nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Puzzles and Curious Problems/38 - The Picnic/Solution/Proof 2

== by : $38$ ==
== Proof ==
From:
:$a + 2 b + 3 c + 4 d = 22$
we can infer that:
:$a + 3 c = 22 - 2 b - 4 d$
is even.
Hence $a$ and $c$ must be of the same parity.
Suppose both $a$ and $c$ are even.
Then $b$ and $d$ are both odd.
Either $a = 2$ and $c = 4$ or vice versa.
In either case:
:$a + 3 c \equiv 2 \pmod 4$
Because $b$ will be odd:
:$2 p \equiv 2 \pmod 4$
and so:
:$a + 3 c = 22 - 2 b - 4 d \equiv 0 \pmod 4$
Thus we have a contradiction.
So $a$ and $c$ must both be odd and therefore $b$ and $d$ must both be even.
$d$ cannot be $4$ since any assignment of $\set {1, 2, 3}$ to $\tuple {a, b, c}$ yields:
:$a + 2 b + 3 c > 22 - \paren {4 \times 4} = 6$
So we must have $d = 2$ and $b = 4$, leading to:
:$a + 3 c = 22 - \paren {2 \times 4} - \paren {4 \times 2} = 6$
from which $a = 3$ and $c = 1$ follow immediately.
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.
== Sources ==
* 
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/38 - The Picnic
Henry Ernest Dudeney/Puzzles and Curious Problems/39 - Surprising Relationship

== by : $39$ ==
;Surprising Relationship
:: "You say that Mr. Tomkins is your uncle?"
:: "Yes, and I am his uncle!"
:: Then -- let me see -- you must be nephew to each other, of course! Funny, isn't it?"
:Can you say quite simply how this might be, without any breach of the marriage law or disregard of the Table of Affinity?
=== Click here for solution 1 ===
=== Click here for solution 2 ===
=== Click here for solution 3 ===
== Historical Note ==
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $39$. -- Surprising Relationship
* : Unclassified Puzzles: $519$. Surprising Relationship
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/39 - Surprising Relationship
Category:Strange Family Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/39 - Surprising Relationship/Historical Note

== Historical Note on by : $39$ - Surprising Relationship ==
The original solution appears to have been the work of himself.
As reported by in ins $1968$ repackaging , the other two solutions originated with .
== Sources ==
* : Unclassified Puzzles: $519$. Surprising Relationship
Category:Historical Notes
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/39 - Surprising Relationship
Henry Ernest Dudeney/Puzzles and Curious Problems/39 - Surprising Relationship/Solution 1

== by : $39$ ==
;Surprising Relationship
== Solution ==
The simplest way for this to happen is as follows.
Two men, $A$ and $B$, married the mother of the other.
Of these marriages, $A$ became Edwin's father and $B$ became Mr. Tomkins' father.
Then $A$ is then the half-brother of Mr. Tomkins, and $B$ is the half-brother of Edwin.
Thus Mr. Tomkins is the uncle of Edwin, who is the nephew of Mr. Tomkins, and Edwin is the uncle of Mr. Tomkins, who is the nephew of Edwin.
== Historical Note ==
== Sources ==
* : Solutions: $39$. -- Surprising Relationship
* : Answers: $519$. Surprising Relationship
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/39 - Surprising Relationship
Henry Ernest Dudeney/Puzzles and Curious Problems/39 - Surprising Relationship/Solution 2

== by : $39$ ==
;Surprising Relationship
== Solution ==
Two women, $V$ and $W$, married the father of the other.
Of these marriages, $W$ became Edwin's mother and $V$ became Mr. Tomkins' mother.
Then $A$ is then the half-brother of Mr. Tomkins, and $B$ is the half-brother of Edwin.
Thus Mr. Tomkins is the uncle of Edwin, who is the nephew of Mr. Tomkins, and Edwin is the uncle of Mr. Tomkins, who is the nephew of Edwin.
== Historical Note ==
== Sources ==
* : Answers: $519$. Surprising Relationship
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/39 - Surprising Relationship
Henry Ernest Dudeney/Puzzles and Curious Problems/39 - Surprising Relationship/Solution 3

== by : $39$ ==
;Surprising Relationship
== Solution ==
A man $G$ marries the mother of a woman $H$, while $H$ marries the father of the $G$.
Of these marriages, $G$ became Edwin's father and $H$ became Mr. Tomkins' mother.
Then $A$ is then the half-brother of Mr. Tomkins, and $B$ is the half-brother of Edwin.
Thus Mr. Tomkins is the uncle of Edwin, who is the nephew of Mr. Tomkins, and Edwin is the uncle of Mr. Tomkins, who is the nephew of Edwin.
== Historical Note ==
== Sources ==
* : Answers: $519$. Surprising Relationship
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/39 - Surprising Relationship
Henry Ernest Dudeney/Puzzles and Curious Problems/3 - Buying Toys

== by : $3$ ==
;Buying Toys
:George and William were sent out to buy toys for the family Christmas tree,
::and, unknown to each other, both went at different times to the same little shop,
::where they had sold all their stock of small toys
::except engines at $4 \oldpence$, balls at $3 \oldpence$ each, dolls at $2 \oldpence$ each, and trumpets at $\tfrac 1 2 \oldpence$ each.
:They both bought some of all, and obtained $21$ articles, spending $2 \shillings$ each.
:But William bought more trumpets than George.
:What were their purchases?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $3$. -- Buying Toys
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/3 - Buying Toys/Solution

== by : $3$ ==
;Buying Toys
== Solution ==
George bought $1$ engine, $1$ ball, $5$ dolls and $14$ trumpets.
William bought $2$ engines, $2$ balls, $1$ doll and $16$ trumpets.
== Proof ==
A glance at the solution in the book clarifies the question, telling us that they obtained $21$ articles each.
Hence we are looking at a variant of the One Hundred Fowls problem.
Recall:
:$1$ shilling ($20 \shillings$) is $12$ (old) pence ($12 \oldpence$)
Let all prices be expressed, therefore, in (old) pence.
Let $e$, $b$, $d$ and $t$ denote the number of engines, balls, dolls and trumpets bought respectively by either George or William.
We have:
Note that all of $e$, $b$ and $d$ need to be strictly positive.
We need to find possible values of $b$ and $d$ so as to make $27 - 5 b - 3 d$ divisible by $7$.
Let $b = 1$.
Then we have:
:$27 - 5 - 3 d = 22 - 3 d = 7 e$
which can be satisfied (only) by $d = 5$, giving:
:$e = 1$, $b = 1$, $d = 5$
Thus:
:$4 e + 3 b + 2 d + \tfrac 1 2 t = 17 + \dfrac t 2 = 24$
hence making $t = 14$.
Let $b = 2$.
Then we have:
:$27 - 10 - 3 d = 17 - 3 d = 7 e$
which can be satisfied (only) by $d = 1$, giving:
:$e = 2$, $b = 2$, $d = 1$
Thus:
:$4 e + 3 b + 2 d + \tfrac 1 2 t = 16 + \dfrac t 2 = 24$
hence making $t = 16$.
Let $b = 3$.
Then we have:
:$27 - 15 - 3 d = 12 - 3 d = 7 e$
which can be satisfied by no value of $d$.
Let $b = 4$.
Then we have:
:$27 - 20 - 3 d = 7 - 3 d = 7 e$
which can be satisfied (only) by $d = 0$.
Similarly, $b - 5$ returns no solutions.
Hence there are two possible purchases of $21$ toys for $2 \shillings$, and those are:
:$1$ engine, $1$ ball, $5$ dolls and $14$ trumpets
and:
:$2$ engines, $2$ balls, $1$ doll and $16$ trumpets.
The answer is completed on taking note that William bought more trumpets than George.
== Sources ==
* : Solutions: $3$. -- Buying Toys
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/40 - An Epitaph (A.D. 1538)

== by : $40$ ==
;An Epitaph (A.D. $1538$)
:Two grandmothers, with their two granddaughters;
:Two husbands, with their two wives;
:Two fathers, with their two daughters;
:Two mothers, with their two sons;
:Two maidens, with their two mothers;
:Two sisters, with their two brothers;
::Yet only six in all lie buried here;
::All born legitimate, from incest clear.
:How might this happen?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $40$. -- An Epitaph (A.D. $1538$)
* : Unclassified Puzzles: $520$. An Epitaph ( $1538$)
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Strange Family Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/40 - An Epitaph (A.D. 1538)/Solution

== by : $40$ ==
;An Epitaph (A.D. $1538$)
== Solution ==
Two widows each had a son.
Each widow married the son of the other.
Each had a daughter in this new relationship.
All those other relationships will be found to result.
== Sources ==
* : Solutions: $40$. -- An Epitaph (A.D. $1538$)
* : Answers: $520$. An Epitaph ( $1538$)
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/41 - Ancient Problem

== by : $41$ ==
;Ancient Problem
:Here is an example of the sort of "Breakfast-time Problem" propounded by in $310$.
:Demochares has lived one-fourth of his life as a boy;
:one-fifth as a youth;
:one-third as a man;
:and has spent thirteen years in his dotage.
:How old is this gentleman?
=== Click here for solution ===
== Historical Note ==
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $41$. -- Ancient Problem
* : Arithmetical and Algebraical Problems: Age Puzzles: $36$. Ancient Problem
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/41 - Ancient Problem/Historical Note

== Historical Note on by : $41$ -- Ancient Problem ==
 points out in his that this puzzle was propounded by in $310$.
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $41$. -- Ancient Problem
* : Arithmetical and Algebraical Problems: Age Puzzles: $36$. Ancient Problem
Category:Historical Notes
Henry Ernest Dudeney/Puzzles and Curious Problems/41 - Ancient Problem/Solution

== by : $41$ ==
;Ancient Problem
== Solution ==
Demochares is $60$ years old.
This is epigram $127$ of book $\text {XIV}$ of The Greek Anthology.
Contemporary middle-aged readers may object to the definition of "dotage" as starting at age $47$.
== Sources ==
* : Solutions: $41$. -- Ancient Problem
* : Answers: Age Puzzles: $36$. Ancient Problem
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/42 - Family Ages

== by : $42$ ==
;Family Ages
:A man and his wife had three children, John, Ben, and Mary,
::and the difference between their parents' ages was the same as between John and Ben and between Ben and Mary.
:The ages of John and Ben, multiplied together, equalled the age of the father,
::and the ages of Ben and Mary multiplied together equalled the age of the mother.
:The combined ages of the family amounted to ninety years.
:What was the age of each person?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $42$. -- Family Ages
* : Arithmetical and Algebraical Problems: Age Puzzles: $37$. Family Ages
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/42 - Family Ages/Solution

== by : $42$ ==
;Family Ages
== Solution ==
The children are triplets, all aged $6$.
The mother and father are both aged $36$.
== Proof ==
Let $j$, $b$ and $m$ denote the ages of John, Ben and Mary respectively.
Let $F$ and $M$ denote the ages of the father and mother respectively.
We have that:
:$\size {F - M} = \size {j - b} = \size {b - m}$
It is immediately apparent that this can be solved simply by:
:$F = M$
and so:
:$j - b - m$
where:
:$j^2 = b^2 = m^2 = F = M$
such that:
:$3 j + 2 F = 90$
The very $3$-ish nature of the final equation tempts us to try $j = 6$, and we see:
:$3 \times 6 + 2 \times 36 = 90$
Note that we are asked to specify the age of each person.
But the only information we have is the difference between the ages, and not which is older than the other.
So if the children have different ages, we cannot tell which is the eldest.
Hence it is impossible in that case to state which child is of which age.
When they are all the same age, the age of each is obvious.
== Sources ==
* : Solutions: $42$. -- Family Ages
* : Answers: $37$. Family Ages
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/43 - Mike's Age

== by : $43$ ==
;Mike's Age
:Pat O'Connor is $1 \tfrac 1 3$ times as old as when he built the pigsty.
:Little Mike, who was $40$ months old when Pat built the sty, is now two years more than half as old as Pat's wife, Biddy, was when Pat built the sty,
::so that when Little Mike is as old as Pat was when he built the sty,
::their three ages combined will amount to just one hundred years.
:How old is Little Mike?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $43$. -- Mike's Age
* : Arithmetical and Algebraical Problems: Age Puzzles: $38$. Mike's Age
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/43 - Mike's Age/Solution

== by : $43$ ==
;Mike's Age
:Pat O'Connor is $1 \tfrac 1 3$ times as old as when he built the pigsty.
:Little Mike, who was $40$ months old when Pat built the sty, is now two years more than half as old as Pat's wife, Biddy, was when Pat built the sty,
::so that when Little Mike is as old as Pat was when he built the sty,
::their three ages combined will amount to just one hundred years.
:How old is Little Mike?
== Solution ==
Mike is currently $10 \tfrac {16} {21}$ years old.
Pat was $22 \tfrac 2 7$ when he built the pigsty, which was $7 \tfrac 3 7$ years ago, and is now $29 \tfrac 5 7$ years old.
Biddy was $17 \tfrac {11} {21}$ when Pat built the pigsty, and is now $24 \tfrac {20} {21}$ years old
== Proof ==
Let $P$ years be Pat's age when he built the pigsty.
Let $M$ years be Mike's age now.
Let $B$ years be Biddy's age when Pat built the pigsty.
First we note that from:
:Pat O'Connor is $1 \tfrac 1 3$ times as old as when he built the pigsty
$P$ built the pigsty $\dfrac P 3$ years ago.
We have:
The remaining ages are calculated hence.
== Sources ==
* : Solutions: $43$. -- Mike's Age
* : Answers: $38$. Mike's Age
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/44 - Their Ages

== by : $44$ ==
;Their Ages
:A man, on being asked the ages of his two sons, stated that
::eighteen more than the sum of their ages is double the age of the elder,
::and six less than the difference of their ages is the age of the younger.
:What are their ages?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $44$. -- Their Ages
* : Arithmetical and Algebraical Problems: Age Puzzles: $39$. Their Ages
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/44 - Their Ages/Solution

== by : $44$ ==
;Their Ages
== Solution ==
The sons are aged $30$ and $12$.
== Proof ==
Let $a$ and $b$ be the ages of the older and younger son respectively.
We have:
== Sources ==
* : Solutions: $44$. -- Their Ages
* : Answers: $39$. Their Ages
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/45 - Brother and Sister

== by : $45$ ==
;Brother and Sister
:A boy on being asked the age of himself and his sister replied:
:"Three years ago I was seven times as old as my sister;
::two years ago I was four times as old;
::last year I was three times as old;
::and this year I am two and one-half times as old."
:What are their ages?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $45$. -- Brother and Sister
* : Arithmetical and Algebraical Problems: Age Puzzles: $40$. Brother and Sister
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/45 - Brother and Sister/Solution

== by : $45$ ==
;Brother and Sister
== Solution ==
The boy is $10$, and his sister is $4$.
== Proof ==
You need only two of the given statements to solve this.
Let $b$ and $s$ be the ages of the brother and sister respectively.
We have:
The final two statements are redundant.
== Sources ==
* : Solutions: $45$. -- Brother and Sister
* : Answers: $40$. Brother and Sister
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/46 - A Square Family

== by : $46$ ==
;A Square Family
:A man had nine children, all born at regular intervals,
::and the sum of the squares of their ages was equal to the square of his own.
:What were the ages of each?
:Every age was an exact number of years.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $46$. -- A Square Family
* : Arithmetical and Algebraical Problems: Age Puzzles: $41$. A Square Family
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/46 - A Square Family/Solution

== by : $46$ ==
;A Square Family
== Solution ==
:$2, 5, 8, 11, 14, 17, 20, 23, 26$.
== Proof ==
Let $a$ years be the age of the youngest child.
Let $d$ years be the interval between consecutive children.
Let $f$ be the age of the father.
We have:
We have that $f^2$ is a multiple of $3$ and so because it is square a multiple of $9$.
So $9 a^2 + 72 a d + 204 d^2$ is also a multiple of $9$ and so $204 d^2$ is a multiple of $9$.
That is, $d^2$ is a multiple of $9$ and so $d$ is a multiple of $3$.
Because of the nature of the question, it can be assumed that $d = 3$, as otherwise there is the unfeasible gap of $48$ years between eldest and youngest child.
So we have:
We now need to find two squares which differ by $60$.
By the Odd Number Theorem we have:
Thus if $14 = a + 12$ then $a = 2$ and we have:
:$48^2 = 2^2 + 5^2 + 8^2 + 11^2 + 14^2 + 17^2 + 20^2 + 23^2 + 26^2$
and the solution is complete.
== Sources ==
* : Solutions: $46$. -- A Square Family
* : Answers: $41$. A Square Family
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/47 - The Quarrelsome Children

== by : $47$ ==
;The Quarrelsome Children
:A man married a widow, and they each already had children.
:Ten years later there was a pitched battle engaging the present family of $12$ children.
:The mother ran to the father and cried,
::"Come at once! Your children and my children are fighting our children!"
:As the parents now had each nine children of their own, how many were born during the ten years?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $47$. -- The Quarrelsome Children
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $184$. The Quarrelsome Children
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/47 - The Quarrelsome Children/Solution

== by : $47$ ==
;The Quarrelsome Children
== Solution ==
:$6$
== Proof ==
Let $A$ and $B$ denote the sets of the children of the man and the widow respectively.
Then
:$A \cap B$ denotes the set of the children of both of them
:$A \cup B$ denotes the set of all the children together.
From the Inclusion-Exclusion Principle:
:$\card {A \cup B} = \card A + \card B - \card {A \cap B}$
where $\card X$ denotes the cardinality of a set $X$.
We have been given that:
:$\card {A \cup B} = 12$
:$\card A = \card B = 9$
which leads to:
:$\card {A \cap B} = 9 + 9 - 12 = 6$
== Sources ==
* : Solutions: $47$. -- The Quarrelsome Children
* : Answers: $184$. The Quarrelsome Children
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/48 - Robinson's Age

== by : $48$ ==
;Robinson's Age
:Robinson said.
::My brother is two years older than I,
::my sister is four years older than he,
::my mother was $20$ when I was born,
::and I was told yesterday that the average age of the four of is is $39$ years.
:What was Robinson's age?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $48$. -- Robinson's Age
* : Arithmetical and Algebraical Problems: Age Puzzles: $45$. Robinson's Age
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/48 - Robinson's Age/Solution

== by : $48$ ==
;Robinson's Age
:Robinson said.
::My brother is two years older than I,
::my sister is four years older than he,
::my mother was $20$ when I was born,
::and I was told yesterday that the average age of the four of is is $39$ years.
:What was Robinson's age?
== Solution ==
:$32$
== Proof ==
The information can be represented as a system of linear simultaneous equations in matrix form as:
:$\begin {pmatrix}
-1 & 1 & 0 & 0 \\
 0 & -1 & -1 & 0 \\
-1 & 0 & 0 & 1 \\
 1 & 1 & 1 & 1 \\
\end {pmatrix} \begin {pmatrix} R \\ B \\ S \\ M \end {pmatrix} = \begin {pmatrix} 2 \\ 4 \\ 20 \\ 4 \times 39 \end {pmatrix}$
where $R$, $B$, $S$ and $M$ are apparent.
It remains to solve this matrix equation.
Conversion to echelon form proceeds as follows:
Hence Robinson's mother is $52$, making Robinson $32$.
His brother is therefore $34$ and his sister $38$.
== Sources ==
* : Solutions: $48$. -- Robinson's Age
* : Answers: $45$. Robinson's Age
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/49 - The Engine-Driver's Name

== by : $49$ ==
;The Engine-Driver's Name
:Three business men -- Smith, Robinson and Jones -- all live in the Leeds-Sheffield district.
:Three railwaymen of similar names live in the same district.
:The business man Robinson and the guard live at Sheffield,
::the business man Jones and the stoker live at Leeds,
::while the business man Smith and the engine-driver live half-way between Leeds and Sheffield.
:The guard's namesake earns $\pounds 1000, 10 \shillings 2 \oldpence$ per annum,
:and the engine-driver earns exactly one-third of the business man living nearest to him.
:Finally, the railwayman Smith beats the stoker at billiards.
:What is the engine driver's name?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Age and Kinship Puzzles: $49$. -- The Engine-Driver's Name
* : Unclassified Puzzles: $521$. The Engineer's Name
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/49 - The Engine-Driver's Name/Historical Note

== Historical Note on by : $49$ - The Engine-Driver's Name ==
This puzzle became one of 's most popular.
Such puzzles became known as Smith-Jones-Robinson puzzles.
This puzzle is mentioned in by , in which himself is mentioned in a footnote.
== Sources ==
* : Answers: $521$. The Engineer's Name
Category:Historical Notes
Henry Ernest Dudeney/Puzzles and Curious Problems/49 - The Engine-Driver's Name/Solution

== by : $49$ ==
;The Engine-Driver's Name
== Solution ==
Smith.
== Proof ==
From the last statement, the stoker is not Smith, who therefore must be Jones or Robinson.
The guard's namesake earns $\pounds 1000, 10 \shillings 2 \oldpence = 240 \, 122 \oldpence$ per annum, which is not divisible by $3$.
Hence the businessman (who is Smith) living near the engine-driver cannot be the guard's namesake.
So the guard is either Jones or Robinson.
So the stoker is not Smith, and the guard is not Smith.
Therefore the engine-driver is Smith.
== Historical Note ==
== Sources ==
* : Solutions: $49$. -- The Engine-Driver's Name
* : Answers: $521$. The Engineer's Name
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions