nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Modern Puzzles/183 - At the Brook/Historical Note

== Historical Note on by : $183$ - At the Brook ==
 reports on his Mathematical Games article in the September $1963$ issue of , where he describes a technique for solving liquid pouring problems using isometric paper.
The subject is also discussed in .
== Sources ==
* : Chapter $4$.
* : Answers: $403$. At the Brook
Category:Historical Notes
Category:Henry Ernest Dudeney/Modern Puzzles/183 - At the Brook
Henry Ernest Dudeney/Modern Puzzles/183 - At the Brook/Solution

== by : $183$ ==
;At the Brook
== Solution ==
Perform the transactions according to the following:
:$\begin {array} {r|c|c|c} & \text {$15$-pint} & \text {$16$-pint} \\ \hline
& 0 & 0 \\
(1): & 15 & 0 \\
(2): & 0 & 15 \\
(3): & 15 & 15 \\
(4): & 14 & 16 \\
(5): & 14 & 0 \\
(6): & 0 & 14 \\
(7): & 15 & 14 \\
(8): & 13 & 16 \\
(9): & 13 & 0 \\
(10): & 0 & 13 \\
(11): & 15 & 13 \\
(12): & 12 & 16 \\
(13): & 12 & 0 \\
(14): & 0 & 12 \\
(15): & 15 & 12 \\
(16): & 11 & 16 \\
(17): & 11 & 0 \\
(18): & 0 & 11 \\
(19): & 15 & 11 \\
(20): & 10 & 16 \\
(21): & 10 & 0 \\
(22): & 0 & 10 \\
(23): & 15 & 10 \\
(24): & 9 & 16 \\
(25): & 9 & 0 \\
(26): & 0 & 9 \\
(27): & 15 & 9 \\
(28): & 8 & 16 \\
\end {array}$
and it is seen that $8$ pints have been measured out in $28$ transactions.
=== General Solution ===
== Historical Note ==
== Sources ==
* : Solutions: $183$. -- At the Brook
* : Answers: $403$. At the Brook
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Category:Henry Ernest Dudeney/Modern Puzzles/183 - At the Brook
Henry Ernest Dudeney/Modern Puzzles/184 - A Prohibition Poser

== by : $184$ ==
;A Prohibition Poser
:Let us now take another step and look at those cases where we are still allowed any amount of waste,
::though the liquid is now limited to a stated quantity.
:The American Prohibition authorities discovered a full barrel of beer,
::and were about to destroy the liquor by letting it run down a drain
::when the owner pointed to two vessels standing by and begged to be allowed to retain in them a small quantity for the immediate consumption of his household.
:One vessel was a $7$-quart and the other a $5$-quart measure.
:The officer was a wag, and, believing it to be impossible, said that if the man could measure an exact quart into each vessel
::(without any pouring back into the barrel) he might do so.
:How was it to be done in the fewest possible transactions without any marking or other tricks?
:Pouring down the drain counts as one transaction.
:Perhaps I should state that an American barrel of beer contains exactly $120$ quarts.
=== Click here for solution $1$ ===
=== Click here for solution $2$ ===
== Sources ==
* : Measuring, Weighing, and Packing Puzzles: $184$. -- A Prohibition Poser
* : Combinatorial & Topological Problems: Liquid Pouring Puzzles: $404$. A Prohibition Poser
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Henry Ernest Dudeney/Modern Puzzles/184 - A Prohibition Poser
Henry Ernest Dudeney/Modern Puzzles/184 - A Prohibition Poser/Solution 1

== by : $184$ ==
;A Prohibition Poser
== Solution ==
Fill the $7$-quart measure and empty it down the drain a total of $14$ times.
Then you will have thrown away $98$ and have $22$ quarts left in the barrel.
This takes $28$ transactions.
Then perform the transactions according to the following:
:$\begin {array} {r|c|c|c} & \text {$7$-quart} & \text {$5$-quart} & \text {Barrel} \\ \hline
& 0 & 0 & 22 \\
(1): & 7 & 0 & 15 \\
(2): & 2 & 5 & 15 \\
(3): & 2 & 0 & 15 \\
(4): & 0 & 2 & 15 \\
(5): & 7 & 2 & 8 \\
(6): & 4 & 5 & 8 \\
(7): & 4 & 0 & 8 \\
(8): & 0 & 4 & 8 \\
(9): & 7 & 4 & 1 \\
(10): & 6 & 5 & 1 \\
(11): & 6 & 0 & 1 \\
(12): & 1 & 5 & 1 \\
(13): & 1 & 0 & 1 \\
(14): & 1 & 1 & 0 \\
\end {array}$
This is a total of $42$ transactions.
== Sources ==
* : Solutions: $184$. -- A Prohibition Poser
* : Answers: $404$. A Prohibition Poser
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Category:Henry Ernest Dudeney/Modern Puzzles/184 - A Prohibition Poser
Henry Ernest Dudeney/Modern Puzzles/184 - A Prohibition Poser/Solution 2

== by : $184$ ==
;A Prohibition Poser
== Solution ==
Fill the $7$-quart measure and empty it down the drain a total of $12$ times.
Fill the $5$-quart measure and empty it down the drain a total of $4$ times.
Then you will have thrown away $104$ and have $16$ quarts left in the barrel.
This takes $32$ transactions.
Then perform the transactions according to the following:
:$\begin {array} {r|c|c|c} & \text {$7$-quart} & \text {$5$-quart} & \text {Barrel} \\ \hline
& 0 & 0 & 16 \\
(1): & 7 & 0 & 9 \\
(2): & 2 & 5 & 9 \\
(3): & 2 & 0 & 9 \\
(4): & 0 & 2 & 9 \\
(5): & 7 & 2 & 2 \\
(6): & 4 & 5 & 2 \\
(7): & 4 & 0 & 2 \\
(8): & 4 & 1 & 1 \\
(9): & 0 & 1 & 1 \\
(10): & 1 & 1 & 0 \\
\end {array}$
This is a total of $42$ transactions.
== Sources ==
* : Solutions: $184$. -- A Prohibition Poser
* : Answers: $404$. A Prohibition Poser
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Category:Henry Ernest Dudeney/Modern Puzzles/184 - A Prohibition Poser
Henry Ernest Dudeney/Modern Puzzles/185 - Prohibition Again

== by : $185$ ==
;Prohibition Again
:Let us now try to discover the fewest possible manipulations under the same conditions as in the last puzzle,
::except that we may now pour back into the barrel.
;The last puzzle
=== Click here for solution ===
== Sources ==
* : Measuring, Weighing, and Packing Puzzles: $185$. -- Prohibition Again
* : Combinatorial & Topological Problems: Liquid Pouring Puzzles: $405$. Prohibition Again
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/185 - Prohibition Again/Solution

== by : $185$ ==
;Prohibition Again
;The last puzzle
== Solution ==
Perform the transactions according to the following:
:$\begin {array} {r|c|c|c} & \text {$7$-quart} & \text {$5$-quart} & \text {Barrel} \\ \hline
& 0 & 0 & 120 \\
(1): & 7 & 0 & 113 \\
(2): & 7 & 5 & 108 \\
(2): & 7 & 5 & 0 \\
(4): & 7 & 0 & 5 \\
(5): & 2 & 5 & 5 \\
(6): & 2 & 0 & 10 \\
(7): & 0 & 2 & 10 \\
(8): & 7 & 2 & 3 \\
(9): & 4 & 5 & 3 \\
(10): & 4 & 0 & 8 \\
(11): & 0 & 4 & 8 \\
(12): & 7 & 4 & 1 \\
(13): & 6 & 5 & 1 \\
(14): & 6 & 0 & 1 \\
(15): & 1 & 5 & 1 \\
(16): & 1 & 0 & 1 \\
(17): & 1 & 1 & 0 \\
\end {array}$
This is a total of $17$ transactions.
== Sources ==
* : Solutions: $185$. -- Prohibition Again
* : Answers: $405$. Prohibition Again
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/186 - The False Scales

== by : $186$ ==
;The False Scales
:A pudding, when placed into one of the pans of a balance, appeared to weigh $4$ ounces more than $\tfrac 9 {11}$ of its true weight,
::but when placed into the other pan it appeared to weigh $3$ pounds more than in the first pan.
:What was its true weight?
=== Click here for solution ===
== Sources ==
* : Measuring, Weighing, and Packing Puzzles: $186$. -- The False Scales
* : Arithmetical and Algebraical Problems: Weight Puzzles: $97$. The False Scales
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Henry Ernest Dudeney/Modern Puzzles/186 - The False Scales
Category:False Balances
Henry Ernest Dudeney/Modern Puzzles/186 - The False Scales/Solution

== by : $186$ ==
;The False Scales
== Solution ==
Let $W$ ounces be the true weight of the pudding.
Note that we are discussing the weight of a pudding.
Hence the everyday avoirdupois scale is to be used.
First recall that there are $16$ ounces (avoirdupois) to the pound (avoirdupois).
Hence $3$ pounds more than in the first pan means $\tfrac 9 {11} W + 4 + 3 \times 16$, that is, $\tfrac 9 {11} W + 52$.
There are two different solutions to this puzzle, depending on how the false balance is faulty.
=== Solution 1: Unequal Weight of Pans ===
=== Solution 2: Unequal Arms ===
== Sources ==
* : Solutions: $186$. -- The False Scales
* : Answers: $97$. The False Scales
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Category:Henry Ernest Dudeney/Modern Puzzles/186 - The False Scales
Henry Ernest Dudeney/Modern Puzzles/186 - The False Scales/Solution 1

== by : $186$ ==
;The False Scales
== Solution ==
Suppose that the false balance has imbalanced pans.
That is, the pans are of different weight.
Then the true weight of the pudding is $154$ ounces.
== Proof ==
Let $W$ ounces be the true weight of the pudding.
We are given that:
:in one pan the apparent weight was $\dfrac {9 W} {11} + 4$
:in the other pan the apparent weight was $\dfrac {9 W} {11} + 52$.
Hence we have:
== Sources ==
* : Solutions: $186$. -- The False Scales
* : Answers: $97$. The False Scales
Category:Henry Ernest Dudeney/Modern Puzzles/186 - The False Scales
Henry Ernest Dudeney/Modern Puzzles/186 - The False Scales/Solution 2

== by : $186$ ==
;The False Scales
== Solution ==
Suppose that the false balance has unequal arms.
That is, the arms are of different length.
Then the true weight of the pudding is $143$ ounces.
== Proof ==
Let $W$ ounces be the true weight of the pudding.
We are given that:
:in one pan the apparent weight was $\dfrac {9 W} {11} + 4$
:in the other pan the apparent weight was $\dfrac {9 W} {11} + 52$.
Hence we have:
Hence the result.
== Sources ==
* : Solutions: $186$. -- The False Scales
* : Answers: $97$. The False Scales
Category:Henry Ernest Dudeney/Modern Puzzles/186 - The False Scales
Henry Ernest Dudeney/Modern Puzzles/187 - Weighing the Goods

== by : $187$ ==
;Weighing the Goods
:A tradesman whose morals had become corrupted during the war by a course of profiteering went to the length of introducing a pair of false scales.
:It will be seen from the diagram that one arm is longer than the other,
::though they are purposely drawn so as to give no clue as to the answer.
:
:As a consequence, it happened that in one of the cases exhibited eight of the little packets
::(it does not matter what they contain)
::exactly balanced three of the canisters,
::while in the other case one packet appeared to be of the same weight as $6$ canisters.
:Now, as the true weight of one canister was known to be exactly one ounce, what was the true weight of the eight packets?
=== Click here for solution ===
== Sources ==
* : Measuring, Weighing, and Packing Puzzles: $187$. -- Weighing the Goods
* : Arithmetical and Algebraical Problems: Weight Puzzles: $98$. Weighing the Goods
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/187 - Weighing the Goods/Solution

== by : $187$ ==
;Weighing the Goods
:A tradesman whose morals had become corrupted during the war by a course of profiteering went to the length of introducing a pair of false scales.
:It will be seen from the diagram that one arm is longer than the other,
::though they are purposely drawn so as to give no clue as to the answer.
:
:As a consequence, it happened that in one of the cases exhibited eight of the little packets
::(it does not matter what they contain)
::exactly balanced three of the canisters,
::while in the other case one packet appeared to be of the same weight as $6$ canisters.
:Now, as the true weight of one canister was known to be exactly one ounce, what was the true weight of the eight packets?
== Solution ==
:$8$ packets weigh $12$ ounces.
== Proof ==
Let the arms of the balance be $x$ and $y$.
Let $P$ and $C$ ounces be the weights of package and canister respectively.
From the laws of physics and the rubric of the question:
We are given that $C = 1$.
So $P = \dfrac 3 2$ and so $8 P = 12$.
== Sources ==
* : Solutions: $187$. -- Weighing the Goods
* : Answers: $98$. Weighing the Goods
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/188 - Monkey and Pulley

== by : $188$ ==
;Monkey and Pulley
:Here is a funny tangle.
:It is a mixture of 's "Monkey and Pulley", 's "How old was Mary?" and some other trifles.
:But it is quite easy if you have a pretty clear head.
:A rope is passed over a pulley.
:It has a weight at one end and a monkey at the other.
:There is the same length of rope on either side and equilibrium is maintained.
:The rope weighs four ounces per foot.
:The age of the monkey and the age of the monkey's mother total four years.
:The weight of the monkey is as many pounds as the monkey's mother is years old.
:The monkey's mother is twice as old as the monkey was
::when the monkey's mother was half as old as the monkey will be
::when the monkey is three times as old as the monkey's mother was
::when the monkey's mother was three times as old as the monkey.
:The weight of the rope and the weight at the end was half as much again as the difference in weight
::between the weight of the weight and the weight and the weight of the monkey.
:Now, what was the length of the rope?
=== Click here for solution ===
== Sources ==
* : Measuring, Weighing, and Packing Puzzles: $188$. -- Monkey and Pulley
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $246$. Monkey and Pulley
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/188 - Monkey and Pulley/Solution

== by : $188$ ==
;Monkey and Pulley
== Solution ==
:$5$ feet.
== Proof ==
First recall that there are $16$ ounces to the pound.
All weights will be discussed hereforth in ounces.
Let $l$ feet be the length of the rope.
Let $m$ years be the age of the monkey.
Let $n$ years be the age of the monkey's mother.
Let $w$ be the weight of the rope.
Let $x$ be the weight of the monkey.
Let $y$ be the weight of the weight.
Let $a$ be the number of years ago when the monkey's mother was $3$ times as old as the monkey.
Let $b$ be the number of years ahead when the monkey will be $3$ times as old as the monkey's mother was at $a$.
Let $c$ be the number of years ago when the monkey's mother was half as old as the monkey will be at $b$.
Now we proceed.
Starting near the bottom and working our way up, we whittle down the variables:
== Sources ==
* : Solutions: $188$. -- Monkey and Pulley
* : Answers: $246$. Monkey and Pulley
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/189 - Weighing the Baby

== by : $189$ ==
;Weighing the Baby
:There was a family group at the automatic weighing machine, trying to weigh the baby.
:Whenever they put the baby on the machine, she always yelled and rolled off,
::while the father was holding off the dog, who always insisted on being included in the operations.
:At last the man, with the baby and Fido, were on the machine together.
:The dial read $180$ pounds.
:The man turned to his wife and said,
::"Baby and I weigh $162$ pounds more than the dog,
::while the dog weighs $70$ per cent less than the baby.
::We must try to work it out at home."
:What was the actual weight of the baby?
=== Click here for solution ===
== Sources ==
* : Measuring, Weighing, and Packing Puzzles: $189$. -- Weighing the Baby
* : Arithmetical and Algebraical Problems: Weight Puzzles: $99$. Weighing the Baby
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/189 - Weighing the Baby/Solution

== by : $189$ ==
;Weighing the Baby
== Solution ==
The baby weighs $30$ pounds.
The man weighs $141$ pounds, and Fido weighs $9$ pounds.
== Proof ==
Let $m$, $b$ and $F$ be the weights of man, baby and Fido respectively.
We have:
== Sources ==
* : Solutions: $189$. -- Weighing the Baby
* : Answers: $99$. Weighing the Baby
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/18 - A Curious Paradox

== by : $18$ ==
;A Curious Paradox
:A man went into a shop to pay a little bill that he owed.
:On placing the money on the counter he found that he had not quite sufficient,
::owing to a small purchase that he had thoughtlessly made on the way.
:"I am so sorry," he said, "but you see I am a little short.
:"Oh, that is all right," replied the tradesman, after looking at the money, "it won't make any difference to me."
:"My good man!" exclaimed the customer ...
:... etc. etc. ...
:"... But really it will not affect my pocket in the slightest."
:Can you explain the mystery?
:It may come to you in a flash.
:The tradesman was certainly correct.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $18$. -- A Curious Paradox
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/18 - A Curious Paradox/Solution

== by : $18$ ==
;A Curious Paradox
== Solution ==
First guess: the small amount difference spans the threshold between tax brackets, which he would have had to pay to the government.
== Proof ==
 explains it in some detail, the gist of which is as follows:
A payment of $\pounds 2$ invoked a cost to the recipient of $1 \oldpence$, leaving him with $\pounds 1 \ 19 \shillings 11 \oldpence$.
Recall:
:$1$ pound sterling ($\pounds 1$) is $20$ shilling ($20 \shillings$)
:$1$ shilling is $12$ (old) pence ($12 \oldpence$)
:The man came to the shop $1 \oldpence$ short, with $\pounds 1 \ 19 \shillings 11 \oldpence$.
At that selling price the tax does not apply.
So paying a penny less is no loss to the seller.
So the initial guess was correct: it was a tax thing.
== Sources ==
* : Solutions: $18$. -- A Curious Paradox
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/190 - Packing Cigarettes

== by : $190$ ==
;Packing Cigarettes
:A manufacturer sends out his cigarettes in boxes of $160$;
::they are packed in $8$ rows of $20$ each, and exactly fill the box.
:Could he, by packing differently, get more cigarettes than $160$ into the box?
:If so, what is the greatest number that he could add?
:At first sight it sounds absurd to expect to get more cigarettes into a box that is already exactly filled,
::but a moment's consideration should give you the key to the paradox.
=== Click here for solution ===
== Sources ==
* : Measuring, Weighing, and Packing Puzzles: $190$. -- Packing Cigarettes
* : Geometrical Problems: Solid Geometry Puzzles: $326$. Packing Cigarettes
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/190 - Packing Cigarettes/Solution

== by : $190$ ==
;Packing Cigarettes
== Solution ==
:
Let the radius of the cross-section of a cigarette be $1$ unit.
It immediately follows that $160$ cigarettes when packed as on the left, in $8$ rows of $20$, is $8 \times 2 = 16$ units high.
When packed as on the right, however, there are $4$ rows of $20$ and $4$ rows of $19$, making $156$ cigarettes.
However, the centres of the cross-sections of the cigarettes in row $n$ are $1 + \paren {n - 1} \sqrt 3$ above the bottom of the pack.
Let us arrange a $9$th row of $20$ on top of that $8$th row.
The top of the top row of cigarettes is now $2 + 8 \sqrt 3 \approx 15.9$ units high, and definitely less than $16$ units.
So we can get $16$ more cigarettes in the box if we pack them in the second way.
== Sources ==
* : Solutions: $190$. -- Packing Cigarettes
* : Answers: $326$. Packing Cigarettes
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/191 - Crossing the Ferry

== by : $191$ ==
;Crossing the Ferry
:Six persons, all related, have to cross a river in a small boat that will only hold two.
:Mr. Webster, who had to plan the little affair, had quarrelled with his father-in-law and his son,
::and, I am sorry to say, Mrs. Webster was not on speaking terms with her mother or her daugther-in-law.
:In fact, the relations were so strained that it was not safe to permit any of the belligerents to pass over together
::or to remain together on the same side of the river.
:And to prevent further discord, no man was to be left with two women or two men with three women.
:How are they to perform the feat in the fewest possible crossings?
:No tricks, such as making use of a rope or current, or swimming across, are allowed.
=== Click here for solution ===
== Sources ==
* : Crossing River Problems: $191$. -- Crossing the Ferry
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $448$. Crossing the Ferry
Category:Henry Ernest Dudeney/Modern Puzzles
Category:River Crossing Problems
Henry Ernest Dudeney/Modern Puzzles/191 - Crossing the Ferry/Solution

== by : $191$ ==
;Crossing the Ferry
== Solution ==
First let us establish who we have here, and assign them labels, according to seniority and gender:
:$(1): \quad$ Mrs. Webster's mother and her husband: let them be labelled $A$ and $X$ (Adrienne and Xavier)
:$(2): \quad$ Mr. Webster and his wife Mrs. Webster: let them be labelled $B$ and $Y$ (Bertram and Yvonne)
:$(3): \quad$ Mr. and Mrs. Webster's son and his wife: let them be labelled $C$ and $Z$ (Colin and Zoe)
Hence the rules state:
:$B$ cannot share a boat, or remain together on the same side as either $A$ or $C$.
:$Y$ cannot share a boat, or remain together on the same side as either $X$ or $Z$.
:Men must not be outnumbered by women at any stage.
Here they are at each stage of the crossing, where the bullet indicates where the boat is:
:$\begin {array} {r c| c l} 
A B C \ X Y Z & \bullet & & \\
A C \ X Z & & \bullet & B \ Y \\
A C \ X Y Z & \bullet & & B\\
A C \ Y & & \bullet & B \ X Z \\
A B C \ Y & \bullet & & X Z \\
B \ Y & & \bullet & A C \ X Z \\
B \ Y Z & \bullet & & A C \ X \\
Y & & \bullet & A B C \ X Z \\
B Y & \bullet & & A C \ X Z \\
& & \bullet & A B C \ X Y Z 
\end{array}$
== Sources ==
* : Solutions: $191$. -- Crossing the Ferry
* : Answers: $448$. Crossing the Ferry
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/192 - Missionaries and Cannibals

== by : $192$ ==
;Missionaries and Cannibals
:There is a strange story of three missionaries and three cannibals,
::who had to cross a river in a small boat that would only carry two men at a time.
:Being acquainted with the peculiar appetites of the cannibals,
::the missionaries could never allow their companions to be in a majority on either side of the river.
:Only one of the missionaries and one of the cannibals could row the boat.
:How did they manage to get across?
=== Click here for solution ===
== Sources ==
* : Crossing River Problems: $192$. -- Missionaries and Cannibals
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $449$. Missionaries and Cannibals
Category:Henry Ernest Dudeney/Modern Puzzles
Category:River Crossing Problems
Henry Ernest Dudeney/Modern Puzzles/192 - Missionaries and Cannibals/Historical Note

== Historical Note on by : $192$ - Missionaries and Cannibals ==
 reports on a graphical technique for solving such questions.
He cites a $1966$ article in , and also .
== Sources ==
* : Chapter $1$. One More River to Cross
* 
* : Answers: $449$. Missionaries and Cannibals
Category:Historical Notes
Henry Ernest Dudeney/Modern Puzzles/192 - Missionaries and Cannibals/Solution

== by : $192$ ==
;Missionaries and Cannibals
== Solution ==
Let the missionaries be labelled $M$, $m$ and $m$.
Let the cannibals be labelled $C$, $c$ and $c$.
In both cases the capital letters indicate who is able to row.
Here they are at each stage of the crossing, where the bullet indicates where the boat is:
:$\begin {array} {l | r c| c l} 
 & M m m \ C c c & \bullet & & \\
1 & M m m \ c & & \bullet & C c \\
2 & M m m \ C c & \bullet & & c\\
3 & M m m & & \bullet & C c c \\
4 & M m m \ C & \bullet & & c c \\
5 & m \ C & & \bullet & M m \ c c \\
6 & M m \ C c & \bullet & & m \ c \\
7 & m \ c & & \bullet & M m \ C c \\
8 & M m \ c c & \bullet & & m \ C \\
9 & c c & & \bullet & M m m \ C \\
10 & C c c & \bullet & & M m m \\
11 & c & & \bullet & M m m \ C c \\
12 & C c & \bullet & & M m m \ c \\
13 & & & \bullet & M m m \ C c c \\
\end{array}$
== Historical Note ==
== Sources ==
* : Solutions: $192$. -- Missionaries and Cannibals
* 
* : Answers: $449$. Missionaries and Cannibals
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions