nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Puzzles and Curious Problems/356 - A Rail Problem

== by : $356$ ==
;A Rail Problem
:There is a garden railing similar to our design.
:
:In each division between two uprights there is an equal number of ornamental rails,
::and a rail is divided in halves and a portion stuck on each side of every upright,
::except that the uprights at the end have not been given half rails.
:Idly counting the rails from one end to another, we found that there were $1223$ rails, counting two halves as one rail.
:We also noticed that the number of those divisions was five more than twice the number of whole rails in a division.
:How many rails were there in each division?
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $356$. -- A Rail Problem
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $256$. A Rail Problem
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/356 - A Rail Problem/Solution

== by : $356$ ==
;A Rail Problem
== Solution ==
There are $51$ divisions of $23$ whole rails each.
== Proof ==
Let $n$ be the number of whole rails in a division.
Let $d$ be the total number of divisions.
Each of the outer divisions has $n + \dfrac 1 2$ rails.
Each of the inner divisions has $n + 1$ rails, that is, $n$ plus the two half rails at either end.
Hence we have:
== Sources ==
* : Solutions: $356$. -- A Rail Problem
* : Answers: $256$. A Rail Problem
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/357 - The Wheel Puzzle

== by : $357$ ==
;The Wheel Puzzle
:
:Place the numbers $1$ to $19$ in the $19$ circles,
::so that wherever there are three in a straight line they shall add up to $30$.
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $357$. -- The Wheel Puzzle
* : Combinatorial & Topological Problems: Magic Star Puzzles: $402$. The Wheel Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/357 - The Wheel Puzzle/Solution

== by : $357$ ==
;The Wheel Puzzle
== Solution ==
Place $10$ in the centre.
Place the numbers $1$ to $9$ around the outside, with $11$ to $19$ opposite them to make each line add up as appropriate.
== Sources ==
* : Solutions: $357$. -- The Wheel Puzzle
* : Answers: $402$. The Wheel Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/358 - Simple Addition

== by : $358$ ==
;Simple Addition
:Can you show that four added to six will make eleven?
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $358$. -- Simple Addition
* : Unclassified Puzzles: $515$. Simple Addition
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/358 - Simple Addition/Solution

== by : $358$ ==
;Simple Addition
== Solution ==
:Express $4$ and $6$ in Roman numerals, perhaps using matches: $\text {IV}$ and $\text {VI}$.
:Then turn $\text {IV}$ upside down and put it below $\text {VI}$, and see it makes $\text {XI}$.
:This is $11$ in Roman numerals.
== Sources ==
* : Solutions: $358$. -- Simple Addition
* : Answers: $515$. Simple Addition
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/359 - Queer Arithmetic

== by : $359$ ==
;Queer Arithmetic
:Can you take away seven-tenths from five so that exactly four remains?
When republished this puzzle in in $1968$, he added the hint that the task should be accomplished "with matches".
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $359$. -- Queer Arithmetic
* : Match Puzzles: $509$. Queer Arithmetic
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/359 - Queer Arithmetic/Solution

== by : $359$ ==
;Queer Arithmetic
== Solution ==
:Arrange $10$ matches to spell the number $\text {FIVE}$.
:Then remove the $7$ of those $10$ that are used to form the $\text F$ and the $\text E$.
:This leaves us with the $3$ matches remaining to form the $\text{IV}$, which is $4$ in Roman numerals.
== Sources ==
* : Solutions: $359$. -- Queer Arithmetic
* : Answers: $509$. Queer Arithmetic
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/35 - The Reapers' Puzzle

== by : $35$ ==
;The Reapers' Puzzle
:Three men were to receive $90 \shillings$ for harvesting a field, conditionally upon the work being done in $5$ days.
:Jake could do it alone in $9$ days, but as Ben was not as good a workman they were compelled to engage Bill for $2$ days,
::in consequence of which Ben got $3 \shillings 9 \oldpence$ less than he would otherwise have received.
:How long would it have taken Ben and Bill together to complete the work?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $35$. -- The Reapers' Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Cistern Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/35 - The Reapers' Puzzle/Solution

== by : $35$ ==
;The Reapers' Puzzle
:Three men were to receive $90 \shillings$ for harvesting a field, conditionally upon the work being done in $5$ days.
:Jake could do it alone in $9$ days, but as Ben was not as good a workman they were compelled to engage Bill for $2$ days,
::in consequence of which Ben got $3 \shillings 9 \oldpence$ less than he would otherwise have received.
:How long would it have taken Ben and Bill together to complete the work?
== Solution ==
:$9 \tfrac {63} {73}$ days.
== Proof ==
Let $a, b, c$ be the rate of working in fields harvested per day of (respectively) Jake, Ben and Bill.
Let $t_a, t_b, t_c$ be the number of days it would take (respectively) Jake, Ben and Bill to harvest the field alone.
Let $s_a, s_b, s_c$ be the number of shillings per day (respectively) Jake, Ben and Bill earn by harvesting fields.
We have that the rate of remuneration is $90 \shillings$ per field in $5$ days.
That is:
:$5 \paren {s_a + s_b + s_c} = 90$
So the rate of remuneration is $18 \shillings$ per field per day.
We have that:
:$t_a = 9$
That is:
:$a = \dfrac 1 9$
Hence after $5$ days, Jake would have got $\dfrac 5 9$ of the field harvested, and would earn $50 \shillings$.
Hence his pay rate is:
:$s_a = 10$
Ben would have then earned $40 \shillings$, but $3 \shillings 9 \oldpence$ of that had to be given to Bill.
So Ben is worth $36 \shillings 3 \oldpence$ for $5$ days' work.
Hence all $90 \shillings$ remuneration is accounted for:
:Jake earns $50 \shillings$ for $5$ days' work
:Ben earns $36 \shillings 3 \oldpence$ for $5$ days' work
:Bill earns $3 \shillings 9 \oldpence$ for $2$ days' work.
Hence we have:
:$s_a = \dfrac 1 5 \times 50 = 10$
:$s_b = \dfrac 1 5 \times 36 \tfrac 1 4 = \dfrac 1 5 \times \dfrac {145} 4 = \dfrac {29} 4 = 7 \tfrac 1 4$
:$s_c = \dfrac 1 2 \times 3 \tfrac 3 4 = \dfrac 1 2 \times \dfrac {15} 4 = \dfrac {15} 8 = 1 \tfrac 1 8$
Hence to earn the full $90 \shillings$ Ben needs to work $90 \div \dfrac {29} 4 = \dfrac {360} {29} = 12 \tfrac {12} {29}$ days.
That is:
:$t_b = 12 \tfrac {12} {29}$
and so:
:$b = \dfrac {29} {360}$
To earn the full $90 \shillings$ Bill needs to work $90 \div \dfrac {15} 8 = 48$ days.
That is:
:$t_c = 48$
and so:
:$c = \dfrac 1 {48}$
Now let $t$ be the number of days it will take Bill and Ben to harvest the field together.
So:
== Sources ==
* : Solutions: $35$. -- The Reapers' Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/360 - Fort Garrisons

== by : $360$ ==
;Fort Garrisons
:Here we have a system of fortifications.
:It will be seen that there are ten forts, connected by lines of outworks,
::and the numbers represent the strength of the small garrisons.
:
:The General wants to dispose these garrisons afresh so that there shall be $100$ men in every one of the five lines of four forts.
:The garrison must be moved bodily -- that is to say, you are not allowed to break them up into other numbers.
:It is an entertaining little puzzle with counters, and not very difficult.
=== Click here for Dudeney's solution ===
=== Click here for full solution ===
== Sources ==
* : Unclassified Problems: $360$. -- Fort Garrisons
* : Combinatorial & Topological Problems: Magic Star Puzzles: $397$. Fort Garrisons
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/360 - Fort Garrisons/Solution

== by : $360$ ==
;Fort Garrisons
== Solution ==
:The diagram shows one way of doing it.
:
== Sources ==
* : Solutions: $360$. -- Fort Garrisons
* : Answers: $397$. Fort Garrisons
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/361 - Constellation Puzzle

== by : $361$ ==
;Constellation Puzzle
:The arrangement of stars in the diagram is known as "The British Constellation".
:It is not given in any star map or books, and it is very difficult to find on the clearest night for the simple reason that it is not visible.
:
:The $21$ stars form seven lines with $5$ stars in every line.
:Can you rearrange these $21$ stars so that they form $11$ straight lines with $5$ stars in every line?
:There are many solutions.
:Try to find a symmetrical one.
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $361$. -- Constellation Puzzle
* : Combinatorial & Topological Problems: Point Alignment Puzzles: $441$. Constellation Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/361 - Constellation Puzzle/Solution

== by : $361$ ==
;Constellation Puzzle
== Solution ==
:Here is one such solution:
:
== Sources ==
* : Solutions: $361$. -- Constellation Puzzle
* : Answers: $441$. Constellation Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/362 - Intelligence Tests

== by : $362$ ==
;Intelligence Tests
:These are examples of the sort of intelligence tests that were given to children in 's day.
:An English officer fell asleep in church during a sermon.
:He was dreaming that the executioner was approaching him to cut off his head,
::and just as the sword was descending on the officer's unhappy neck
::his wife lightly touched her husband on the back of his neck with her fan to awaken him.
:The shock was too great, and the officer fell forward dead.
:Now, there is something wrong with this.
:What is it?
:Another such question goes along these lines:
:If we sell apples by the cubic inch,
::how can we really find the exact number of cubic inches in, say, a dozen dozen apples?
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $362$. -- Intelligence Tests
* : Unclassified Puzzles: $533$. Intelligence Tests
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/362 - Intelligence Tests/Solution

== by : $362$ ==
;Intelligence Tests
== Solution ==
:As the dreamer never woke from his dream, he cannot possibly have been able to relate it to anyone.
:So nobody could have known anything about it.
:Hence the story must have been a complete invention.
:As for the question about the apples: dunk them in water and measure the runoff.
== Sources ==
* : Solutions: $362$. -- Intelligence Tests
* : Answers: $533$. Intelligence Tests
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/363 - At the Mountain Top

== by : $363$ ==
;At the Mountain Top
:"When I was in Italy I was taken to the top of a mountain
::and shown that a mug would hold less liquor at the top of the mountain than in the valley beneath.
:Can you tell me," asked Professor Rackbrane, "what mountain this might be that has so strange a property?"
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $363$. -- At the Mountain Top
* : Unclassified Puzzles: $534$. At the Mountain Top
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/363 - At the Mountain Top/Solution

== by : $363$ ==
;At the Mountain Top
== Solution ==
:The surface of water, or other liquid, is always spherical, and the greater the sphere is the less is its convexity.
:The spherical surface of the water must, therefore, be less above the brim of the vessel
::and consequently it will hold less at the top of a mountain than at the bottom.
:This applies to any mountain whatsoever.
== Sources ==
* : Solutions: $363$. -- At the Mountain Top
* : Answers: $534$. At the Mountain Top
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/364 - Cupid's Arithmetic

== by : $364$ ==
;Cupid's Arithmetic
:Dora Crackham one morning produced a slip of paper bearing the jumble of figures shown in our diagram.
:
:She said that a young mathematician had this poser presented to him by his betrothed when she was in a playful mood.
:"What am I to do with it?" asked George.
:"Just interpret its meaning," she replied. "If it is properly regarded it should not be difficult to decipher."
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $364$. -- Cupid's Arithmetic
* : Unclassified Puzzles: $535$. Cupid's Arithmetic
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/364 - Cupid's Arithmetic/Solution

== by : $364$ ==
;Cupid's Arithmetic
== Solution ==
:All the young mathematician had to do was to reverse the paper and hold it up to the light,
::or hold it in front of a mirror,
::where he would immediately see that his betrothed's curious jumble of figures will read:
:"Kiss me, dearest."
== Sources ==
* : Solutions: $364$. -- Cupid's Arithmetic
* : Answers: $535$. Cupid's Arithmetic
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/365 - Tangrams

== by : $365$ ==
;Tangrams
:Those readers who were interested in the article on "Tangrams" in ,
::may be glad to have a further collection of examples of the strikingly realistic figures and designs that can be produced by combining these curious shaped pieces.
:In the diagram the square is shown cut into the $7$ pieces.
:If you mark the point $B$, midway between $A$ and $C$, on one side of a square of any size,
::and $D$, midway between $C$ and $E$, on an adjoining side, the direction of the cuts is obvious.
:
=== Examples ===
== Historical Note ==
== Sources ==
* : Unclassified Problems: $365$. -- Tangrams
* : Unclassified Puzzles: $536$. Tangrams
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/365 - Tangrams/Examples

== by : $365$ ==
;Examples of Patterns Created using Tangrams
:In the examples given below, two complete sets of seven pieces have been combined.
: $\qquad$ 
:::A man riding a bicycle, and a man pushing a wheelbarrow
: $\qquad$ 
:::A boy riding a donkey, and a motor-car
: $\qquad$ 
:::A house, and a dog
: $\qquad$ 
:::A horse, and the British lion
: $\qquad$ 
:::Two cricketers in motion
== Historical Note ==
== Sources ==
* : Unclassified Problems: $365$. -- Tangrams
* : Unclassified Puzzles: $536$. Tangrams
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/36 - The Flagons of Wine

== by : $36$ ==
;The Flagons of Wine
:A quart of Burgundy costs $4 \shillings 9 \oldpence$, but $3 \oldpence$ is returnable on the empty flagon,
::so that the Burgundy seems to be worth $4 \shillings 6 \oldpence$
:For $12$ of the capsules with which each of the quart flagons is sealed, a free flagon of the same value is obtained.
:What is the value of a single capsule?
:Obviously a twelfth of $4 \shillings 6 \oldpence$ which is $4 \tfrac 1 2 \oldpence$
:But the free flagon also has a capsule worth $4 \tfrac 1 2 \oldpence$, so that this full flagon appears to be worth $4 \shillings 10 \tfrac 1 2 \oldpence$,
::which makes the capsule worth a twelfth of $4 \shillings 10 \tfrac 1 2 \oldpence$, or $4 \tfrac 7 8 \oldpence$,
::and so on ad infinitum, with an ever-increasing value.
:Where is the fallacy, and what is the real worth of a capsule?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $36$. -- The Flagons of Wine
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/36 - The Flagons of Wine/Solution

== by : $36$ ==
;The Flagons of Wine
== Solution ==
It is assumed that all empties are returned.
Then $13$ flagons of wine plus $1$ flagon and $1$ cap will cost $12 \times 4 \shillings 6 \oldpence$, less $3 \oldpence$ which makes $53 \shillings 9 \oldpence$ plus $12$ caps.
Without the capsule system, these $13$ flagons of wine would cost $13 \times 4 \shillings 6 \oldpence$ which is $58 \shillings 6 \oldpence$.
Therefore the difference, $4 \shillings 9 \oldpence$ represents the $12$ caps, each of which is worth $4 \tfrac 3 4 \oldpence$
For $12$ more flagons of wine plus $1$ flagon and $1$ cap, I pay $11 \times 4 \shillings 6 \oldpence$ less $3 \oldpence$ which is $49 \shillings 3 \oldpence$
The cost of these $12$, without caps, would be $54 \shillings$, which shows again the difference of $4 \shillings 9 \oldpence$, or $4 \tfrac 3 4 \oldpence$ a cap.
The cost of the caps will be the same for every dozen flagons of wine that we buy.
The stated error consists partly in overlooking the $3 \oldpence$, which is the value of the empty flagon when returned.
== Sources ==
* : Solutions: $36$. -- The Flagons of Wine
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/37 - A Wages Paradox

== by : $37$ ==
;A Wages Paradox
:"I want a rise, sir," said the office-boy.
:"That's nonsense," said the employer.
:"If I give you a rise you will really be getting less wages per week than you are getting now."
:The boy pondered over this, but was unable to see how such a thing could happen.
:Can you explain it?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $37$. -- A Wages Paradox
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/37 - A Wages Paradox/Solution

== by : $37$ ==
;A Wages Paradox
:"I want a rise, sir," said the office-boy.
:"That's nonsense," said the employer.
:"If I give you a rise you will really be getting less wages per week than you are getting now."
:The boy pondered over this, but was unable to see how such a thing could happen.
:Can you explain it?
== Solution ==
The unconvincing solution given by is as follows.
The office-boy was earning $\pounds 3, 10 \shillings$ per week.
This was to be increased to $\pounds 200$ per annum, paid monthly by cheque.
It is decided that the first month of the new arrangement was that he was to continue to be paid week by week.
There are $5$ weeks in the first month.
As the total paid in the first month is $\pounds 16, 13 \shillings 4 \oldpence$, it follows that the weekly sum paid on each of the five pay days would be $\pounds 3, 6 \shillings 8 \oldpence$ only.
This is less by $3 \shillings 4 \oldpence$ than under the old arrangement.
So he will receive less per week, though earning more per annum.
This state of affairs happens, of course, only when there are $5$ paydays in the month.
== Sources ==
* : Solutions: $37$. -- A Wages Paradox
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/38 - The Picnic

== by : $38$ ==
;The Picnic
:Four married couples had a picnic together, and their refreshments included $32$ bottles of lemonade.
:Mary only disposed of one bottle,
::Anne had two,
::Jane swallowed the contents of three,
::and Elizabeth emptied four bottles.
:The husbands were more thirsty,
::except John MacGregor, who drank the same quantity as his better half.
:Lloyd Jones drank twice as much as his wife,
::William Smith three times as much as his wife,
::and Patrick Dolan four times as much as his wife demanded.
:The puzzle is to find the surnames of the ladies.
:Which man was married to which woman?
=== Click here for solution ===
== Also see ==
* Henry Ernest Dudeney: Puzzles and Curious Problems $51$ - Sharing the Apples
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $38$. -- The Picnic
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/38 - The Picnic
Henry Ernest Dudeney/Puzzles and Curious Problems/38 - The Picnic/Solution

== by : $38$ ==
;The Picnic
== Solution ==
:Mary is the wife of William Smith
:Anne is the wife of Patrick Dolan
:Jane is the wife of John MacGregor
:Elizabeth is the wife of Lloyd Jones.
== Proof 1 ==
== Proof 2 ==
So:
:Jane drank the same quantity as her husband John MacGregor, that is, $3$ bottles
:Lloyd Jones drank twice as much as the $4$ bottles drunk by his wife Elizabeth
:William Smith drank three times as much as his wife Mary, who drank just $1$ bottle
:Patrick Dolan drank four times as much as his wife Anne's $2$ bottles.
== Also see ==
* Henry Ernest Dudeney: Puzzles and Curious Problems $51$ - Sharing the Apples
== Sources ==
* : Solutions: $38$. -- The Picnic
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/38 - The Picnic
Henry Ernest Dudeney/Puzzles and Curious Problems/38 - The Picnic/Solution/Proof 1

== by : $38$ ==
== Proof ==
Let us set up the following system of linear simultaneous equations in matrix form:
:$\begin {pmatrix}
 1 & 1 & 1 & 1 \\
 1 & 2 & 3 & 4 \\
\end {pmatrix} \begin {pmatrix} a \\ b \\ c \\ d \end {pmatrix} = \begin {pmatrix} 10 \\ 22 \end {pmatrix}$
where:
:$\set {a, b, c, d} = \set {1, 2, 3, 4}$
Conversion to echelon form proceeds as follows:
The first of these can be rewritten in the form:
:$a = c + 2 d - 2$
from which it is immediate that $a$ and $c$ are of the same parity.
Hence $b$ and $d$ are also of the same parity as each other.
This reduces the possible permutations of $\tuple {a, b, c, d}$ that need to be investigated.
Let $a = 1$, and so $c = 3$:
Then:
So $a \ne 1$.
Let $a = 2$, and so $c = 4$:
Then:
So $a \ne 2$.
Let $a = 3$, and so $c = 1$:
Then:
So $a = 3$, $c = 1$, $d = 2$ is a possible solution, giving $b = 4$.
We check whether $a + 2 b + 3 c + 4 d = 22$, and find:
:$a + 2 b + 3 c + 4 d = 3 + 2 \times 4 + 3 \times 1 + 4 \times 2 = 22$
showing that this is indeed a valid solution.
Let $a = 4$, and so $c = 2$:
Then:
So $a = 4$, $c = 2$, $d = 3$ is a possible solution, giving $b = 1$.
We check whether $a + 2 b + 3 c + 4 d = 22$, but find:
:$a + 2 b + 3 c + 4 d = 4 + 2 \times 1 + 3 \times 2 + 4 \times 3 = 24$
showing that this is not a valid solution after all.
So we have eliminated all possible solutions except:
:$\tuple {a, b, c, d} = \tuple {3, 4, 1, 2}$
So:
:Jane drank the same quantity as her husband John MacGregor, that is, $3$ bottles
:Lloyd Jones drank twice as much as the $4$ bottles drunk by his wife Elizabeth
:William Smith drank three times as much as his wife Mary, who drank just $1$ bottle
:Patrick Dolan drank four times as much as his wife Anne's $2$ bottles.
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/38 - The Picnic