Henry Ernest Dudeney/Modern Puzzles/Problems Concerning Games

Henry Ernest Dudeney: Modern Puzzles: Problems Concerning Games

$193$ - A Domino Square

Select any eighteen dominoes you please from an ordinary box,

and arrange them any way you like in a square so that no number shall be repeated in any row or any column.

The example given is imperfect, for it will be seen that though no number is repeated in any one of the columns yet three of the rows break the condition.

There are two $4$'s and two blanks in the first row, two $5$s and two $6$'s in the third row, and two $3$'s in the fourth row.

Can you form an arrangement without such errors?

Blank counts as a number.

$194$ - A Domino Star

Place the $28$ dominoes, as shown in the diagram, so as to form a star with alternate rays of $4$ and $3$ dominoes.

Every ray must contain $21$ pips (in the example only one ray contains this number)

and the central numbers must be $1$, $2$, $3$, $4$, $5$, $6$, and two blanks, as at present, and these may be in any order.

In every ray the dominoes must be placed according to the ordinary rule, $6$ against $6$, blank against blank, and so on.

$195$ - Domino Groups

If the dominoes are laid out in the manner shown in the diagram and I then break the line into $4$ lengths of $7$ dominoes each,

it will be found that the sum of the pips in the first group is $49$, in the second $34$, in the third $46$, and in the fourth $39$.

Now I want to play them out so that all the four groups of seven when the line is broken shall contain the same number of pips.

Can you find a way of doing it?

$196$ - Les Quadrilles

It is required to arrange a complete set of $28$ dominoes so as to form the figure shown in the diagram,

with all the numbers forming a series of squares.

Thus, in the upper two rows we have a square of blanks, and a square of four $3$'s, and a square of $4$'s, and a square of $1$'s and so on.

This is, in fact, a perfect solution under the conditions usually imposed,

but what I now ask for is an arrangement with no blanks anywhere on the outer edge.

At present every number from blank to $6$ inclusive will be found somewhere on the margin.

Can you construct an arrangement with all the blanks inside?

$197$ - A Puzzle with Cards

Take from the pack the $13$ cards forming the suit of diamonds and arrange them in this order face downwards with the $3$ at the top and $5$ at the bottom:

$3$, $8$, $7$, ace, queen, $6$, $4$, $2$, jack, king, $10$, $9$, $5$.

Now play them out in a row on the table in this way.

As you spell "ace" transfer for each letter a card from the top to the bottom of the pack -- A-C-E -- and play the fourth card on to the table.

Then spell T-W-O, while transferring three more cards to the bottom, and place the next card on the table.

Then spell T-H-R-E-E, while transferring five to the bottom, and so on until all are laid out in a row,

and you will find that they will be all in regular order.

Of course, you will spell out the knave as J-A-C-K.

Can you arrange the whole pack so that they will play out correctly in order,

first all the diamonds, then the hearts, then the spades, and lastly the clubs?

$198$ - A Card Trick

Take an ordinary pack of playing-cards and regard all the court cards as tens.

Now, look at the top card -- say it is a seven -- place it on the table face downwards and play more cards on top of it, counting up to twelve.

Thus, the bottom card being seven, the next will be eight, the next nine, and so on, making six cards in that pile.

Then look again at the top card of pack -- say it is a queen -- then count $10$, $11$, $12$ (three cards in all) and complete the second pile.

Continue this, always counting up to twelve, and if at last you have not put sufficient cards to complete a pile, put these apart.

Now, if I am told how many piles have been made and how many unused cards remain over,

I can at once tell you the sum of all the bottom cards in the piles.

I simply multiply by $13$ the number of piles less $4$, and add the number of cards left over.

Thus, if there were $6$ piles and $4$ cards over, then $13$ times $2$ (i.e. $6$ less $4$) added to $5$ equals $31$, the sum of the bottom cards.

Why is this?

This is the question.

$199$ - Golf Competition Puzzle

I was asked to construct some schedules for players in American golf competitions.

The conditions are:

$(1)$ Every player plays every player once, and once only.

$(2)$ There are half as many links as players, and every player plays twice on every links except one, on which he plays but once.

$(3)$ All the players play simultaneously in every round, and the last round is the one in which every player is playing on a links for the first time.

I have written out schedules for a long series of even numbers of players up to $26$,

but the problem is too difficult for this page except for in its most simple form -- for six players.

Can the reader, calling the players $A$, $B$, $C$, $D$, $E$, and $F$,

and pairing these in all possible ways, such as $AB$, $CD$, $EF$, $AF$, $BD$, $CE$, etc.,

complete this table for six players?

$\qquad \begin {array} {r | c |} \text {Rounds} & 1 & 2 & 3 & 4 & 5 \\ \hline \text{$1$st Links} & & & & & \\ \hline \text{$2$nd Links} & & & & & \\ \hline \text{$3$rd Links} & & & & & \\ \hline \end{array}$

$200$ - Cricket Scores

In a country match Great Muddleton, who went in first, made a score of which they were proud.

Then Little Wurzleford had their innings and scored a quarter less.

The Muddletonians in their next attempt made a quarter less than their opponents,

who, curiously enough, were only rewarded on their second attempt by a quarter less than their last score.

Thus, every innings was a quarter less fruitful in runs than the one that preceded it.

Yet the Muddletonians won the match by $50$ runs.

Can you give the exact score for every one of the four innings?

$201$ - Football Results

The following table appeared in a newspaper:

$\qquad \begin {array} {r | cccc |cc|c} & \text {Played} & \text {Won} & \text {Lost} & \text {Drawn} & \text {Goals For} & \text {Against} & \text {Points} \\ \hline \text{Scotland} & 3 & 3 & 0 & 0 & 7 & 1 & 6 \\ \hline \text{England} & 3 & 1 & 1 & 1 & 2 & 3 & 3 \\ \hline \text{Wales} & 3 & 1 & 1 & 1 & 3 & 3 & 3 \\ \hline \text{Ireland} & 3 & 0 & 3 & 0 & 1 & 6 & 0 \\ \hline \end{array}$

It is known that Scotland had beaten England $3 - 0$.

It is now possible to find the scores in the other five matches from the table.

How many goals were won, drawn, or lost by each side in every match?