nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Modern Puzzles/177 - The Six-Pointed Star

== by : $177$ ==
;The Six-Pointed Star
:We have considered the question of the five-pointed star.
:We shall now find the six-pointed star even more interesting.
:In this case we can always use the twelve consecutive numbers $1$ to $12$ and the sum of the four numbers in every line will always be $26$.
:The numbers at the six points of the star may add up to any even number from $24$ to $54$ inclusive, except $28$ and $50$, which are impossible.
:It will be seen in the example that the six points add up to $24$.
:
:If for every number in its present position you substitute its difference from $13$ you will get another solution, its complementary,
::with the points adding up to $54$, which is $78$ less $24$.
:The two complementary totals will always sum to $78$.
:I will give the total number of different solutions and point out some of the pretty laws which govern the problem,
::but I will leave the reader this puzzle to solve.
:There are six arrangements, and six only, in which all the lines of four and the six points also add up to $26$.
:Can you find one or all of them?
=== Click here for solution ===
== Sources ==
* : Magic Star Problems: $177$. -- The Six-Pointed Star
* : Combinatorial & Topological Problems: Magic Star Puzzles: $394$. The Six-Pointed Star
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/177 - The Six-Pointed Star/Historical Note

== Historical Note on by : $177$ - The Six-Pointed Star ==
 reported this problem in his Mathematical Games column in in the December $1965$ issue.
In correspondence, he received confirmation from and that there are in fact $80$ patterns for this six-pointed star, rather than the $74$ offered by .
== Sources ==
* : Answers: $395$. The Seven-Pointed Star
* : $5$: Magic Stars and Polyhedrons
Category:Historical Notes
Henry Ernest Dudeney/Modern Puzzles/177 - The Six-Pointed Star/Mistake

== Source Work ==
:
:Magic Star Problems
::$177$ -- The Six-Pointed Star
:
:Combinatorial & Topological Problems
::Magic Star Puzzles
:::$394$. The Six-Pointed Star
== Mistake ==
:There are $37$ solutions in all, or $74$ if we count complementaries.
:$32$ of these are regular, and $5$ are irregular.
:Of the $37$ solutions, $6$ have their points summing to $26$. These are as follows:
::$\begin {array} {rrrrrrrrrrrr}
10 & 6 & 2 & 3 & 1 & 4 & 7 & 9 & 5 & 12 & 11 & 8 \\
9 & 7 & 1 & 4 & 3 & 2 & 6 & 11 & 5 & 10 & 12 & 8 \\
5 & 4 & 6 & 8 & 2 & 1 & 9 & 12 & 3 & 11 & 7 & 10 \\
5 & 2 & 7 & 8 & 1 & 3 & 11 & 10 & 4 & 12 & 6 & 9 \\
10 & 3 & 1 & 4 & 2 & 6 & 9 & 8 & 7 & 12 & 11 & 5 \\
8 & 5 & 3 & 1 & 2 & 7 & 10 & 4 & 11 & 9 & 12 & 6 \\
\end {array}$
:...
:Also note that where the $6$ points add to $24$, $26$, $30$, $32$, $34$, $36$ or $38$, the respective number of solutions is $3$, $6$, $2$, $4$, $7$, $6$ and $9$, making $37$ in all.
== Correction ==
There are in fact $80$ solutions, as noted by in .
== Sources ==
* : Solutions: $177$. -- The Six-Pointed Star
* : Answers: $394$. The Six-Pointed Star
* : $5$: Magic Stars and Polyhedrons
Category:Mistakes/Modern Puzzles
Category:Mistakes/536 Puzzles & Curious Problems
Henry Ernest Dudeney/Modern Puzzles/177 - The Six-Pointed Star/Solution

== by : $177$ ==
;The Six-Pointed Star
== Solution ==
's solution is as follows:
:
:$(1): \quad$ In every solution, the sum of the numbers in triangle $ABC$ of Figure $\text I$ must equal the sum of the numbers in the triangle $DEF$.
:::This can be anywhere from $12$ to $27$ inclusive, except $14$ and $25$, which are impossible.
:::We need only obtain solutions for $12$, $13$, $15$, $16$, $17$, $18$ and $19$, because those of the complementaries may be derived by substituting for every number its difference from $13$.
:$(2): \quad$ Every arrangement is composed of three independent diamonds $AGHF$, $DKBL$ and $EMCI$, each of which must add up to $26$.
:$(3): \quad$ The sum of the numbers in opposite external triangles will always be equal, for example, $AIK$ equals $LMF$.
:$(4): \quad$ If the difference between $26$ and its triangle sum $ABC$ is added to any number at a point, say $A$, it will give the sum of the two numbers in the relative positions of $L$ and $M$.
:::Thus for example in Figure $\text {II}$: $10 + 13 = 11 + 12$, and $6 + 13 = 8 + 11$.
:
:$(5): \quad$ There are $6$ pairs summing the $13$, which are: $12 + 1$, $11 + 2$, $10 + 3$, $9 + 4$, $8 + 5$, $7 + 6$.
:::One or two such pairs may occur among the numbers at the points, but never three.
:::The relative positions of these pairs determine the type of solution.
:::In the regular type, as in Figure $\text {II}$, $A$ and $F$ and also $G$ and $H$, as indicated by the dotted lines, always sum to $13$, but this class can be subdivided.
:::Figures $\text {III}$ and $\text {IV}$ are examples of the two irregular types.
: $\qquad$ 
:There are $37$ solutions in all, or $74$ if we count complementaries.
:$32$ of these are regular, and $5$ are irregular.
:Of the $37$ solutions, $6$ have their points summing to $26$. These are as follows:
::$\begin {array} {rrrrrrrrrrrr}
10 & 6 & 2 & 3 & 1 & 4 & 7 & 9 & 5 & 12 & 11 & 8 \\
9 & 7 & 1 & 4 & 3 & 2 & 6 & 11 & 5 & 10 & 12 & 8 \\
5 & 4 & 6 & 8 & 2 & 1 & 9 & 12 & 3 & 11 & 7 & 10 \\
5 & 2 & 7 & 8 & 1 & 3 & 11 & 10 & 4 & 12 & 6 & 9 \\
10 & 3 & 1 & 4 & 2 & 6 & 9 & 8 & 7 & 12 & 11 & 5 \\
8 & 5 & 3 & 1 & 2 & 7 & 10 & 4 & 11 & 9 & 12 & 6 \\
\end {array}$
:The first of these is Figure $\text {II}$, and the last but one is Figure $\text {III}$.
:A reference to those diagrams shows how to write out the star.
:The first four are of the regular type and the last two are irregular.
:Also note that where the $6$ points add to $24$, $26$, $30$, $32$, $34$, $36$ or $38$, the respective number of solutions is $3$, $6$, $2$, $4$, $7$, $6$ and $9$, making $37$ in all.
There are also another $6$ solutions which did not find.
== Historical Note ==
== Sources ==
* : Solutions: $177$. -- The Six-Pointed Star
* : Answers: $394$. The Six-Pointed Star
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/178 - The Seven-Pointed Star

== by : $178$ ==
;The Seven-Pointed Star
:We have already dealt briefly with stars of five and six points.
:The case of the seven-pointed star is particularly interesting.
:All you have to do is place the numbers $1$, $2$, $3$, up to $14$ in the fourteen discs so that every line of four disks shall add up to $30$.
:
:If you make a rough diagram and use numbered counters, you will find it difficult to break away from the fascination of the thing.
:Possibly, however, not a single reader will hit upon a simple method of solution:
::his answer, when found, will be obtained by mere patience and luck.
:Yet, like those of the large majority of the puzzles given in these pages,
::the solution is subject to law, if you can unravel it.
=== Click here for solution ===
== Sources ==
* : Magic Star Problems: $178$. -- The Seven-Pointed Star
* : Combinatorial & Topological Problems: Magic Star Puzzles: $395$. The Seven-Pointed Star
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/178 - The Seven-Pointed Star/Historical Note

== Historical Note on by : $178$ - The Seven-Pointed Star ==
 reported this problem in his Mathematical Games column in in the December $1965$ issue.
In correspondence, he received confirmation from that there are in fact $72$ patterns for this seven-pointed star, rather than the $56$ offered by .
This was confirmed by the independent work of and .
In $1966$, used a computer at the in Ontario to confirm the $72$ solutions.
== Sources ==
* : Answers: $395$. The Seven-Pointed Star
* : $5$: Magic Stars and Polyhedrons
Category:Historical Notes
Henry Ernest Dudeney/Modern Puzzles/178 - The Seven-Pointed Star/Mistake

== Source Work ==
:
:Magic Star Problems
::$178$ -- The Seven-Pointed Star
:
:Combinatorial & Topological Problems
::Magic Star Puzzles
:::$395$. The Seven-Pointed Star
== Mistake ==
:There are $56$ different arrangements, counting complements.
:Class $\text I$ is those as above, where pairs in the positions $7 - 8$, $13 - 2$, $3 - 12$, $14 - 1$ all add to $15$, and there are $20$ such cases.
:Class $\text {II}$ includes cases where pairs in the positions $7 - 2$, $8 - 13$, $3 - 1$, $12 - 14$ all add to $15$, and there are $20$ such cases.
:Class $\text {III}$ includes cases where pairs in the positions $7 - 8$, $13 - 2$, $3 - 1$, $12 - 14$ all add to $15$, and there are $16$ such cases.
== Correction ==
There are in fact $72$ solutions, as noted by in .
== Sources ==
* : Solutions: $178$. -- The Seven-Pointed Star
* : Answers: $395$. The Seven-Pointed Star
* : $5$: Magic Stars and Polyhedrons
Category:Mistakes/Modern Puzzles
Category:Mistakes/536 Puzzles & Curious Problems
Henry Ernest Dudeney/Modern Puzzles/178 - The Seven-Pointed Star/Solution

== by : $178$ ==
;The Seven-Pointed Star
== Solution ==
's solution is as follows:
:
:Place $5$ in the top point.
:Let the $4$ numbers in the horizontal line, that is $7$, $11$, $9$, $3$, be such that:
::the two outside numbers sum to $10$
::the inner numbers sum to $20$
::the difference between the two outer numbers is twice the difference between the two inner numbers.
:Then their complements with $15$ are placed in the relative positions indicated by the dotted lines.
:The remaining $4$ numbers, that is $13$, $2$, $14$, $1$ are easily adjusted.
:From this arrangement we can get $3$ others.
::$(1): \quad$ Change the $13$ with the $1$ and the $14$ with the $2$.
::$(2)$ and $(3): \quad$ Substitute in each of those two above arrangements its difference from $15$.
:There are $56$ different arrangements, counting complements.
:Class $\text I$ is those as above, where pairs in the positions $7 - 8$, $13 - 2$, $3 - 12$, $14 - 1$ all add to $15$, and there are $20$ such cases.
:Class $\text {II}$ includes cases where pairs in the positions $7 - 2$, $8 - 13$, $3 - 1$, $12 - 14$ all add to $15$, and there are $20$ such cases.
:Class $\text {III}$ includes cases where pairs in the positions $7 - 8$, $13 - 2$, $3 - 1$, $12 - 14$ all add to $15$, and there are $16$ such cases.
There are also another $16$ solutions which did not find.
== Historical Note ==
== Sources ==
* : Solutions: $178$. -- The Seven-Pointed Star
* : Answers: $395$. The Seven-Pointed Star
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/179 - Two Eight-Pointed Stars

== by : $179$ ==
;Two Eight-Pointed Stars
:The puzzles of stars with five, six, and seven points that I have given lead up to the eight-pointed star.
:The star may be formed in two different ways, as shown in our diagram, and the first example is a solution.
:The numbers $1$ to $16$ are so placed that every straight line of four adds up to $34$.
:If you substitute for every number its difference from $17$ you will get the complementary solution.
: $\qquad$ 
:Let the reader try to discover some of the other solutions, and he will find it a very hard nut, even with this one to help him.
:But I will present the puzzle in an easy and entertaining form.
:When you know how, every arrangement in the first star can be transferred to the second one automatically.
:Every line of four numbers in the one case will appear in the other, only the order of the numbers will have to be changed.
:Now, with this information given, it is not a difficult puzzle to find a solution for the second star.
=== Click here for solution ===
== Sources ==
* : Magic Star Problems: $179$. -- Two Eight-Pointed Stars
* : Combinatorial & Topological Problems: Magic Star Puzzles: $396$. Two Eight-Pointed Stars
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/179 - Two Eight-Pointed Stars/Historical Note

== Historical Note on by : $179$ - Two Eight-Pointed Stars ==
In $1963$, analysed this problem and found $112$ solutions.
He was also able to estimate that the $9$-pointed star has more than $2000$ distinct patterns.
== Sources ==
* : Answers: $396$. Two Eight-Pointed Stars
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/179 - Two Eight-Pointed Stars/Solution

== by : $179$ ==
;Two Eight-Pointed Stars
== Solution ==
 presents this solution:
:
:If you find any solution to one of the stars, you can immediately transfer it to the other by noting the relative positions in the case given.
:No attempt at this stage has been made to find out how many stars there are of order $8$.
== Historical Note ==
== Sources ==
* : Solutions: $179$. -- Two Eight-Pointed Stars
* : Answers: $396$. Two Eight-Pointed Stars
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/17 - Name their Wives

== by : $17$ ==
;Name their Wives
:A man left a legacy of $\pounds 1000$ to $3$ relatives and their wives.
:The wives received together $\pounds 396$.
:Jane received $\pounds 10$ more than Catherine,
::and Mary received $\pounds 10$ more than Jane.
:John Smith was given just as much as his wife,
:Henry Snooks got half as much again as his wife,
:and Tom Crowe received twice as much as his wife.
:What was the Christian name of each man's wife?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $17$. -- Name their Wives
* : Arithmetical and Algebraical Problems: Money Puzzles: $10$. Name their Wives
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/17 - Name their Wives/Solution

== by : $17$ ==
;Name their Wives
== Solution ==
:Catherine is Mrs. John Smith
:Jane is Mrs. Henry Snooks
:Mary is Mrs. Tom Crowe.
== Proof ==
Let $C$, $M$ and $J$ be the sizes of legacies of Catherine, Mary and Jane respectively.
We are given that:
Now we note that the total inherited by the husbands is $1000 - 396 = 604$.
Using the method of false position, we multiply $C$, $J$ and $M$ by $1$, $\dfrac 3 2$ and $2$ respectively, and sum them to see what we get.
Thus we find:
:$122 + \dfrac 3 2 \times 132 + 2 \times 142 = 604$
and marvel that the first guess was spot on.
So:
:Catherine is Mrs. John Smith
:Jane is Mrs. Henry Snooks
:Mary is Mrs. Tom Crowe.
== Sources ==
* : Solutions: $17$. -- Name their Wives
* : Answers: $10$. Name their Wives
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/180 - The Damaged Measure

== by : $180$ ==
;The Damaged Measure
:Here is a new puzzle that is interesting, and it reminds one, though it is really very different,
::of the classical puzzle by concerning the weight that was broken in pieces
::which would then allow of any weight in pounds being determined from one pound up to the total weight of all the pieces.
:A young man has a yardstick from which $3$ inches have been broken off,
::so that it is only $33$ inches in length.
:Some of the graduation marks are also obliterated, so that only eight of these marks are legible;
::yet he is able to measure any given number of inches from $1$ inch up to $33$ inches.
:Where are these marks placed?
:As an example, I give in the diagram the case of a $13$-inch rod with $4$ markings.
:
:If I want to measure $4$ inches, I take $1$ and $3$; for $8$ inches, $6$ and $2$; for $10$ inches, $3$, $1$ and $6$; and so on.
:Of course, the exact measure must be taken at once on the rod;
::otherwise the single mark of $1$ inch repeated a sufficient number of times would measure any length, which would make the puzzle absurd.
=== Click here for solution ===
== Sources ==
* : Measuring, Weighing, and Packing Puzzles: $180$. -- The Damaged Measure
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $453$. The Damaged Measure
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/180 - The Damaged Measure/Historical Note

== Historical Note on by : $180$ - The Damaged Measure ==
 reported in $1968$ that no general rule had yet been found for the ruler problem.
However, considerable progress had been made since first proposed it.
In particular, discovered in $1956$ that eight marks are sufficient to mark a $36$-inch ruler so that all integer lengths from $1$ to $36$ can be measured, outdoing 's solution by $3$ inches.
== Sources ==
* 
* : Answers: $453$. The Damaged Measure
Category:Historical Notes
Henry Ernest Dudeney/Modern Puzzles/180 - The Damaged Measure/Solution

== by : $180$ ==
;The Damaged Measure
== Solution ==
Divide the $33$-inch measure into the following sections:
:$1$, $3$, $1$, $9$, $2$, $7$, $2$, $6$, $2$
Another solution is:
:$1$, $1$, $1$, $1$, $6$, $6$, $6$, $6$, $5$
According to himself, there are $16$ different ways of doing it.
== Historical Note ==
== Sources ==
* : Solutions: $180$. -- The Damaged Measure
* : Answers: $453$. The Damaged Measure
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/181 - The Six Cottages

== by : $181$ ==
;The Six Cottages
:A circular road, $27$ miles long, surrounds a tract of wild and desolate country,
::and on this road are $6$ cottages so placed that one cottage or another is at a distance of one, two, three up to $26$ miles inclusive from some other cottage.
:Thus, Brown may be a mile from Stiggins, Jones two miles from Rogers, Wilson three miles from Jones, and so on.
:Of course, they can walk in either direction if required.
:Can you place the cottages at distances that will fulfil the conditions?
:
:The illustration is intended to give no clue as to the relative distances.
=== Click here for solution ===
== Sources ==
* : Measuring, Weighing, and Packing Puzzles: $181$. -- The Six Cottages
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $454$. The Six Cottages
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/181 - The Six Cottages/Historical Note

== Historical Note on by : $181$ - The Six Cottages ==
This problem is merely a circular version of the previous problem, The Damaged Measure.
It turns out that the road could have been $31$ miles long so that all integer distances between $1$ and $30$ could be measured.
== Sources ==
* 
* : Answers: $454$. The Six Cottages
Category:Historical Notes
Henry Ernest Dudeney/Modern Puzzles/181 - The Six Cottages/Solution

== by : $181$ ==
;The Six Cottages
== Solution ==
The distances between the houses need to be, in the following order:
:$1, 1, 4, 4, 3, 14$
around the circular road.
== Historical Note ==
== Sources ==
* : Solutions: $181$. -- The Six Cottages
* : Answers: $454$. The Six Cottages
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/182 - A New Domino Puzzle

== by : $182$ ==
;A New Domino Puzzle
:Two dominoes have been placed together so that by taking the pips in unbroken conjunction
::I can get all the numbers from $1$ to $9$ inclusive.
:
:Thus, $1$, $2$ and $3$ can be taken alone;
::then $1$ and $3$ make $4$; $3$ and $2$ make $5$; $3$ and $3$ make $6$;
::$1$, $3$ and $3$ make $7$; $3$, $3$ and $2$ make $8$, and $1$, $3$, $3$ and $2$ make $9$.
:It would not have been allowed to take the $1$ and the $2$ to make $3$, nor to take the first $3$ and the $2$ to make $5$.
:The numbers would not have been in conjunction.
:Now try to arrange four dominoes so that you can make the pips in this way sum to any number from $1$ to $24$ inclusive.
:The dominoes need not be placed $1$ against $1$, $2$ against $2$, and so on, as in play.
=== Click here for solution ===
== Sources ==
* : Measuring, Weighing, and Packing Puzzles: $182$. -- A New Domino Puzzle
* : Domino Puzzles: $481$. A New Domino Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Dominoes
Henry Ernest Dudeney/Modern Puzzles/182 - A New Domino Puzzle/Historical Note

== Historical Note on by : $182$ - A New Domino Puzzle ==
 found an improved solution which gives sums from $1$ to $29$:
:
He also provides a $3$-domino solution which gives sums from $1$ to $17$:
:
== Sources ==
* : Answers: $481$. A New Domino Puzzle
Category:Historical Notes
Henry Ernest Dudeney/Modern Puzzles/182 - A New Domino Puzzle/Solution

== by : $182$ ==
;A New Domino Puzzle
== Solution ==
:
== Historical Note ==
== Sources ==
* : Solutions: $182$. -- A New Domino Puzzle
* : Answers: $481$. A New Domino Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/183 - At the Brook

== by : $183$ ==
;At the Brook
:In introducing liquid measuring puzzles in my book , I have said,
::"It is the general opinion that puzzles of this class can only be solved by trial,
::but I think formulæ can be constructed for the solution generally of certain related cases.
::It is a practically unexplored field for investigation."
:So far as I know, the hint has not been taken and the field is still unexplored,
::so I recently took advantage of a little unexpected leisure to look into the matter.
:The result, as I thought probable, was that I struck some new and very interesting things.
:Let us take the simplest possible case of a man who goes to a brook with only two vessels with which to measure a given quantity of water.
:When we are dealing, say, with a barrel of wine we may have complications arising from the barrel being full or empty,
::from its capacity and contents being known or unknown,
::from waste of wine being permitted or not permitted,
::and from pouring back into the barrel being allowed.
:All these points are eliminated.
:Is it then possible that any puzzle remains? Let us see.
:A man goes to the brook with two measures of $15$ pints and $16$ pints.
:How is he to measure exactly $8$ pints of water, in the fewest possible transactions?
:Filling or emptying a vessel or pouring any quantity from one vessel to another counts as a transaction.
:The puzzle is not difficult, but I think the reader will find it very entertaining and instructive.
:I need hardly add that no tricks, such as marking or tilting the vessels, are allowed.
=== Click here for solution ===
== Sources ==
* : Measuring, Weighing, and Packing Puzzles: $183$. -- At the Brook
* : Combinatorial & Topological Problems: Liquid Pouring Puzzles: $403$. At the Brook
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Henry Ernest Dudeney/Modern Puzzles/183 - At the Brook
Henry Ernest Dudeney/Modern Puzzles/183 - At the Brook/General Solution

== by : $183$ ==
;At the Brook
== General Solution ==
 starts by demonstrating that any integer quantity from $1$ to $16$ pints can be measured using these two containers.
Perform the transactions according to the one of the following two procedures:
:$\begin {array} {r|c|c|c} & \text {$15$-pint} & \text {$16$-pint} \\ \hline
 & 0 & 0 \\
* & 0 & 16 \\
* & 15 & 1 \\
 & 0 & 1 \\
 & 1 & 0 \\
 & 1 & 16 \\
* & 15 & 2 \\
 & 0 & 2 \\
 & 2 & 0 \\
 & 2 & 16 \\
* & 15 & 3 \\
 & 0 & 3 \\
 & 3 & 0 \\
 & 3 & 16 \\
* & 15 & 4 \\
 & 0 & 4 \\
 & 4 & 0 \\
 & 4 & 16 \\
* & 15 & 5 \\
 & 0 & 5 \\
 & 5 & 0 \\
 & 5 & 16 \\
* & 15 & 6 \\
 & 0 & 6 \\
 & 6 & 0 \\
 & 6 & 16 \\
* & 15 & 7 \\
 & 0 & 7 \\
 & 7 & 0 \\
\end {array}
\qquad \begin {array} {r|c|c|c} & \text {$15$-pint} & \text {$16$-pint} \\ \hline
 & 0 & 0 \\
* & 15 & 0 \\
 & 0 & 15 \\
 & 15 & 15 \\
* & 14 & 16 \\
 & 14 & 0 \\
 & 0 & 14 \\
 & 15 & 14 \\
* & 13 & 16 \\
 & 13 & 0 \\
 & 0 & 13 \\
 & 15 & 13 \\
* & 12 & 16 \\
 & 12 & 0 \\
 & 0 & 12 \\
 & 15 & 12 \\
* & 11 & 16 \\
 & 11 & 0 \\
 & 0 & 11 \\
 & 15 & 11 \\
* & 10 & 16 \\
 & 10 & 0 \\
 & 0 & 10 \\
 & 15 & 10 \\
* & 9 & 16 \\
 & 9 & 0 \\
 & 0 & 9 \\
 & 15 & 9 \\
* & 8 & 16 \\
\end {array}$
In the first column, the asterisks mark where the quantities from $1$ pint to $8$ pints have been measured, and a continuance of the pattern for another few transactions.
In the second column, the asterisks mark where the quantities from $15$ pints down to $8$ pints have been measured.
It is seen that fewer transactions are needed to get to $8$ by filling the $15$-pint container first, while to get to $7$ it is quicker to start by filling the $16$-pint container.
== Historical Note ==
== Sources ==
* : Solutions: $183$. -- At the Brook
* : Answers: $403$. At the Brook
Category:Henry Ernest Dudeney/Modern Puzzles/183 - At the Brook