nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Puzzles and Curious Problems/284 - Lamp Signalling/Solution

== by : $284$ ==
;Lamp Signalling
:Two spies on the opposite sides of a river devised a method for signalling by night.
:They each put up a stand, like the diagram, and each had three lamps which could show either white, red or green.
:
:They constructed a code in which every different signal meant a sentence.
:Note that a single lamp on any one of the hooks could only mean the same thing,
::that two lamps hung on the upper hooks $1$ and $2$ could not be distinguished from two on, for example, $4$ and $5$.
:However, two red lamps on $1$ and $5$ could be distinguished from two on $1$ and $6$,
::and two on $1$ and $2$ from two on $1$ and $3$.
:Remembering the variations of colour as well as of position, what is the greatest number of signals that could be sent?
== Solution ==
:$471$.
== Proof ==
With $3$ lamps all of one colour, we can make $15$ distinct messages for each colour.
Because there are $3$ colours, that makes $3 \times 15 = 45$ messages.
With $2$ of one colour and $1$ of another, we can make the same number: $15$ distinct messages.
Each of these admits $3$ variations of colour order, making $45$.
There are $6$ ways of choosing $1$ lamp of one colour and $2$ of another colour.
Hence there are $6 \times 45 = 270$ messages made with $3$ lamps of $2$ colours.
With $1$ lamp of each colour we can make $6 \times 15 = 90$ messages.
With $2$ lamps of one colour we can make $7$ messages.
Because there are $3$ colours, that makes $3 \times 7 = 21$ messages.
With $1$ lamp of one colour and $1$ lamp of another colour, we can get $14$ messages.
There are $6$ ways of choosing $1$ lamp of one colour and $1$ lamp of another colour.
Hence there are $6 \times 14 = 42$ messages made with $2$ lamps of $2$ colours.
With $1$ lamp only, you can make only $1$ message.
As there are $3$ colours, that makes $3$ more messages.
Hence the total:
:$45 + 270 + 90 + 21 + 42 + 3 = 471$
== Sources ==
* : Solutions: $284$. -- Lamp Signalling
* : Answers: $463$. Lamp Signalling
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/285 - The Teashop Check

== by : $285$ ==
;The Teashop Check
:We give an example of the check supposed to be used at certain popular teashops.
:The waitress punches holes in the tickets to indicate the amount of the purchase.
::$\boxed {\begin{array} {rcl} \\
\tfrac 1 2 \oldpence & --- & \bullet \\
1 \oldpence & --- \\
1 \tfrac 1 2 \oldpence & --- \\
2 \oldpence & --- \\
2 \tfrac 1 2 \oldpence & --- \\
3 \oldpence & --- & \bullet \\
4 \oldpence & --- \\
6 \oldpence & --- \\
7 \oldpence & --- \\
8 \oldpence & --- \\
1 \shillings & --- \\
 & & \\
\end{array} }$
:Thus, in the example, the two holes indicate that the customer has to pay $3 \tfrac 1 2 \oldpence$
:But the girl might, if she had chosen, have punched in any one of three other ways --
::$2 \tfrac 1 2 \oldpence$ and $1 \oldpence$, or $2 \oldpence$ and $1 \tfrac 1 2 \oldpence$, or $2 \oldpence$, $1 \oldpence$ and $\tfrac 1 2 \oldpence$
:On one occasion a waitress said, "I can punch this ticket in any one of $10$ different ways, and no more."
:Her coworker, whose customer owed a different amount, said, "Same here."
:What were the amounts of the purchases of each of their customers?
:Only one hole is allowed to be punched against any given amount.
=== Click here for solution ===
== Sources ==
* : Combination and Group Problems: $285$. -- The Teashop Check
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/285 - The Teashop Check/Solution

== by : $285$ ==
;The Teashop Check
== Solution ==
One of the customers' bills was $7 \oldpence$, and the other's was $3 \shillings 4 \tfrac 1 2 \oldpence$
== Proof ==
First note that the total of all the prices is $3 \shillings 11 \tfrac 1 2 \oldpence$
This is what you would get if all the holes were punched.
$7 \oldpence$ can be punched as follows:
For every one of these punch patterns, there is a directly complementary one which selects all the opposite cash values.
These all add up to $3 \shillings 11 \tfrac 1 2 \oldpence - 7 \oldpence$, or $3 \shillings 4 \tfrac 1 2 \oldpence$
Hence the result.
== Sources ==
* : Solutions: $285$. -- The Teashop Check
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/286 - Unlucky Breakdowns

== by : $286$ ==
;Unlucky Breakdowns
:On a day of great festivities, a large crowd gathered for a day's outing and pleasure.
:They all agreed to pile into a bunch of wagons, each of which was to carry the same number of people.
:But ten of the wagons broke down half way, so each of the other wagons then had to carry one more person than had been planned.
:As they were about to start back, it was discovered that $15$ more of these wagons had become unserviceable,
::and so there were three more people in each working wagon on the way back than started out.
:How many people were there in the party?
=== Click here for solution ===
== Sources ==
* : Combination and Group Problems: $286$. -- Unlucky Breakdowns
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $247$. Unlucky Breakdowns
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/286 - Unlucky Breakdowns/Solution

== by : $286$ ==
;Unlucky Breakdowns
== Solution ==
There were $900$ people, all travelling in $100$ wagons, $9$ to a wagon.
Then they were down to $90$ wagons, in which they travelled $10$ to a wagon.
Then they were down to $75$ wagons, in which they travelled $12$ to a wagon.
== Proof ==
Let $P$ be the number of people going on the outing.
Let $W$ be the number of wagons.
Let $n$ be the number of people per wagon who started out.
We have:
== Sources ==
* : Solutions: $286$. -- Unlucky Breakdowns
* : Answers: $247$. Unlucky Breakdowns
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/287 - The Handcuffed Prisoners

== by : $287$ ==
;The Handcuffed Prisoners
:Nine dangerous convicts needed to be guarded.
:Every day except Sunday they were taken out for exercise, handcuffed together in groups of three, as in the diagram:
:
:On no day in any one week were the same two men to be handcuffed together.
:If will be seen how they were sent out on Monday.
:Can you arrange the nine men in triplets for the remaining $5$ days?
:It will be seen that No. $1$ cannot be handcuffed to No. $2$ again, but $1$ and $3$ can subsequently be so.
=== Click here for solution ===
== Also see ==
* Fifteen Schoolgirls Puzzle, of which this is a relative
== Sources ==
* : Combination and Group Problems: $287$. -- The Handcuffed Prisoners
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $464$. The Handcuffed Prisoners
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/287 - The Handcuffed Prisoners/Solution

== by : $287$ ==
;The Handcuffed Prisoners
:Nine dangerous convicts needed to be guarded.
:Every day except Sunday they were taken out for exercise, handcuffed together in groups of three, as in the diagram:
:
:On no day in any one week were the same two men to be handcuffed together.
:If will be seen how they were sent out on Monday.
:Can you arrange the nine men in triplets for the remaining $5$ days?
:It will be seen that No. $1$ cannot be handcuffed to No. $2$ again, but $1$ and $3$ can subsequently be so.
== Solution ==
The following is a solution.
Each block of three is an arrangement of the prisoners such that each prisoner will have been handcuffed to every other prisoner once and only once:
$\quad \begin{array} \\ 1-2-3 \\ 4-5-6 \\ 7-8-9 \end{array}
\qquad
\begin{array} \\ 2-6-8 \\ 5-9-1 \\ 3-7-4 \end{array}
\qquad
\begin{array} \\ 6-1-7 \\ 9-4-2 \\ 8-3-5 \end{array}
\qquad
\begin{array} \\ 1-4-8 \\ 2-5-7 \\ 6-9-3 \end{array}
\qquad
\begin{array} \\ 7-2-9 \\ 3-6-4 \\ 8-1-5 \end{array}
\qquad
\begin{array} \\ 4-3-1 \\ 5-8-2 \\ 9-7-6 \end{array}$
== Also see ==
* Fifteen Schoolgirls Puzzle, of which this is a relative
== Sources ==
* : Solutions: $287$. -- The Handcuffed Prisoners
* : Answers: $464$. The Handcuffed Prisoners
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/288 - Seating the Party

== by : $288$ ==
;Seating the Party
:On a family outing, Dora asked in how many ways they could all be seated.
:There were $6$ of them, three of each gender, and $6$ seats:
::one beside the driver, two with their backs to the driver, and two behind them, facing the driver,
:No two of the same gender were allowed to sit side by side.
:The only people who were able to drive were the men.
:So, how many ways could they all be seated?
=== Click here for solution ===
== Sources ==
* : Combination and Group Problems: $288$. -- Seating the Party
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $465$. Seating the Party
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/288 - Seating the Party/Solution

== by : $288$ ==
;Seating the Party
== Solution ==
The number of ways of seating the party is $144$.
== Proof ==
There are three choices of drivers and three choices of women to sit beside him.
Anyone else can sit with their back directly to the driver; there are four choices remaining.
The one beside them must be of a different gender; there are two choices remaining.
The last two people have two ways to fill in the remaining seats.
Therefore the number of ways to seat the party is:
:$3 \times 3 \times 4 \times 2 \times 2 = 144$
== Sources ==
* : Solutions: $288$. -- Seating the Party
* : Answers: $465$. Seating the Party
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/289 - Magic Square Trick

== by : $289$ ==
;Magic Square Trick
:Here is an advertising trick that appeared in America many years ago.
:Place in the empty squares such figures (different in every case, and no two squares containing the same figure)
::so that they shall add up to $15$ in as many straight directions as possible.
$\qquad \begin{array} {|c|c|c|}
\hline \ \ & \ \ & \ \ \\
\hline \ \ & 5 & \ \ \\
\hline \ \ & \ \ & \ \ \\
\hline \end{array}$
:A large prize was offered, but no correct solutions received.
To explain further:
:Each cell is expected to contain a different number of instances of the same digit, combined to form an expression;
:Each cell contains a different digit, of whatever multiplicity;
:The results of evaluation of these expressions should add to $15$ in as many directions as possible;
:The individual digits themselves, were they to occur singly (as the $5$ does in the example), themselves form a magic square of order $3$.
=== Click here for solution ===
== Sources ==
* : Magic Square, Measuring, Weighing, and Packing Problems: $289$. -- Magic Square Trick
* : Combinatorial & Topological Problems: Magic Square Puzzles: $389$. Magic Square Trick
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/289 - Magic Square Trick/Historical Note

== Historical Note on by : $289$ - Magic Square Trick ==
 reports on simpler constructions for some of the results, suggested by :
but none of these come close to fixing the flaw in the basic premise of the solution of this puzzle.
== Sources ==
* : Answers: $389$. Magic Square Trick
Category:Historical Notes
Henry Ernest Dudeney/Puzzles and Curious Problems/289 - Magic Square Trick/Solution

== by : $289$ ==
;Magic Square Trick
== Solution ==
The solution given by appears to be flawed:
$\qquad \begin{array} {|c|c|c|}
\hline \dfrac {\paren {8 \times 8} - 8} 8 + \dfrac 8 {8 + 8} & 1 + 1 + 1 + 1 + 1 & \dfrac {6 + 6} 6 \times \dfrac 6 {6 + 6} \\
\hline 3 - 3 & 5 & \dfrac {\paren {7 \times 7} + 7 + 7 + 7} 7 \\
\hline \dfrac {\paren {4 \times 4} + 4 + 4 + 4} 4 + \dfrac 4 {4 + 4} & \dfrac {9 + 9 + 9 + 9 + 9} 9 \ & 2 + \dfrac 2 {2 + 2} \\
\hline \end{array}$
These evaluate to:
$\qquad \begin{array} {|c|c|c|}
\hline 7 \tfrac 1 2 & 5 & 2 \tfrac 1 2 \\
\hline 0 & 5 & 10 \\
\hline 7 \tfrac 1 2 & 5 & 2 \tfrac 1 2 \\
\hline \end{array}$
which do indeed add to $15$ along all rows, columns and diagonals.
Nothing, of course, was said about these numbers needing all to be different.
Of course, the individual digits themselves form the traditional magic square of order $3$:
$\qquad \begin{array} {|c|c|c|}
\hline 8 & 1 & 6 \\
\hline 3 & 5 & 7 \\
\hline 4 & 9 & 2 \\
\hline \end{array}$
However, it is seen that the multiplicities of the digits of these entries themselves are:
$\qquad \begin{array} {|c|c|c|}
\hline 7 & 5 & 6 \\
\hline 2 & 1 & 6 \\
\hline 9 & 6 & 4 \\
\hline \end{array}$
which as can be seen are not all different.
And, trivially, if these all do not have to be different, then why are we wasting our time with this when they could all be given a multiplicity of $1$?
== Historical Note ==
== Sources ==
* : Solutions: $289$. -- Magic Square Trick
* : Answers: $389$. Magic Square Trick
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/28 - A Farthings Puzzle

== by : $28$ ==
;A Farthings Puzzle
:Find a sum of money expressed in pounds, shillings and pence
::which, when you take the currency indicators and punctuation away, reads the number that you get when you reduce the sum to farthings.
:For example, $\pounds 10, 10 \shillings 7 \oldpence$ is $10 \, 108$ farthings which is exactly one farthing out.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $28$. -- A Farthings Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/28 - A Farthings Puzzle/Solution

== by : $28$ ==
;A Farthings Puzzle
== Solution ==
:$\pounds 12, 12 \shillings 8 \oldpence$ which is $12 \, 128$ farthings.
== Sources ==
* : Solutions: $28$. -- The Farthings Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/290 - A Four-Figure Magic Square

== by : $290$ ==
;A Four-Figure Magic Square
:In this square, as every cell contains the same number -- $1234$ -- the three columns, three rows and two long diagonals naturally add up alike.
$\qquad \begin{array} {|c|c|c|}
\hline 1234 & 1234 & 1234 \\
\hline 1234 & 1234 & 1234 \\
\hline 1234 & 1234 & 1234 \\
\hline \end{array}$
:The puzzle is to form and place nine different $4$-figure numbers (using the same figures) so that they shall form a perfect magic square.
:That is, the numbers all together must contain nine of each of $1$, $2$, $3$ and $4$, and they must be proper numbers without using fractions or any other trick like that.
=== Click here for solution ===
== Sources ==
* : Magic Square, Measuring, Weighing, and Packing Problems: $290$. -- A Four-Figure Magic Square
* : Combinatorial & Topological Problems: Magic Square Puzzles: $390$. A Four-Figure Magic Square
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/290 - A Four-Figure Magic Square/Solution

== by : $290$ ==
;A Four-Figure Magic Square
== Solution ==
$\qquad \begin{array} {|c|c|c|}
\hline 2243 & 1341 & 3142 \\
\hline 3141 & 2242 & 1343 \\
\hline 1342 & 3143 & 2241 \\
\hline \end{array}$
The magic constant is $6276$.
== Sources ==
* : Solutions: $290$. -- A Four-Figure Magic Square
* : Answers: $390$. A Four-Figure Magic Square
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/291 - Progressive Squares

== by : $291$ ==
;Progressive Squares
:This is a magic square, adding up to $287$ in every row, every column, and each of the two diagonals.
:
:If we remove the outer margin of numbers we have another square giving sums of $205$.
:If we again remove the margin there is left a magic square adding up to $123$.
:Now fill up the vacant spaces in the diagram with such numbers from $1$ to $81$ inclusive as have not already been given,
::so that there shall be formed a magic square adding up to $369$ in each of twenty directions.
=== Click here for solution ===
== Sources ==
* : Magic Square, Measuring, Weighing, and Packing Problems: $291$. -- Progressive Squares
* : Combinatorial & Topological Problems: Magic Square Puzzles: $391$. Progressive Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/292 - Conditional Magic Square

== by : $292$ ==
;Conditional Magic Square
:Can you form a magic square with all the columns rows, and two long diagonals, adding up alike,
::with numbers $1$ to $25$ inclusive, placing only the odd numbers on the shaded squares in the diagram,
::and the even numbers on the other squares?
:
:There are a good many solutions.
:Can you find one of them?
=== Click here for solution ===
== Sources ==
* : Magic Square, Measuring, Weighing, and Packing Problems: $292$. -- Conditional Magic Square
* : Combinatorial & Topological Problems: Magic Square Puzzles: $392$. Conditional Magic Square
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/292 - Conditional Magic Square/Solution

== by : $292$ ==
;Conditional Magic Square
== Solution ==
Here is one solution:
:
== Sources ==
* : Solutions: $292$. -- Conditional Magic Square
* : Answers: $392$. Conditional Magic Square
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/293 - The Twenty Pennies

== by : $293$ ==
;The Twenty Pennies
:If sixteen pennies are arranged in the form of a square
::there will be the same number of pennies in every row, column and each of the long diagonals.
:Can you do the same with twenty pennies?
=== Click here for solution ===
== Sources ==
* : Magic Square, Measuring, Weighing, and Packing Problems: $293$. -- The Twenty Pennies
* : Combinatorial & Topological Problems: Point Alignment Puzzles: $436$. The Twenty Pennies
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/293 - The Twenty Pennies/Solution

== by : $293$ ==
;The Twenty Pennies
== Solution ==
Start with the square of $16$ pennies in $4$ rows of $4$.
Place:
:one penny on the first penny in the first row
:one penny on the third penny in the second row
:one penny on the fourth penny in the third row
:one penny on the second penny in the fourth row.
== Sources ==
* : Solutions: $293$. -- The Twenty Pennies
* : Answers: $436$. The Twenty Pennies
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/294 - The Keg of Wine

== by : $294$ ==
;The Keg of Wine
:A man had a $10$-gallon keg of wine and a jug.
:One day he drew off a jugful of wine and filled up the keg with water.
:Later on, when the wine and water had got thoroughly mixed, he drew off another jugful, and again filled up the keg with water.
:The keg then contained equal quantities of wine and water.
:What was the capacity of the jug?
=== Click here for solution ===
== Sources ==
* : Magic Square, Measuring, Weighing, and Packing Problems: $294$. -- The Keg of Wine
* : Combinatorial & Topological Problems: Liquid Pouring Puzzles: $406$. The Keg of Wine
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/294 - The Keg of Wine/Solution

== by : $294$ ==
;The Keg of Wine
== Solution ==
Approximately $2.93$ gallons.
== Proof ==
Let $J$ gallons be the volume of the jug.
At the first drawing there will be $10 - J$ gallons of wine in the keg.
After mixing it with water, the concentration of wine per unit volume is $\dfrac {10 - J} {10}$.
The second drawing takes $J \dfrac {10 - J} {10}$ gallons of wine out of the keg.
Hence there remains $10 - J - J \dfrac {10 - J} {10}$ gallons of wine in the keg.
Then the keg is refilled.
Because there are equal quantities of wine and water in the keg, that means there must be $5$ gallons of each.
That is:
This evaluates to approximately $10 - 7.07$, or about $2.93$ gallons.
== Sources ==
* : Solutions: $294$. -- The Keg of Wine
* : Answers: $406$. The Keg of Wine
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/295 - Blending the Teas

== by : $295$ ==
;Blending the Teas
:A grocer buys two kinds of tea --
::one at $2 \shillings 8 \oldpence$ per pound,
::and the other, a better quality, at $3 \shillings 4 \oldpence$ per pound.
:He mixes together some of each, which he proposes to sell at $3 \shillings 7 \oldpence$ a pound,
::and so make a profit of $25$ per cent on the cost.
:How many pounds of each kind must he use to make a mixture of $100$ pounds weight?
=== Click here for solution ===
== Sources ==
* : Magic Square, Measuring, Weighing, and Packing Problems: $295$. -- Blending the Teas
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $249$. Blending the Teas
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/295 - Blending the Teas/Solution

== by : $295$ ==
;Blending the Teas
== Solution ==
:$70$ pounds of cheaper tea and $30$ pounds of the more expensive one.
== Proof ==
To simplify the analysis, we express all prices in pence:
:$2 \shillings 8 \oldpence$ per pound is $2 \times 12 + 8 = 32 \oldpence$ per pound
:$3 \shillings 4 \oldpence$ per pound is $3 \times 12 + 4 = 40 \oldpence$ per pound
:$3 \shillings 7 \oldpence$ per pound is $3 \times 12 + 7 = 43 \oldpence$ per pound
:$25 \%$ profit on $32 \oldpence$ per pound is $8 \oldpence$, hence the cheaper brand will be selling at $40 \oldpence$ per pound
:$25 \%$ profit on $40 \oldpence$ per pound is $10 \oldpence$, hence the dearer brand will be selling at $50 \oldpence$ per pound.
Let $a$ be the weight of cheaper tea going into making the blend.
Let $b$ be the weight of dearer tea going into making the blend.
We have:
== Sources ==
* : Solutions: $295$. -- Blending the Teas
* : Answers: $249$. Blending the Teas
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/296 - Water Measurement

== by : $296$ ==
;Water Measurement
:A maid was sent to the brook with two vessels that exactly measured $7$ pints and $11$ pints exactly.
:She had to bring back exactly $2$ pints of water.
:What is the smallest possible number of transactions necessary?
:A "transaction" is filling a vessel, of emptying it, or pouring from one vessel to another.
=== Click here for solution ===
== Sources ==
* : Magic Square, Measuring, Weighing, and Packing Problems: $296$. -- Water Measurement
* : Combinatorial & Topological Problems: Liquid Pouring Puzzles: $407$. Water Measurement
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/296 - Water Measurement/Solution

== by : $296$ ==
;Water Measurement
== Solution ==
:$14$ transactions are needed.
== Proof ==
Let $V_7$ and $V_{11}$ denote the $7$-pint and $11$-pint vessels respectively.
In the below, the successive lines indicate the status of each of $V_7$ and $V_{11}$:
:$\begin{array}{c|c} V_7 & V_{11} \\ \hline
7 & 0 \\
0 & 7 \\
7 & 7 \\
3 & 11 \\
3 & 0 \\
0 & 3 \\
7 & 3 \\
0 & 10 \\
7 & 10 \\
6 & 11 \\
6 & 0 \\
0 & 6 \\
7 & 6 \\
2 & 11 \\
\end{array}$
and it is seen that the required $2$ pints are in the $7$-pint vessel.
== Sources ==
* : Solutions: $296$. -- Water Measurement
* : Answers: $407$. Water Measurement
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions