nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Puzzles and Curious Problems/272 - A Madam Problem/Solution

== by : $272$ ==
;A Madam Problem
== Solution ==
:$80$.
== Proof ==
There are $4$ instances of $\text M$, and each is symmetrically disposed.
Hence it is sufficient to count the number of ways you can form $\text {MADAM}$ from one of these and multiply by $4$.
So to proceed.
Starting with any given $\text M$, you can go:
:$1$: Straight down the spoke of the wheel to the $\text M$ opposite
:$2$: Around the edge clockwise to the next $\text M$
:$3$: Around the edge anticlockwise to the next $\text M$
:$4$: Down the spoke to the $\text D$ and back again to the $\text M$ you started at
:$5$: Around the edge clockwise to the $\text D$ and back again to the $\text M$ you started at
:$6$: Around the edge anticlockwise to the $\text D$ and back again to the $\text M$ you started at
:$7$: Down the spoke of the wheel to the $\text D$ and then up the spoke to the left to the $\text M$
:$8$: Down the spoke of the wheel to the $\text D$ and then up the spoke to the right to the $\text M$
:$9$: Around the edge clockwise to the $\text D$, then along the diagonal via the $\text A$, to the adjacent $\text M$
:$10$: Around the edge clockwise to the $\text D$, then along the diagonal to the $\text A$ back to the starting $\text M$
:$11$: Around the edge anticlockwise to the $\text D$, then along the diagonal via the $\text A$, to the adjacent $\text M$
:$12$: Around the edge anticlockwise to the $\text D$, then along the diagonal to the $\text A$ back to the starting $\text M$
:$13$: Down the spoke to the $\text A$, then back up to the clockwise $\text D$ on the rim, and around the rim to the adjacent $\text M$
:$14$: Down the spoke to the $\text A$, then back up to the clockwise $\text D$ on the rim, down the diagonal to the next $\text A$, and then up to the adjacent $\text M$
:$15$: Down the spoke to the $\text A$, then back up to the anticlockwise $\text D$ on the rim, and around the rim to the adjacent $\text M$
:$16$: Down the spoke to the $\text A$, then back up to the anticlockwise $\text D$ on the rim, down the diagonal to the next $\text A$, and then up to the adjacent $\text M$
:$17$: Down the spoke to the $\text A$, then back up to the clockwise $\text D$ on the rim, then back around the rim to the $\text M$ you started at
:$18$: Down the spoke to the $\text A$, then back up to the clockwise $\text D$ on the rim, then back the way you came to the $\text M$ you started at
:$19$: Down the spoke to the $\text A$, then back up to the anticlockwise $\text D$ on the rim, then back around the rim to the $\text M$ you started at
:$20$: Down the spoke to the $\text A$, then back up to the anticlockwise $\text D$ on the rim, then back the way you came to the $\text M$ you started at
That is $20$.
Hence the total number of ways is $80$.
== Sources ==
* : Solutions: $272$. -- A Madam Problem
* : Answers: $429$. A Madam Problem
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/273 - City Luncheons

== by : $273$ ==
;City Luncheons
:The clerks attached to the firm of Pilkins and Popinjay arranged that three of them would lunch together every day at a particular table
::so long as they could avoid the same three men sitting down twice together.
:The same number of clerks of Messrs. Radson, Robson, and Ross decided to do precisely the same, only with four men at a time instead of three.
:On working it out they found that Radson's staff could keep it up exactly three times as many days as their neighbours.
:What is the least number of men there could have been in each staff?
=== Click here for solution ===
== Sources ==
* : Combination and Group Problems: $273$. -- City Luncheons
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $457$. City Luncheons
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/273 - City Luncheons/Solution

== by : $273$ ==
;City Luncheons
== Solution ==
There are $15$ members of staff at each firm.
== Proof ==
Let $n$ be the number of staff at both Pilkins and Popinjay and Radson, Robson, and Ross.
The number of ways to seat $3$ people together out of $n$ is the binomial coefficient $\dbinom n 3 = \dfrac {n \paren {n - 1} \paren {n - 2} } {3 \times 2 \times 1}$.
The number of ways to seat $4$ people together out of $r$ is the binomial coefficient $\dbinom n 4 = \dfrac {n \paren {n - 1} \paren {n - 2} \paren {n - 3} } {4 \times 3 \times 2 \times 1}$.
Hence we want to find $n$ such that:
:$3 \dfrac {n \paren {n - 1} \paren {n - 2} } {3 \times 2 \times 1} = \dfrac {n \paren {n - 1} \paren {n - 2} \paren {n - 3} } {4 \times 3 \times 2 \times 1}$
This simplifies down to:
:$12 = n - 3$
from which the answer is that $n = 15$.
We see that:
:$\dbinom {15} 3 = \dfrac {15 \times 14 \times 13} {3 \times 2 \times 1} = 5 \times 7 \times 13 = 455$
and:
:$\dbinom {15} 4 = \dfrac {15 \times 14 \times 13 \times 12} {4 \times 3 \times 2 \times 1} = 5 \times 7 \times 13 \times 3 = 1365 = 455 \times 3$
== Sources ==
* : Solutions: $273$. -- City Luncheons
* : Answers: $457$. City Luncheons
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/274 - Halfpennies and Tray

== by : $274$ ==
;Halfpennies and Tray
:What is the greatest number of halfpennies that can be laid flat on a circular tray
::(with a small brim to prevent overlapping the edge)
::of exactly $9$ inches in diameter, inside measurements?
:No halfpenny may rest, however slightly, on another.
:Of course, everybody should know that a halfpenny is exactly one inch in diameter.
=== Click here for solution ===
== Sources ==
* : Combination and Group Problems: $274$. -- Halfpennies and Tray
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/274 - Halfpennies and Tray/Solution

== by : $274$ ==
;Halfpennies and Tray
== Solution ==
The tentative answer given by is $64$, which he reports as having been achieved by a certain Dr. G.T. Bennett.
There may be room for further investigation.
== Sources ==
* : Solutions: $274$. -- Halfpennies and Tray
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/275 - The Necklace Problem

== by : $275$ ==
;The Necklace Problem
:How many different necklaces can be made with $8$ beads, where each bead may be either black or white,
::the beads being indistinguishable except by colour?
=== Click here for solution ===
== Sources ==
* : Combination and Group Problems: $275$. -- The Necklace Problem
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $458$. The Necklace Problem
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/275 - The Necklace Problem/Historical Note

== Historical Note on by : $275$ - The Necklace Problem ==
 discusses the general problem, using a recursive formula to give the number of different necklaces with $n$ beads of $m$ different colours, in his $1966$ collection .
== Sources ==
* : $20$: The Calculus of Finite Differences
* : Answers: $458$. The Necklace Problem
Category:Historical Notes
Henry Ernest Dudeney/Puzzles and Curious Problems/276 - An Effervescent Puzzle

== by : $276$ ==
;An Effervescent Puzzle
:In how many ways can the letters in the word $\text {EFFERVESCES}$ be arranged in a line without two $\text E$s ever appearing together?
:Of course, two occurrences of the same letter, such as $\text {F F}$, have no separate identity,
::so that to interchange them will make no difference.
:When the reader has done that, he should try the case where the letters have to be arranged differently in a circle, with no two $\text E$s together.
:We are here, of course, only concerned with their positions on the circumference, and you must always read in a clockwise direction.
=== Click here for solution ===
== Sources ==
* : Combination and Group Problems: $276$. -- An Effervescent Puzzle
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $459$. An Effervescent Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/276 - An Effervescent Puzzle/Solution

== by : $276$ ==
;An Effervescent Puzzle
== Solution ==
The answer to the first question is: $88 \, 200$ ways.
The answer to the second question is: $6300$ ways.
== Proof ==
First we ignore the $\text E$'s and arrange the rest.
Consider the remaining letters as:
:$\text F_1 \text F_2 \text {R V S}_1 \text {C S}_2$
There are $7!$ ways to arrange these letters.
Since the $\text F$'s and $\text S$'s are supposed to be indistingishable, we should divide the number of ways they can swap around to undo double-counting.
Therefore there are:
:$\dfrac {7!}{2! 2!}$
ways to arrange these letters.
Now we insert the $\text E$'s inbetween these letters, or before and after these letters.
There are $8$ available spaces to put these $4$ $\text E$'s in.
There are, therefore:
:$\dfrac {7!}{2! 2!} \times \dbinom 8 4 = 88 \, 200$
ways to arrange these letters in a line where no $\text E$'s are directly next to each other.
For the second question, we first arrange the letters, without $\text E$'s, on a single line.
As above, there are:
:$\dfrac {7!}{2! 2!}$
ways to arrange these letters.
Now we insert the $\text E$'s inbetween these letters.
However, unlike when we did this on a line, the two ends, when wrapped in a circle, are actually the same space, so two $\text E$'s cannot occupy them at the same time.
Out the the $\dbinom 8 4$ possibilities, $\dbinom 6 2$ of them would be in this undesired arrangement.
(We cannot use $\dbinom 7 4$, i.e. consider the ends as the same space, as they are distinct spaces when we consider shifting.)
There are, therefore:
:$\dfrac {7!}{2! 2!} \times \paren {\dbinom 8 4 - \dbinom 6 2} = 69 \, 300$
ways to arrange these letters in a line where no $\text E$'s are directly next to each other.
Now we wrap this line around in a circle.
In a circle, shifting all letters over by one space would not make a difference.
There are $11$ ways to shift the letters for each arrangement (including not shifting).
So, when wrapped around in a circle, there are:
:$\dfrac {69 \, 300} {11} = 6300$
ways to arrange these letters where no $\text E$'s are directly next to each other.
== Sources ==
* : Solutions: $276$. -- An Effervescent Puzzle
* : Answers: $459$. An Effervescent Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/277 - Tessellated Tiles

== by : $277$ ==
;Tessellated Tiles
:Here we have $20$ tiles, all coloured with the same four colours.
:
:The puzzle is to select any $16$ of these tiles that you choose and arrange them in the form of a square,
::always placing same colours together -- white against white, red against red, and so on.
=== Click here for solution ===
== Sources ==
* : Combination and Group Problems: $277$. -- Tessellated Tiles
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $460$. Tessellated Tiles
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/277 - Tessellated Tiles/Solution

== by : $277$ ==
;Tessellated Tiles
== Solution ==
:Discard the first tile in each of the four horizontal rows, and arrange the rest as follows:
:
== Sources ==
* : Solutions: $277$. -- Tessellated Tiles
* : Answers: $460$. Tessellated Tiles
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/278 - The Thirty-Six Letter Puzzle

== by : $278$ ==
;The Thirty-Six Letter Puzzle
:If you try to fill up this square by repeating the letters $A$, $B$, $C$, $D$, $E$, $F$,
::so that no $A$ shall be in a line across, downwards, or diagonally, with another $A$,
::no $B$ with another $B$, no $C$ with another $C$, and so on,
::you will find that it is impossible to get in all the $36$ letters under these conditions.
:
:The puzzle is to place as many letters as possible.
=== Click here for solution ===
== Sources ==
* : Combination and Group Problems: $278$. -- The Thirty-Six Letter Puzzle
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $461$. The Thirty-Six Letter Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/278 - The Thirty-Six Letter Puzzle/Solution

== by : $278$ ==
;The Thirty-Six Letter Puzzle
== Solution ==
:
== Sources ==
* : Solutions: $278$. -- The Thirty-Six Letter Puzzle
* : Answers: $461$. The Thirty-Six Letter Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/279 - Roses, Shamrocks, and Thistles

== by : $279$ ==
;Roses, Shamrocks, and Thistles
:Place the numbers $1$ to $12$ (one number in every design) so that they shall add up to the same sum in the following $7$ different ways --
::viz., each of the two centre columns, each of the two central rows,
::the four roses together, the four shamrocks together, and the four thistles together.
:
=== Click here for solution ===
== Sources ==
* : Combination and Group Problems: $279$. -- Roses, Shamrocks, and Thistles
* : Combinatorial & Topological Problems: Magic Star Puzzles: $400$. Roses, Shamrocks, and Thistles
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/279 - Roses, Shamrocks, and Thistles/Solution

== by : $279$ ==
;Roses, Shamrocks, and Thistles
== Solution ==
:It is clear that the sum obtained in the different ways is $26$.
:Here is one of many arrangements that solve the puzzle.
:
== Sources ==
* : Solutions: $279$. -- Roses, Shamrocks, and Thistles
* : Answers: $400$. Roses, Shamrocks, and Thistles
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/27 - Buying Tobacco

== by : $27$ ==
;Buying Tobacco
:A box of $50$ cigarettes cost the same in shillings and pence as some tobacco bought in pence and shillings.
:The change out of a $10 \shillings$ note was the same as the cost of the cigarettes.
:What did the cigarettes cost?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $27$. -- Buying Tobacco
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/27 - Buying Tobacco/Solution

== by : $27$ ==
;Buying Tobacco
:A box of $50$ cigarettes cost the same in shillings and pence as some tobacco bought in pence and shillings.
:The change out of a $10 \shillings$ note was the same as the cost of the cigarettes.
:What did the cigarettes cost?
== Solution ==
The cigarettes cost $2 \shillings 5 \oldpence$, while the tobacco cost $5 \shillings 2 \oldpence$
== Proof ==
Recall that $1 \shillings$ is worth $12 \oldpence$
Suppose the cigarettes cost $x \shillings y \oldpence$
Then the tobacco would cost $y \shillings x \oldpence$
In pence, the cigarettes cost $\paren {12 x + y} \oldpence$ while the tobacco cost $\paren {12 y + x} \oldpence$
We are given that:
:$10 \times 12 - \paren {12 x + y} - \paren {12 y + x} = \paren {12 x + y}$
which can be rewritten as:
:$25 x + 14 y = 120$
Taking $\bmod 5$, we have:
:$14 y \equiv 0 \pmod 5$
which by Euclid's Lemma implies that $y$ is divisible by $5$.
$y \ne 0$ as $120$ is not divisible by $25$.
$y < 10$ as $14 y > 120$ otherwise.
Therefore $y = 5$, and $x = 2$ follows.
== Sources ==
* : Solutions: $27$. -- Buying Tobacco
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/280 - The Ten Barrels

== by : $280$ ==
;The Ten Barrels
:A merchant had ten barrels of sugar, which he placed in the form of a pyramid, as shown.
:Every barrel bore a different number, except one, which was not marked.
:
:It will be seen that he had accidentally arranged them so that the numbers in the three sides added up alike --
::that is, to $16$.
:Can you arrange them so that the three sides shall sum to the smallest number possible?
:Of course the central barrel (which happens to be $7$ in the diagram) does not come into the count.
=== Click here for solution ===
== Sources ==
* : Combination and Group Problems: $280$. -- The Ten Barrels
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $462$. The Ten Barrels
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/280 - The Ten Barrels/Solution

== by : $280$ ==
;The Ten Barrels
== Solution ==
The two basic solutions are as follows:
: $\qquad$ 
Changing the positions of the side numbers, while not changing the numbers contained on a side, gives another $8$ solutions in each case, not counting reflections or rotations as different.
== Sources ==
* : Solutions: $280$. -- The Ten Barrels
* : Answers: $462$. The Ten Barrels
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/281 - A Match Puzzle

== by : $281$ ==
;A Match Puzzle
:The $16$ squares of a chessboard are enclosed by $16$ matches.
:It is required to place an odd number of matches inside the square so as to enclose $4$ groups of $4$ squares each.
:
:There are $4$ distinct ways to do this, up to reflection and rotation.
=== Click here for solution ===
== Sources ==
* : Combination and Group Problems: $281$. -- A Match Puzzle
* : Match Puzzles: $497$. A Match Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/281 - A Match Puzzle/Solution

== by : $281$ ==
;A Match Puzzle
== Solution ==
In each case $11$ matches are used:
: $\qquad$ 
: $\qquad$ 
== Sources ==
* : Solutions: $281$. -- A Match Puzzle
* : Answers: $497$. A Match Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/282 - The Magic Hexagon

== by : $282$ ==
;The Magic Hexagon
:In the diagram it will be seen how the numbers from $1$ to $19$ are arranged so that all $12$ lines add up to $23$.
:
:Six of the lines are the six sides, and the other six lines radiate from the centre.
:Can you find a different arrangement that will still add up to $23$ in all the $12$ directions?
=== Click here for solution ===
== Sources ==
* : Combination and Group Problems: $282$. -- The Magic Hexagon
* : Combinatorial & Topological Problems: Magic Star Puzzles: $401$. The Magic Hexagon
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/282 - The Magic Hexagon/Solution

== by : $282$ ==
;The Magic Hexagon
== Solution ==
:This is the only correct answer:
:
== Sources ==
* : Solutions: $282$. -- The Magic Hexagon
* : Answers: $401$. The Magic Hexagon
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/283 - Pat in Africa

== by : $283$ ==
;Pat in Africa
:Many years ago, when the world was different, a team of explorers consisting of $5$ men from Western Civilization and $5$ natives
::fell into the hands of a hostile local chief, who, after receiving a number of gifts, consented to let them go,
::but only after half of them had been flogged by the head of the security services.
:The Westerners cruelly hatched a plot to make the flogging fall upon the $5$ natives.
:They were all to be arranged in a circle, and Pat, in position no. $1$, was given a number to count round and round in the clockwise direction.
:In the diagram, $W$ represents a Westerner, and $N$ represents a native.
:
:When that number fell on a man, he was to be taken out for flogging,
::while the counting went on from where it left off until another man fell out,
::and so on until the five men had been selected for punishment.
:If Pat had remembered the number correctly, and had begun at the right man,
::the flogging would all have fallen upon the $5$ natives.
:But Pat was humane at heart, and did not hold with the casual cruelty of his fellows,
::and so deliberately used the wrong number and started at the wrong man,
::with the result that the Westerners all got the flogging instead.
:Can you find:
::$(1)$ the number Pat selected, and the man he started the count at,
::$(2)$ the number he had been expected to use, and the man he was supposed to have begun at?
:The smallest possible number is required in each case.
Note that the inherent racism and colonialism in the language and sentiments of this puzzle as it originally appeared have been softened for modern audiences.
=== Click here for solution ===
== Sources ==
* : Combination and Group Problems: $283$. -- Pat in Africa
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $248$. Pat in Africa
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/283 - Pat in Africa/Solution

== by : $283$ ==
;Pat in Africa
== Solution ==
Pat was expected to start at $9$ and count $29$.
This would have counted out all the natives.
However, instead he started the count at $1$ (himself) and counted $11$.
== Proof ==
We are given the order of the men:
:$W_1, N_2, W_3, N_4, N_5, W_6, N_7, W_8, N_9, W_{10}$
Counting $11$ from $W_1$ (which is Pat) takes out $W_1$ himself, leaving:
:$N_2, W_3, N_4, N_5, W_6, N_7, W_8, N_9, W_{10}$
Counting $11$ from $N_2$ takes out $W_3$, leaving:
:$N_2, N_4, N_5, W_6, N_7, W_8, N_9, W_{10}$
Counting $11$ from $N_4$ takes out $W_6$, leaving:
:$N_2, N_4, N_5, N_7, W_8, N_9, W_{10}$
Counting $11$ from $N_7$ takes out $W_{10}$, leaving:
:$N_2, N_4, N_5, N_7, W_8, N_9$
Counting $11$ from $N_2$ takes out $W_8$, leaving:
:$N_2, N_4, N_5, N_7, N_9$
So all Westerners are taken out for a flogging, while the natives escape.
Let us count $29$ starting at $N_9$.
Counting $29$ from $N_9$ takes out $N_7$, leaving:
:$W_1, N_2, W_3, N_4, N_5, W_6, W_8, N_9, W_{10}$
Counting $29$ from $W_8$ takes out $N_9$, leaving:
:$W_1, N_2, W_3, N_4, N_5, W_6, W_8, W_{10}$
Counting $29$ from $W_{10}$ takes out $N_4$, leaving:
:$W_1, N_2, W_3, N_5, W_6, W_8, W_{10}$
Counting $29$ from $N_5$ takes out $N_5$, leaving:
:$W_1, N_2, W_3, W_6, W_8, W_{10}$
Counting $29$ from $W_6$ takes out $N_2$, leaving:
:$W_1, W_3, W_6, W_8, W_{10}$
So in this case, all the natives are flogged and the Westerners escape.
== Sources ==
* : Solutions: $283$. -- Pat in Africa
* : Answers: $248$. Pat in Africa
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:Josephus Problem
Henry Ernest Dudeney/Puzzles and Curious Problems/284 - Lamp Signalling

== 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?
=== Click here for solution ===
== Sources ==
* : Combination and Group Problems: $284$. -- Lamp Signalling
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $463$. Lamp Signalling
Category:Henry Ernest Dudeney/Puzzles and Curious Problems