nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Modern Puzzles/80 - The Flocks of Sheep/Solution

== by : $80$ ==
;The Flocks of Sheep
== Solution ==
Adam had $60$ sheep.
Ben had $50$ sheep.
Claude had $40$ sheep.
Dan had $30$ sheep.
== Proof ==
Let $A$, $B$, $C$ and $D$ be the number of sheep owned by Adam, Ben, Claude and Dan respectively.
Let $B_1$ and $C_1$ be the number of sheep owned by Ben and Claude respectively after Claude gives Ben a quarter of his.
Let $B_2$ and $A_2$ be the number of sheep owned by Ben and Adam respectively after Adam gives Ben a third of his.
Let $B_3$ and $C_3$ be the number of sheep owned by Ben and Claude respectively after Ben gives Claude a quarter of his.
Let $C_4$ and $D_4$ be the number of sheep owned by Claude and Dan respectively after Claude gives Dan a fifth of his.
Let $A_5$, $B_5$, $C_5$ and $D_5$ be the number of sheep owned by Adam, Ben, Claude and Dan respectively at the end of the series of exchanges.
We have:
and so on.
== Sources ==
* : Solutions: $80$. -- The Flocks of Sheep
* : Answers: $162$. The Flocks of Sheep
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/81 - Pussy and the Mouse

== by : $81$ ==
;Pussy and the Mouse
:"There's a mouse in one of these barrels," said the dog.
:"Which barrel?" asked the cat.
:"Why, the five hundredth barrel."
:"What do you mean, the five hundredth? There are only five barrels in all."
:"It's the five hundredth if you count backwards and forwards this way."
:And the dog explained that if you count like this:
 1 2 3 4 5
 9 8 7 6
 10 11 12 13
:so that the seventh barrel would be the one marked $3$ and the twelfth barrel the one numbered $4$.
The story goes on laboriously to its inevitable conclusion that the mouse escapes before the cat has finished counting, until:
:Now, which was the five hundredth barrel?
:Can you find a quick way of arriving at the answer without making the actual count?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $81$. -- Pussy and the Mouse
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $164$. Pussy and the Mouse
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/81 - Pussy and the Mouse/Solution

== by : $81$ ==
;Pussy and the Mouse
== Solution ==
Barrel no. $4$.
== Proof ==
Let $n$ be the count.
Let $m$ be the number of the barrel the mouse is in.
By the method of counting, all we are concerned about here is the value of $n \pmod 8$.
If $n \pmod 8 = 0$, then $m = 2$.
If $1 \le n \pmod 8 \le 5$, then $m = n \pmod 8$.
Otherwise $m = 10 - n \pmod 8$.
Here $n = 500$, and so $n \pmod 8 = 4$.
== Sources ==
* : Solutions: $81$. -- Pussy and the Mouse
* : Answers: $164$. Pussy and the Mouse
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/82 - Army Figures

== by : $82$ ==
;Army Figures
:A certain division in an army was composed of a little over twenty thousand men, made up of five brigades.
:It was know that one third of the first brigade,
::two-sevenths of the second brigade,
::seven-twelfths of the third,
::nine-thirteenths of the fourth,
::and fifteen-twenty-seconds of the fifth brigades happened in every case to be the same number of men.
:Can you discover how many men there were in every brigade?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $82$. -- Army Figures
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $165$. Army Figures
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/82 - Army Figures/Solution

== by : $82$ ==
;Army Figures
== Solution ==
The $5$ brigades contained respectively $5670$, $6615$, $3240$, $2730$ and $2772$ men.
== Proof ==
Let $a$, $b$, $c$, $d$ and $e$ be the number of men in each of the first, second, third, fourth and fifth brigades respectively.
We are given that:
:$\dfrac a 3 = \dfrac {2 b} 7 = \dfrac {7 c} {12} = \dfrac {9 d} {13} = \dfrac {15 e} {22}$
Our first task is to place all these over a common denominator.
To do this we calculate:
:$\lcm \set {3, 7, 12, 13, 22} = 3 \times 7 \times 4 \times 13 \times 11 = 12012$
which gives us:
:$\dfrac {4004 a} {12012} = \dfrac {3432 b} {12012} = \dfrac {7007 c} {12012} = \dfrac {8316 d} {12012} = \dfrac {8190 e} {12012}$
That is:
:$(1) \quad 4004 a = 3432 b = 7007 c = 8316 d = 8190 e$
Now we calculate:
Dividing by each number in turn, we have:
That is:
Thus we have found a set of numbers which fulfil the conditions for $(1)$:
:$\set {1890, 2205, 1080, 910, 924}$
and hence so does:
:$\set {1890 k, 2205 k, 1080 k, 910 k, 924 k}$
for $k \in \Z_{\ge 1}$.
We have:
:$1890 + 2205 + 1080 + 910 + 924 = 7009$
Setting $k = 3$ we get:
:$5670 + 6615 + 3240 + 2730 + 2772 = 21027$
which fulfils the condition:
:A certain division in an army was composed of a little over twenty thousand men ...
Hence we can state:
== Sources ==
* : Solutions: $82$. -- Army Figures
* : Answers: $165$. Army Figures
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/83 - A Critical Vote

== by : $83$ ==
;A Critical Vote
:A meeting of a charitable society was held to decide whether the members should expand their operations.
:It was arranged that during the count those in favour of the motion should remain standing,
::and those who voted against should sit down.
:"Ladies and gentlemen," said the chairman in due course, "I have the pleasure to announce that the motion is carried by a majority exactly equal to exactly a quarter of the opposition."
:"Excuse me, sir," called somebody from the back, "but some of us over here could not sit down, because there are not enough chairs."
:"Then those who wanted to sit down but couldn't are to hold up their hands ... I find there are a dozen of you, so the motion is lost by a majority of one."
:Now, how many people voted at that meeting?
Note that the language and concepts have been updated to accommodate modern sensibilities.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $83$. -- A Critical Vote
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $166$. A Critical Vote
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/83 - A Critical Vote/Solution

== by : $83$ ==
;A Critical Vote
== Solution ==
There were $207$ people at the meeting, of which $103$ voted in favour and $104$ against.
== Proof ==
Let $n$ be the total number of people at the meeting.
Let $a$ be the number of people in favour of the motion.
Let $b$ be the number of people against the motion.
Let $m$ be the number of people who wanted to vote against the motion but could not, as they could not sit down.
We have:
== Sources ==
* : Solutions: $83$. -- A Critical Vote
* : Answers: $166$. A Critical Vote
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/84 - The Three Brothers

== by : $84$ ==
;The Three Brothers
:The discussion arose before one of the tribunals as to which of a tradesman's three sons could best be spared for service in the Army.
:"All I know as to their capabilities," said the father, "is this:
::Arthur and Benjamin can do a certain quantity of work in eight days,
::which Arthur and Charles will do in nine days,
::and which Benjamin and Charles will take ten days over."
:Of course, it was at once seen that as longer time was taken over the job whenever Charles was one of the pair,
::he must be the slowest worker.
:This was all they wanted to know, but it is an interesting puzzle to ascertain just how long each son would be required to do that job alone.
:Can you discover?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $84$. -- The Three Brothers
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $167$. The Three Brothers
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Cistern Problems
Henry Ernest Dudeney/Modern Puzzles/84 - The Three Brothers/Solution

== by : $84$ ==
;The Three Brothers
:The discussion arose before one of the tribunals as to which of a tradesman's three sons could best be spared for service in the Army.
:"All I know as to their capabilities," said the father, "is this:
::Arthur and Benjamin can do a certain quantity of work in eight days,
::which Arthur and Charles will do in nine days,
::and which Benjamin and Charles will take ten days over."
:Of course, it was at once seen that as longer time was taken over the job whenever Charles was one of the pair,
::he must be the slowest worker.
:This was all they wanted to know, but it is an interesting puzzle to ascertain just how long each son would be required to do that job alone.
:Can you discover?
== Solution ==
Arthur, Benjamin and Charles would take respectively $14 \tfrac {34} {49}$ days, $17 \tfrac {23} {41}$ days, and $23 \tfrac 7 {31}$ days.
== Proof ==
Let $a, b, c$ be the rate of working in jobs per day of (respectively) Arthur, Benjamin and Charles.
Let $t_a, t_b, t_c$ be the number of days it would take (respectively) Arthur, Benjamin and Charles to do the job alone.
We have:
and so:
Hence and so.
== Also see ==
* Henry Ernest Dudeney: Puzzles and Curious Problems: $138$ - The Three Workmen, exactly the same problem but with different names.
== Sources ==
* : Solutions: $84$. -- The Three Brothers
* : Answers: $167$. The Three Brothers
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/85 - The House Number

== by : $85$ ==
;The House Number
:A man said the house of his friend was in a long street,
::numbered on his side one, two, three, and so on,
::and that all the numbers on one side of him added up exactly the same as all the numbers on the other side of him.
:He said he knew there were more than fifty houses on that side of the street,
::but not as many as five hundred.
:Can you discover the number of that house?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $85$. -- The House Number
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $168$. The House Number
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/85 - The House Number/Solution

== by : $85$ ==
;The House Number
== Solution ==
The man lived at no. $204$ in a street of $288$ houses.
== Proof ==
Let there be $m$ houses in the street, where we are told $50 < m < 500$.
Let the man live at no. $n$.
We have that:
Trying out a few values of $m$, we see:
:$\begin{array} {r|r} m & n \\ \hline
 1 & 1 \\
 8 & 6 \\
49 & 35 \\
288 & 204 \\
1681 & 1189 \\
\end{array}$
Formally, we have:
From Pell's Equation: $x^2 - 2 y^2 = 1$, we have:
Hence we can create the solutions:
:$\begin{array} {r|r|r|r} x & y & m = \dfrac {x - 1} 2 & n = \dfrac y 2 \\ \hline
 3 & 2 & 1 & 1 \\
17 & 12 & 8 & 6 \\
99 & 70 & 49 & 35\\
577 & 408 & 288 & 204 \\
3363 & 2378 & 1681 & 1189 & \\
\end{array}$
Because there are between $50$ and $500$ houses in the street, we know the man lives at no. $204$ in a $288$-house street.
== Sources ==
* : Solutions: $85$. -- The House Number
* : Answers: $168$. The House Number
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/86 - A New Street Puzzle

== by : $86$ ==
;A New Street Puzzle
:Brown lived in a street which contained more than twenty houses, but fewer than five hundred,
::all numbered one, two, three, four, etc., throughout.
:Brown discovered that all the numbers from one upwards to his own number inclusive summed to exactly half the sum of all the numbers from one up to, and including, the last house.
:Now what was the number of his house?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $86$. -- A New Street Puzzle
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $169$. A New Street Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/86 - A New Street Puzzle/Solution

== by : $86$ ==
;A New Street Puzzle
== Solution ==
Brown lived at No. $84$ on a street of $119$ houses.
== Proof ==
Let there be $m$ houses in the street, where we are told $20 < m < 500$.
Let the man live at no. $n$.
We have that:
Hence this problem is equivalent to finding the answers to Pell's equation:
:$2 x^2 - 1 = y^2$
or equivalently:
:$y^2 - 2 x^2 = -1$
From Solution to Pell's Equation, the solutions to this are:
:${p_r}^2 - 2 {q_r}^2 = \paren {-1}^r$ for $r = 1, 2, 3, \ldots$
where $\dfrac {p_r} {q_r}$ are the convergent of the Continued Fraction Expansion of $\sqrt 2$.
It is the odd integer values of $r$ that we need in order to make the equal to $-1$.
From Continued Fraction Expansion of Root 2:
Hence we have $x$ and $y$, and thence $n = \dfrac {x - 1} 2$ and $m = \dfrac {y - 1} 2$ as:
:$\begin{array} {r|r} x & y & n & m \\ \hline
 1 & 1 & 0 & 0 \\
 5 & 7 & 2 & 3 \\
29 & 41 & 14 & 20 \\
169 & 239 & 84 & 119 \\
985 & 1393 & 492 & 696 \\
\end{array}$
Because there are between $20$ and $500$ houses in the street, we know the man lives at no. $84$ in a $119$-house street.
== Historical Note ==
== Sources ==
* : Solutions: $86$. -- A New Street Puzzle
* : Answers: $169$. A New Street Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/87 - Another Street Puzzle

== by : $87$ ==
;Another Street Puzzle
:A long street in Brussels has all the odd numbers of the houses on one side and all the even numbers on the other.
:$(1)$ If a man lives in an odd-numbered house and all the numbers on one side of him, added together, equal the numbers on the other side,
:::how many houses are there, and what is the number of his house?
:$(2)$ If a man lives on the even side and all the numbers on one side of him equal those on the other side,
:::how many houses are there, and what is his number?
:We will assume that there are more than fifty houses on each side of the street and fewer than five hundred.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $87$. -- Another Street Puzzle
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $170$. Another Street Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/87 - Another Street Puzzle/Solution

== by : $87$ ==
;Another Street Puzzle
== Solution ==
:$(1): \quad$ The man lives at no. $239$ in a street with $169$ odd-numbered houses.
:$(2): \quad$ The man lives at no. $408$ in a street with $208$ even-numbered houses.
== Proof ==
;Odd-Numbered Side
Let there be $m$ houses in the street, where we are told $50 < m < 500$.
Let the man live at no. $n = 2 k + 1$.
We have that:
Hence this problem is equivalent to finding the answers to Pell's equation:
:$n^2 - 2 m^2 = -1$
From Solution to Pell's Equation, the solutions to this are:
:${p_r}^2 - 2 {q_r}^2 = \paren {-1}^r$ for $r = 1, 2, 3, \ldots$
where $\dfrac {p_r} {q_r}$ are the convergent of the Continued Fraction Expansion of $\sqrt 2$.
It is the odd integer values of $r$ that we need in order to make the equal to $-1$.
From Continued Fraction Expansion of Root 2:
Hence we have $n$ and $m$ as:
:$\begin{array} {r|r} m & n \\ \hline
 1 & 1 \\
 5 & 7 \\
29 & 41 \\
169 & 239 \\
985 & 1393 \\
\end{array}$
Because there are between $20$ and $500$ houses in the street, we know the man lives at no. $239$ in a street with $169$ houses on the odd-number side.
;Even-Numbered Side
Let there be $m$ houses in the street, where we are told $50 < m < 500$.
Let the man live at no. $n$.
Let us suppose all the house numbers were divided by $2$.
Then the man is living on a street where the houses go $1$, $2$, $3$, $\ldots$, $m$.
This is then the same problem as $85$ - The House Number in the same collection.
In this case, the man lives at no. $204$ in a street of $288$ houses.
So, when we multiply the house numbers by $2$ so they are back up to what they were, we find that the man lives at no. $408$.
== Sources ==
* : Solutions: $87$. -- Another Street Puzzle
* : Answers: $170$. Another Street Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/88 - Correcting an Error

== by : $88$ ==
;Correcting an Error
:Hermione was given a certain number to multiply by $409$,
::but she made a blunder that is very common with mudbloods when learning the elements of simple arithmetic:
::she placed the first figure of her product by $4$ below the second figure from the right instead of below the third.
:We have all done that as youngsters (speak for yourself, Harry, old boy) when there has happened to be a $0$ in the multiplier.
:The result of Hermione's mistake was that her answer was wrong by $328,320$, entirely in consequence of that little slip.
:Now, what was the multiplicand?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $88$. -- Correcting an Error
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $171$. Correcting an Error
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/88 - Correcting an Error/Solution

== by : $88$ ==
;Correcting an Error
== Solution ==
:$912$
== Proof ==
Let $x$ be the multiplicand.
Misplacing the partial product as described causes the multiplier to be $49$ and not $409$.
Hence we have:
:$409 x - 49 x = 328320$
which after some simple algebra gives:
:$x = 912$
Easy peasy lemon squeezy.
== Sources ==
* : Solutions: $88$. -- Correcting an Error
* : Answers: $171$. Correcting an Error
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/89 - The Seventeen Horses

== by : $89$ ==
;The Seventeen Horses
:"I suppose you all know this old puzzle," said Jeffries.
:"A farmer left seventeen horses to be divided among his three sons in the following proportions:
::To the eldest, one-half;
::to the second, one-third;
::and to the youngest, one-ninth.
:How should they be divided?
:"Yes; I think we all know that," said Robinson, "but it can't be done.
::The answer always given is a fallacy."
:(Considerable pointless argument ensues.)
:... The terms of the will can be exactly carried out, without any mutilation of a horse.
:... How should the horses be divided in strict accordance with the directions?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $89$. -- The Seventeen Horses
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $172$. The Seventeen Horses
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Seventeen Horses
Henry Ernest Dudeney/Modern Puzzles/89 - The Seventeen Horses/Solution

== by : $89$ ==
;The Seventeen Horses
== Solution ==
In Seventeen Horses, this is analysed in some detail.
The telling point is:
:The fact that the ratios do not add up to $1$ tells us straight away that those ratios are not in fact the actual fractions of the inheritance that each will receive.
:It is noted that the question carefully does not state that one son receives $\dfrac 1 2$, the next son $\dfrac 1 3$ and the third $\dfrac 1 9$, because that is not what actually happens here.
This, effectively, is how himself answers the question.
== Sources ==
* : Solutions: $89$. -- The Seventeen Horses
* : Answers: $172$. The Seventeen Horses
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/8 - Buying Buns

== by : $8$ ==
;Buying Buns
:Buns were being sold at three prices:
::one a penny,
::two a penny,
::and three a penny.
:Some children (there were as many boys as girls) were given sevenpence to spend on these buns, each receiving exactly alike.
:How many buns did each receive?
:Of course no buns were divided.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $8$. -- Buying Buns
* : Arithmetical and Algebraical Problems: Money Puzzles: $5$. Buying Buns
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/8 - Buying Buns/Mistake

== Source Work ==
:
:Arithmetical and Algebraical Problems
::Money Puzzles
:::$8$ -- Buying Buns
:
:Arithmetical and Algebraical Problems
::Money Puzzles
:::$5$ -- Buying Buns
== Mistake ==
::Buns were being sold at three prices:
::one a penny,
::two a penny,
::and three a penny.
:Some children (there were as many boys as girls) were given sevenpence to spend on these buns, each receiving exactly alike.
:How many buns did each receive?
:Of course no buns were divided.
The solution offered:
:There must have been three boys and three girls,
::each of whom received two buns at three a penny
::and one bun at two a penny,
::the cost of which would be exactly sevenpence.
== Correction ==
There are in fact $12$ solutions.
The full analysis can be found here.
== Sources ==
* : Solutions: $8$. -- Buying Buns
* : Answers: $5$. Buying Buns
Category:Mistakes/Modern Puzzles
Category:Mistakes/536 Puzzles & Curious Problems
Henry Ernest Dudeney/Modern Puzzles/8 - Buying Buns/Solution

== by : $8$ ==
;Buying Buns
== Solution ==
The solution given by is:
:There must have been three boys and three girls,
::each of whom received two buns at three a penny
::and one bun at two a penny,
::the cost of which would be exactly sevenpence.
However, there are in fact $12$ solutions.
Let them be presented in the form $\tuple {a, b, c}$ where $a$, $b$ and $c$ are the numbers of penny buns, two-for-a-penny buns and three-for-a-penny buns received by each child:
:$14$ children in total:
::$\tuple {0, 1, 0}$
:$6$ children in total (the solution given):
::$\tuple {0, 1, 2}$
:$2$ children in total:
::$\tuple {0, 1, 9}$
::$\tuple {0, 3, 6}$
::$\tuple {0, 5, 3}$
::$\tuple {0, 7, 0}$
::$\tuple {1, 1, 6}$
::$\tuple {1, 3, 3}$
::$\tuple {1, 5, 0}$
::$\tuple {2, 1, 3}$
::$\tuple {2, 3, 0}$
::$\tuple {3, 1, 0}$
== Proof ==
Let $2 T$ be the total number of children.
Let $a$ be the number of penny buns bought by each child.
Let $b$ be the number of two-for-a-penny buns bought by each child.
Let $c$ be the number of three-for-a-penny buns bought by each child.
All of $a$, $b$, $c$ and $T$ are natural numbers where $T > 0$.
Hence we have:
Thus $T$ is a divisor of $21$, and so can be either $1$, $3$, $7$ or $21$.
$21$ is clearly off the table, because that would give:
:$21 \paren {6 a + 3 b + 2 c} = 21$
and so:
:$6 a + 3 b + 2 c = 1$
which cannot be satisfied in natural numbers.
Let $T = 7$.
Then we have:
:$6 a + 3 b + 2 c = 3$
We note that $b$ must be odd in order to make $3$.
So whatever happens, $b \ge 1$.
When $b = 1$ we have:
:$6 a + 3 \times 1 + 2 c = 3$
allowing the solution:
:$\tuple {a, b, c} = \tuple {0, 1, 0}$
That is all you get for $T = 7$.
So, there can be $14$ children each receiving a single two-for-a-penny bun.
That is, $14$ buns are bought, and each child receives one.
Let $T = 3$.
Then we have:
:$6 a + 3 b + 2 c = 7$
Setting $b = 1$ again gives:
:$6 a + 3 \times 1 + 2 c = 7$
This leads to:
:$6 a + 2 c = 4$
for which the only solution is $c = 2$.
This allows the solution:
:$\tuple {a, b, c} = \tuple {0, 1, 2}$
So there can be $6$ children each receiving $1$ two-for-a-penny bun, and $2$ three-for-a-penny buns.
Finally let $T = 1$.
This leads to:
:$6 a + 3 b + 2 c = 21$
which gives rise to the solutions for $\tuple {a, b, c}$ as:
:$\tuple {0, 1, 9}$
:$\tuple {0, 3, 6}$
:$\tuple {0, 5, 3}$
:$\tuple {0, 7, 0}$
:$\tuple {1, 1, 6}$
:$\tuple {1, 3, 3}$
:$\tuple {1, 5, 0}$
:$\tuple {2, 1, 3}$
:$\tuple {2, 3, 0}$
:$\tuple {3, 1, 0}$
So $2$ children have a total of $10$ different ways they can pool their $7 \oldpence$ to buy their buns and share them equitably.
== Sources ==
* : Solutions: $8$. -- Buying Buns
* : Answers: $5$. Buying Buns
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions