nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Puzzles and Curious Problems/297 - Mixing the Wine

== by : $297$ ==
;Mixing the Wine
:A glass is one-third full of wine,
::and another glass, with equal capacity, is one-fourth full of wine.
:Each is filled with water and their contents mixed in a jug.
:Half of the mixture is poured into one of the glasses.
:What proportion of this is wine and what part water?
=== Click here for solution ===
== Sources ==
* : Magic Square, Measuring, Weighing, and Packing Problems: $297$. -- Mixing the Wine
* : Combinatorial & Topological Problems: Liquid Pouring Puzzles: $408$. Mixing the Wine
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/297 - Mixing the Wine/Solution

== by : $297$ ==
;Mixing the Wine
== Solution ==
:$7$ parts wine to $17$ parts water.
== Proof ==
Let $c$ be the capacity of one glass.
After they have been filled with water:
:the first one contains $\dfrac {4 c} {12}$ of wine
:the second one contains $\dfrac {3 c} {12}$ of wine.
In the jug together, there is $\dfrac {7 c} {12}$ of wine out of a total of $2 c$ of liquid.
When poured into one of the glasses, there is then $\dfrac 1 2 \times \dfrac {7 c} {12} = \dfrac {7 c} {24}$ of wine.
As this is in a container of capacity $c$, the ratio of wine to water is $\dfrac {7 c} {24} \div \dfrac {24 - 7 c} {24} = \dfrac 7 {17}$.
== Sources ==
* : Solutions: $297$. -- Mixing the Wine
* : Answers: $408$. Mixing the Wine
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/298 - The Stolen Balsam

== by : $298$ ==
;The Stolen Balsam
:Three men robbed a gentleman of a vase containing $24$ ounces of balsam.
:While running away, they met in the forest a glass seller, of whom, in a great hurry, they purchased three vessels.
:On reaching a place of safety they wished to divide the booty,
::but they found that their vessels contained $5$, $11$, and $13$ ounces respectively.
:How could they divide the balsam into equal portions?
=== Click here for solution ===
== Sources ==
* : Magic Square, Measuring, Weighing, and Packing Problems: $298$. -- The Stolen Balsam
* : Combinatorial & Topological Problems: Liquid Pouring Puzzles: $409$. The Stolen Balsam
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/298 - The Stolen Balsam
Henry Ernest Dudeney/Puzzles and Curious Problems/298 - The Stolen Balsam/Historical Note

== Historical Note on by : $298$ - The Stolen Balsam ==
The $6$-transaction solution was the one provided by .
The improved $5$-transaction solution was discovered by .
== Sources ==
* : Answers: $409$. The Stolen Balsam
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/298 - The Stolen Balsam
Category:Historical Notes
Henry Ernest Dudeney/Puzzles and Curious Problems/298 - The Stolen Balsam/Solution

== by : $298$ ==
;The Stolen Balsam
== Solution ==
=== Solution $1$ ===
=== Solution $2$ ===
== Historical Note ==
== Sources ==
* : Solutions: $298$. -- The Stolen Balsam
* : Answers: $409$. The Stolen Balsam
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/298 - The Stolen Balsam
Henry Ernest Dudeney/Puzzles and Curious Problems/298 - The Stolen Balsam/Solution 1

== by : $298$ ==
;The Stolen Balsam
== Solution ==
One of several ways is as follows:
:$\begin {array} {r|cccc}
& \mathrm {oz.} & \mathrm {oz.} & \mathrm {oz.} & \mathrm {oz.} \\ \hline 
\text {The vessels can hold} & 24 & 13 & 11 & 5 \\
\text {Their contents at the start} & 24 & 0 & 0 & 0 \\
\text {Make their contents} & 0 & 8 & 11 & 5 \\
\text {Then} & 16 & 8 & 0 & 0 \\
\text {Then} & 16 & 0 & 8 & 0 \\
\text {Then} & 3 & 13 & 8 & 0 \\
\text {Then} & 3 & 8 & 8 & 5 \\
\text {Then} & 8 & 8 & 8 & 0 \\
\end {array}$
== Historical Note ==
== Sources ==
* : Solutions: $298$. -- The Stolen Balsam
* : Answers: $409$. The Stolen Balsam
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/298 - The Stolen Balsam
Henry Ernest Dudeney/Puzzles and Curious Problems/298 - The Stolen Balsam/Solution 2

== by : $298$ ==
;The Stolen Balsam
== Solution ==
A solution using only $5$ transactions is as follows:
:$\begin {array} {r|cccc}
& \mathrm {oz.} & \mathrm {oz.} & \mathrm {oz.} & \mathrm {oz.} \\ \hline 
\text {The vessels can hold} & 24 & 13 & 11 & 5 \\
\text {Their contents at the start} & 24 & 0 & 0 & 0 \\
\text {Make their contents} & 8 & 0 & 11 & 5 \\
\text {Then} & 8 & 11 & 0 & 5 \\
\text {Then} & 8 & 13 & 3 & 0 \\
\text {Then} & 8 & 8 & 3 & 5 \\
\text {Then} & 8 & 8 & 8 & 0 \\
\end {array}$
== Historical Note ==
== Sources ==
* : Answers: $409$. The Stolen Balsam
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/298 - The Stolen Balsam
Henry Ernest Dudeney/Puzzles and Curious Problems/299 - The Weight of the Fish

== by : $299$ ==
;The Weight of the Fish
:A man caught a fish.
:The tail weighed $9$ ounces.
:The head weighed as much as the tail and half the body,
::and the body weighed as much as the head and tail together.
:What is the weight of the fish?
=== Click here for solution ===
== Sources ==
* : Magic Square, Measuring, Weighing, and Packing Problems: $299$. -- The Weight of the Fish
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $250$. The Weight of the Fish
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/299 - The Weight of the Fish/Solution

== by : $299$ ==
;The Weight of the Fish
== Solution ==
:$72$ ounces, or $4 \tfrac 1 2$ pounds.
== Proof ==
Let $t$, $h$ and $b$ denote the weights in ounces of tail, head and body respectively.
We have:
The weight in pounds can be calculated from the conversion factor of $16$ ounces to the pound.
== Sources ==
* : Solutions: $299$. -- The Weight of the Fish
* : Answers: $250$. The Weight of the Fish
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/29 - The Shopkeeper's Puzzle

== by : $29$ ==
;The Shopkeeper's Puzzle
:A shopkeeper uses a code word where each letter stands for the digits from $0$ to $9$.
:What is the code used to encode this addition sum?
 GAUNT
+ OILER
 ------
 RGUOEI
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $29$. -- The Shopkeeper's Puzzle
* : Arithmetical and Algebraical Problems: Cryptarithm Puzzles: $154$. The Shopkeeper's Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Cryptarithms
Henry Ernest Dudeney/Puzzles and Curious Problems/29 - The Shopkeeper's Puzzle/Solution

== by : $29$ ==
;The Shopkeeper's Puzzle
:A shopkeeper uses a code word where each letter stands for the digits from $0$ to $9$.
:What is the code used to encode this addition sum?
 GAUNT
+ OILER
 ------
 RGUOEI
== Solution ==
REGULATION
== Proof ==
As this is an addition sum with a $6$ digit sum, the first digit of the sum is $1$.
Hence $R = 1$.
Because $N + E = E$, either $N = 0$ or $N = 9$ and there was a carry from the right.
But the only way there can be a carry from the right is from $T + R = I$, where $R = 1$.
Hence $T$ would have to be $9$.
But that would make both $T = 9$ and $N = 9$, which cannot happen.
Hence $N = 0$.
We have that $G + O + 1 = G$, so $O = 9$.
Thus we have so far:
 1 2 3 4 5 6 7 8 9 0
--------------------
 R O N
and it is apparent that the only anagram of GAUNT OILER with this pattern is REGULATION.
Hence we have:
 1 2 3 4 5 6 7 8 9 0
--------------------
 R E G U L A T I O N
and the completed cryptarithm:
 36407
+ 98521
 ------
 134928
== Sources ==
* : Solutions: $29$. -- The Shopkeeper's Puzzle
* : Answers: $154$. The Shopkeeper's Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/2 - A Legacy Puzzle

== by : $2$ ==
;A Legacy Puzzle
:A man left legacies to his three sons and to a hospital, amounting in all to $\pounds 1,320$.
:If he had left the hospital legacy also to his first son, that son would have received as much as the other two sons together.
:If he had left it to his second son, that son would have received twice as much as the other two sons together.
:If he had left the hospital legacy to his third son, he would have received then thrice as much as the first son and second son together.
:Find the amount of each legacy.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $2$. -- A Legacy Puzzle
* : Arithmetical and Algebraical Problems: Money Puzzles: $13$. A Legacy Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/2 - A Legacy Puzzle
Henry Ernest Dudeney/Puzzles and Curious Problems/2 - A Legacy Puzzle/Solution

== by : $2$ ==
;A Legacy Puzzle
:A man left legacies to his three sons and to a hospital, amounting in all to $\pounds 1,320$.
:If he had left the hospital legacy also to his first son, that son would have received as much as the other two sons together.
:If he had left it to his second son, that son would have received twice as much as the other two sons together.
:If he had left the hospital legacy to his third son, he would have received then thrice as much as the first son and second son together.
:Find the amount of each legacy.
== Solution ==
:The legacy to the first son was $\pounds 55$,
::to the second son $\pounds 275$,
::to the third son $\pounds 385$,
::and to the hospital $\pounds 605$,
::making $\pounds 1320$ in all.
== Proof ==
Let $a$, $b$ and $c$ pounds be the legacies of the first, second and third sons respectively.
Let $h$ be the legacy of the hospital.
We have:
We set up this system of linear simultaneous equations in matrix form as:
$\quad \begin {pmatrix}
1 & 1 & 1 & 1 \\
1 & -1 & -1 & 1 \\
-2 & 1 & -2 & 1 \\
-3 & -3 & 1 & 1 \\
\end {pmatrix} \begin {pmatrix} a \\ b \\ c \\ h \end {pmatrix} = \begin {pmatrix} 1320 \\ 0 \\ 0 \\ 0 \end {pmatrix}$
It remains to solve this matrix equation.
In reduced echelon form, this gives:
from which the legacies can be read off directly.
== Sources ==
* : Solutions: $2$. -- A Legacy Puzzle
* : Answers: $13$. A Legacy Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/2 - A Legacy Puzzle
Henry Ernest Dudeney/Puzzles and Curious Problems/2 - A Legacy Puzzle/Working

== Working for by : $2$ -- A Legacy Puzzle ==
The simultaneous equations in matrix form:
$\quad \begin {pmatrix}
1 & 1 & 1 & 1 \\
1 & -1 & -1 & 1 \\
-2 & 1 & -2 & 1 \\
-3 & -3 & 1 & 1 \\
\end {pmatrix} \begin {pmatrix} a \\ b \\ c \\ h \end {pmatrix} = \begin {pmatrix} 1320 \\ 0 \\ 0 \\ 0 \end {pmatrix}$
when converted to reduced echelon form, gives:
$\quad \begin {pmatrix} 
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end {pmatrix} \begin {pmatrix} a \\ b \\ c \\ h \end {pmatrix} = \begin {pmatrix} 55 \\ 275 \\ 385 \\ 605 \\ \end {pmatrix}$
== Proof ==
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/2 - A Legacy Puzzle
Henry Ernest Dudeney/Puzzles and Curious Problems/300 - Fresh Fruits

== by : $300$ ==
;Fresh Fruits
:Some fresh fruit was being weighed for some domestic purpose.
:It was found that the apples, pears and plums exactly balanced each other as follows:
::One pear and three apples weigh the same as $10$ plums;
::and one apple and six plums weigh the same as one pear.
:How many plums alone would weigh the same as one pear?
=== Click here for solution ===
== Sources ==
* : Magic Square, Measuring, Weighing, and Packing Problems: $300$. -- Fresh Fruits
* : Arithmetical and Algebraical Problems: Weight Puzzles: $100$. Fresh Fruits
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/300 - Fresh Fruits/Solution

== by : $300$ ==
;Fresh Fruits
== Solution ==
One pear weighs the same as $7$ plums.
== Proof ==
Let $a$, $p$ and $m$ denote the weights of one apple, one pear and one plum respectively.
We have:
== Sources ==
* : Solutions: $300$. -- Fresh Fruits
* : Answers: $100$. Fresh Fruits
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/301 - Weighing the Tea

== by : $301$ ==
;Weighing the Tea
:A grocer proposed to put up $20$ pounds of China tea into $2$-pound packets,
::but the weights had been misplaced by somebody, and he could only find the $5$-pound and the $9$-pound weights.
:What is the quickest way for him to do the business?
:We will say at once that only nine weighings are really necessary.
=== Click here for solution ===
== Sources ==
* : Magic Square, Measuring, Weighing, and Packing Problems: $301$. -- Weighing the Tea
* : Arithmetical and Algebraical Problems: Weight Puzzles: $101$. Weighing the Tea
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/301 - Weighing the Tea/Solution

== by : $301$ ==
;Weighing the Tea
== Solution ==
Having weighed a quantity of tea, we can then use that as a further weight to make further weighings.
We will refer to each weight merely by its numerical weight in pounds for simplicity.
:$(1): \quad$ With the $5$-pound weight and the $9$-pound in opposite pans, weigh a $4$-pound weight of tea.
:$(2): \quad$ With the $4$-pound weight of tea, weigh another $4$-pound weight of tea.
:$(3): \quad$ With the $4$-pound weight of tea, weigh a third $4$-pound weight of tea.
:$(4): \quad$ With the $4$-pound weight of tea, weigh a fourth $4$-pound weight of tea.
::::Note that the remaining tea in the chest will also make a $4$-pound weight of tea.
:$(5)$ to $(9): \quad$ Divide each of the $4$-pound weights of tea equally between the two pans of the balance.
== Sources ==
* : Solutions: $301$. -- Weighing the Tea
* : Answers: $101$. Weighing the Tea
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/302 - Delivering the Milk

== by : $302$ ==
;Delivering the Milk
:A milkman one morning was driving to his dairy with two $10$-gallon cans full of milk,
::when he was stopped by two countrywomen, who implored him to sell them a quart of milk each.
:Mrs. Green had a jug holding exactly $5$ pints, and Mrs. Brown a jug holding exactly $4$ pints,
::but the milkman had no measure whatsoever.
:How did he manage to put an exact quart into each of the jugs?
:It was the second quart that gave all the difficulty.
:But he contrived to do it in as few as nine transactions --
::and by a "transaction" we mean the pouring from a can into a jug, or from one jug to another, or from a jug back to the can.
:How did he do it?
=== Click here for solution ===
== Sources ==
* : Magic Square, Measuring, Weighing, and Packing Problems: $302$. -- Delivering the Milk
* : Combinatorial & Topological Problems: Liquid Pouring Puzzles: $410$. Delivering the Milk
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/302 - Delivering the Milk/Solution

== by : $302$ ==
;Delivering the Milk
== Solution ==
To simplify understanding, let us convert all measurements to pints.
Let $C_1$ and $C_2$ denote the two $10$-gallon cans, each of which hold $80$ pints.
Let $J_5$ and $J_4$ denote the $5$ pint and $4$ pint jugs repectively.
Proceed as follows:
:$(1): \quad$ Fill $J_5$ from $C_1$.
:$(2): \quad$ Fill $J_4$ from $J_5$.
:$(3): \quad$ Empty $J_4$ back into $C_1$.
:$(4): \quad$ Empty $J_5$ into $J_4$.
:$(5): \quad$ Fill $J_5$ from $C_1$.
:$(6): \quad$ Fill $J_4$ from $J_5$.
:$(7): \quad$ Empty $J_4$ back into $C_1$.
At this stage, $J_5$ contains the first required quart, $C_2$ is still full, $J_4$ is empty, and $C_1$ is $1$ quart short of full.
:$(8): \quad$ Fill $J_4$ from $C_2$.
:$(9): \quad$ Fill $C_1$ from $J_4$.
As $C_1$ was $1$ quart, that is $2$ pints, short of full, $J_4$ now holds the second required $2$ quarts.
== Sources ==
* : Solutions: $302$. -- Delivering the Milk
* : Answers: $410$. Delivering the Milk
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/303 - Crossing the River

== by : $303$ ==
;Crossing the River
:During the Turkish stampede in Thrace, a small detachment found itself confronted by a wide and deep river.
:However, they discovered a boat in which two children were rowing about.
:It was so small that it would only carry the two children, or one grown up person.
:How did the officer get himself and his $357$ soldiers across the river and leave the two children finally in joint possession of their boat?
:And how many times need the boat pass from shore to shore?
=== Click here for solution ===
== Sources ==
* : Crossing River Problem, and Problems Concerning Games and Puzzle Games: $303$. -- Crossing the River
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $450$. Crossing the River
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:River Crossing Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/303 - Crossing the River/Solution

== by : $303$ ==
;Crossing the River
== Solution ==
In order to get one soldier across the river, proceed as follows:
:Two children both cross the river in the boat.
:One child returns with the boat.
:One soldier crosses in the boat.
:The child on the opposite shore returns with the boat.
Hence we are in the same situation as before, but with one soldier on the opposite bank.
There is a total of $358$ people to get across the river.
It will therefore take $358 \times 4 = 1432$ crossings to get all the men over, leaving the children together with their boat.
== Also see ==
* $19$ - De Viro et Muliere Ponderantibus Plaustrum from by , of which this is a variant.
== Sources ==
* : Solutions: $303$. -- Crossing the River
* : Answers: $450$. Crossing the River
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/304 - Grasshoppers' Quadrille

== by : $304$ ==
;Grasshoppers' Quadrille
:It is required to make the white men change places with the black men in the fewest possible moves.
:There is no diagonal play, nor are there captures.
:
:The white men can only move to the right or downwards, and the black men to the left or upwards,
::but they may leap over one of the opposite colour, as in draughts.
:It is quite easy when once you have hit on the method of solution.
=== Click here for solution ===
== Sources ==
* : Crossing River Problem, and Problems Concerning Games and Puzzle Games: $304$. -- Grasshoppers' Quadrille
* : Geometrical Problems: Moving Counter Puzzles: $381$. Grasshoppers' Quadrille
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/304 - Grasshoppers' Quadrille/Solution

== by : $304$ ==
;Grasshoppers' Quadrille
== Solution ==
The counters can be exchanged in $120$ moves.
== Proof ==
Consider the central column containing $3$ white and $3$ black counters.
These can be made to change places in $15$ moves.
Number the $7$ squares downwards from $1$ to $7$.
Now play:
:$3 - 4$, $5 - 3$, $6 - 5$, $4 - 6$, $2 - 4$, $1 - 2$, $3 - 1$, $5 - 3$, $7 - 5$, $6 - 7$, $4 - 6$, $2 - 4$, $3 - 2$, $5 - 3$, $4 - 5$
of which $6$ are simple moves, and $9$ are jumps.
Now there are $7$ horizontal rows of $3$ white and $3$ black counters, if we exclude that central column.
Each of these can be similarly interchanged in $15$ moves.
For each row, this can be done at any stage where there is a vacant space in the central column.
For example, the central row can be done straight away, as the central space starts empty.
So, as there are $7$ rows and $1$ column, the counters can be exchanged in $8 \times 15 = 120$ moves.
== Sources ==
* : Solutions: $304$. -- Grasshoppers' Quadrille
* : Answers: $381$. Grasshoppers' Quadrille
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/305 - Domino Frames

== by : $305$ ==
;Domino Frames
:Take an ordinary set of $28$ dominoes and return double $3$, double $4$, double $5$, and double $6$ to the box as not wanted.
:Now, with the remainder form three square frames, in the manner shown, so that the pips in every side shall add up alike.
:
:In the example given the sides sum to $15$.
:If this were to stand, the sides of the other two frames must also sum to $15$.
:But you can take any number you like, and it will be seen that it is not required to place $6$ against $6$, $5$ against $5$, and so on, as in play.
=== Click here for solution ===
== Sources ==
* : Crossing River Problem, and Problems Concerning Games and Puzzle Games: $305$. -- Domino Frames
* : Domino Puzzles: $486$. Domino Frames
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Dominoes
Henry Ernest Dudeney/Puzzles and Curious Problems/305 - Domino Frames/Solution

== by : $305$ ==
;Domino Frames
== Solution ==
:The three diagrams show a solution.
:
:
:
:The sum of all the pips is $132$.
:One-third of this is $44$.
:First divide the dominoes into any three groups of $44$ pips each.
:Then, if we decide to try $12$ for the sum of the sides, $4$ times $12$ being $4$ more than $44$,
::we must arrange in every case that the four corners in a frame shall sum to $4$.
:The rest is done by trial and exchanges from one group to another of dominoes containing an equal number of pips.
== Sources ==
* : Solutions: $305$. -- Domino Frames
* : Answers: $486$. Domino Frames
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/306 - A Puzzle in Billiards

== by : $306$ ==
;A Puzzle in Billiards
:Alfred Addlestone can give Benjamin Bounce $20$ points in $100$, and beat him;
::Bounce can give Charlie Cruikshank $25$ points in $100$, and beat him.
:Now, how many points can Addlestone give Cruikshank in order to beat him in a game of $200$ up?
:Of course we assume that the players play constantly with the same relative skill.
=== Click here for solution ===
== Sources ==
* : Crossing River Problem, and Problems Concerning Games and Puzzle Games: $306$. -- A Puzzle in Billiards
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/306 - A Puzzle in Billiards/Solution

== by : $306$ ==
;A Puzzle in Billiards
== Solution ==
:$82$ points.
== Proof ==
Let $A$, $B$ and $C$ denote Addlestone, Bounce and Cruikshank respectively
From the definition of the problem, $A$ can score $100$ while $B$ can score only $79$.
Similarly, $B$ can score $100$ while $C$ can score only $74$.
Multiply $79$ by $74$, double, and divide by $100$, and you get $116.92$.
So $C$ can score $117$ (there are no fractional points) while $A$ can make $200$.
Therefore $A$ can give $C$ $82$ points and win.
There is a case for saying that $A$ can give $C$ $83$ points and still win -- it depends on how fractional points are interpreted.
However, it is certain that $A$ can win having given $C$ those $82$.
== Sources ==
* : Solutions: $306$. -- A Puzzle in Billiards
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/307 - Scoring at Billiards

== by : $307$ ==
;Scoring at Billiards
:What is the highest score that you can make in two consecutive shots at billiards?
=== Click here for solution ===
== Sources ==
* : Crossing River Problem, and Problems Concerning Games and Puzzle Games: $307$. -- Scoring at Billiards
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/307 - Scoring at Billiards/Solution

== by : $307$ ==
;Scoring at Billiards
== Solution ==
:$18$.
== Proof ==
:$(1): \quad$ Cannon $(2)$, pot the red $(3)$, and in-off red $(3)$, leaving white up, total $8$.
:$(2): \quad$ Cannon $(2)$, pot the white $(2)$, pot the red $(3)$, and in-off red $(3)$, total $10$.
== Sources ==
* : Solutions: $307$. -- Scoring at Billiards
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions