles: $159$. Letter-Figure Puzzle
Category:Historical Notes
Henry Ernest Dudeney/Puzzles and Curious Problems/104 - Letter-Figure Puzzle/Solution

== by : $104$ ==
;Letter-Figure Puzzle
== Solution ==
 A B C D E F H J K L
---------------------
 1 3 5 7 8 9 6 4 2 0
== Proof ==
Trivially from $(8)$ and $(9)$
:$L = 0$
Hence immediately from $(1)$:
:$A = 1$
Hence:
:$10 < A C < 20$
From $(2)$ we inspect pairs ending in the same digit whose divisor is that same digit twice, and this leads us to:
:$C = 5$
Hence:
== Sources ==
* : Solutions: $104$. -- Letter-Figure Puzzle
* : Answers: $159$. Letter-Figure Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/105 - Equal Fractions

== by : $105$ ==
;Equal Fractions
:Can you construct three ordinary vulgar fractions
::(say, $\tfrac 1 2$, $\tfrac 1 3$, or $\tfrac 1 4$, or anything up to $\tfrac 1 9$ inclusive)
::all of the same value, using in every group all the nine digits once, and once only?
:The fractions may be formed in one of the following ways:
::$\dfrac a b = \dfrac c d = \dfrac {e f} {g h j}$, or $\dfrac a b = \dfrac c {d e} = \dfrac {f g} {h j}$.
:We have only found five cases, but the fifth contains a simple little trick that may escape the reader.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $105$. -- Equal Fractions
* : Arithmetical and Algebraical Problems: Digital Puzzles: $133$. Equal Fractions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/105 - Equal Fractions/Solution

== by : $105$ ==
;Equal Fractions
== Solution ==
The examples given by are:
:$\dfrac 2 4 = \dfrac 3 6 = \dfrac {79} {158}$
:$\dfrac 3 6 = \dfrac 7 {14} = \dfrac {29} {58}$
:$\dfrac 3 6 = \dfrac 9 {18} = \dfrac {27} {54}$
:$\dfrac 2 6 = \dfrac 3 9 = \dfrac {58} {174}$
The fifth case which he offers up is:
:$\dfrac {\cdotp 2} 1 = \dfrac {\cdotp 6} 3 = \dfrac {97} {485}$
where it is noted that the first two fractions are not actually proper fractions.
It is worth calling into question 's definition of a vulgar fraction, as he is unclear.
== Sources ==
* : Solutions: $105$. -- Equal Fractions
* : Answers: $133$. Equal Fractions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/106 - Digits and Primes

== by : $106$ ==
;Digits and Primes
:Using the $9$ digits once, and once only,
::can you find prime numbers that will add up to the smallest total possible?
:Here is an example. The four primes contain all the $9$ digits once, and once only, and add up to $450$,
::but this total can be considerably reduced.
:It is quite an easy puzzle.
 6 1
 2 8 3
 4 7
+ 5 9
------
 4 5 0
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $106$. -- Digits and Primes
* : Arithmetical and Algebraical Problems: Digital Puzzles: $134$. Digits and Primes
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/106 - Digits and Primes/Solution

== by : $106$ ==
;Digits and Primes
== Solution ==
The solution given by is:
 8 9
 6 1
 4 7
 5
 3
+ 2
 -----
 2 0 7
but we also have:
 8 9
 6 7
 4 1
 5
 3
+ 2
 -----
 2 0 7
== Proof ==
From the start we make the assumption that we can do this using primes less than $100$.
We have that $4$, $6$ and $8$ must appear in the tens column.
Hence we must use one element of each of the following sets:
:$\set {41, 43, 47}$
:$\set {61, 67}$
:$\set {83, 89}$
By inspection we can see that $41$, $67$ and $89$ use up $6$ digits each, while the remaining digits are primes on their own.
There can be no sum lower than this.
We also note that $41 + 67 = 47 + 61$, giving a second solution.
== Sources ==
* : Solutions: $106$. -- Digits and Primes
* : Answers: $134$. Digits and Primes
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/107 - A Square of Digits

== by : $107$ ==
;A Square of Digits
$\qquad \begin{array}{|c|c|c|}
\hline 2 & 1 & 8 \\
\hline 4 & 3 & 9 \\
\hline 6 & 5 & 7 \\
\hline
\end{array} \qquad \begin{array}{|c|c|c|}
\hline 2 & 7 & 3 \\
\hline 5 & 4 & 6 \\
\hline 8 & 1 & 9 \\
\hline
\end{array} \qquad \begin{array}{|c|c|c|}
\hline 3 & 2 & 7 \\
\hline 6 & 5 & 4 \\
\hline 9 & 8 & 1 \\
\hline
\end{array}$
:The $9$ digits may be arranged in a square in many ways,
::so that the numbers formed in the first row and second row will sum to the third row.
:We give $3$ examples, and it will be found that the difference between the first total, $657$, and the second total, $819$,
::is the same as the difference between the second, $819$, and the third, $981$ --
::that is, $162$.
:Now, can you form $8$ such squares, every one containing the $9$ digits,
::so that the common difference between the $8$ totals is throughout the same?
:Of course it will not be $162$.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $107$. -- A Square of Digits
* : Arithmetical and Algebraical Problems: Digital Puzzles: $135$. A Square of Digits
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/107 - A Square of Digits/Solution

== by : $107$ ==
;A Square of Digits
== Solution ==
:$\begin{array}{|c|c|c|}
\hline 2 & 4 & 3 \\
\hline 6 & 7 & 5 \\
\hline 9 & 1 & 8 \\
\hline
\end{array} \qquad \begin{array}{|c|c|c|}
\hline 3 & 4 & 1 \\
\hline 5 & 8 & 6 \\
\hline 9 & 2 & 7 \\
\hline
\end{array} \qquad \begin{array}{|c|c|c|}
\hline 1 & 5 & 4 \\
\hline 7 & 8 & 2 \\
\hline 9 & 3 & 6 \\
\hline
\end{array} \qquad \begin{array}{|c|c|c|}
\hline 3 & 1 & 7 \\
\hline 6 & 2 & 8 \\
\hline 9 & 4 & 5 \\
\hline
\end{array}$
:$\begin{array}{|c|c|c|}
\hline 2 & 1 & 6 \\
\hline 7 & 3 & 8 \\
\hline 9 & 5 & 4 \\
\hline
\end{array} \qquad \begin{array}{|c|c|c|}
\hline 2 & 1 & 5 \\
\hline 7 & 4 & 8 \\
\hline 9 & 6 & 3 \\
\hline
\end{array} \qquad \begin{array}{|c|c|c|}
\hline 3 & 1 & 8 \\
\hline 6 & 5 & 4 \\
\hline 9 & 7 & 2 \\
\hline
\end{array} \qquad \begin{array}{|c|c|c|}
\hline 2 & 3 & 5 \\
\hline 7 & 4 & 6 \\
\hline 9 & 8 & 1 \\
\hline
\end{array}$
The common difference is $9$.
== Sources ==
* : Solutions: $107$. -- A Square of Digits
* : Answers: $135$. A Square of Digits
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/108 - The Nine Digits

== by : $108$ ==
;The Nine Digits
:It will be found that $32 \, 547 \, 891$ multiplied by $6$ (thus using all the $9$ digits once, and once only)
::gives the product $195 \, 287 \, 346$ (also containing all the $9$ digits once, and once only).
:Can you find another number to be multiplied by $6$ under the same conditions?
:Remember that the $9$ digits must appear once, and once only, in the sums multiplied and in the product.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $108$. -- The Nine Digits
* : Arithmetical and Algebraical Problems: Digital Puzzles: $136$. The Nine Digits
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Penholodigital Integers
Henry Ernest Dudeney/Puzzles and Curious Problems/108 - The Nine Digits/Solution

== by : $108$ ==
;The Nine Digits
== Solution ==
 offers up:
:$94 \, 857 \, 312 \times 6 = 569 \, 143 \, 872$
 reports that has found $2$ more:
:$89 \, 745 \, 321 \times 6 = 538 \, 471 \, 926$
:$98 \, 745 \, 231 \times 6 = 592 \, 471 \, 386$
However, an exhaustive search using a simple computer program reveals that there are in fact $87$ solutions in total:
:Try it online!
== Sources ==
* : Solutions: $108$. -- The Nine Digits
* : Answers: $136$. The Nine Digits
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/109 - Perfect Squares

== by : $109$ ==
;Perfect Squares
:Find $4$ numbers such that the sum of every two and the sum of all four may be perfect squares.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Digital Puzzles: $109$. -- Perfect Squares
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $188$. Perfect Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/109 - Perfect Squares/Solution

== by : $109$ ==
;Perfect Squares
== Solution ==
The smallest such set seems to be:
:$10 \, 430, 3970, 2114, 386$
We have:
== Proof ==
== Sources ==
* : Solutions: $109$. -- Perfect Squares
* : Answers: $188$. Perfect Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/10 - Mental Arithmetic

== by : $10$ ==
;Mental Arithmetic
:If a tobacconist offers a cigar at $7 \tfrac 3 4 \oldpence$,
::but says we can have the box of $100$ for $65 \shillings$,
::shall we save much by buying the box?
:In other words, what would $100$ at $7 \tfrac 3 4 \oldpence$ cost?
:By a little rule that we shall give the calculation takes only a few moments.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $10$. -- Mental Arithmetic
Category:Henry Ernest Dudeney/Puzzles and Curious Problems