|
PROBLEMS
CONCERNING GAMES.
"The little pleasure of the game."
MATTHEW PRIOR.
Every game lends itself to the propounding of a variety of puzzles. They can
be made, as we have seen, out of the chessboard and the peculiar moves of the
chess pieces. I will now give just a few examples of puzzles with playing cards
and dominoes, and also go out of doors and consider one or two little posers in
the cricket field, at the football match, and the horse race and motor-car
race.
378.—DOMINOES
IN PROGRESSION.
It will be seen that I have played six dominoes, in the illustration, in
accordance with the ordinary rules of the game, 4 against 4, 1 against 1, and so
on, and yet the sum of the spots on the successive dominoes, 4, 5, 6, 7, 8, 9,
are in arithmetical progression; that is, the numbers taken in order have a
common difference of 1. In how many different ways may we play six dominoes,
from an ordinary box of twenty-eight, so that the numbers on them may lie in
arithmetical progression? We must always play from left to right, and numbers in
decreasing arithmetical progression (such as 9, 8, 7, 6, 5, 4) are not
admissible.
Solution 379.—THE
FIVE DOMINOES.
Here is a new little puzzle that is not difficult, but will probably be found
entertaining by my readers. It will be seen that the five dominoes are so
arranged in proper sequence (that is, with 1 against 1, 2 against 2, and so on),
that the total number of pips on the two end dominoes is five, and the sum of
the pips on the three dominoes in the middle is also five. There are just three
other arrangements giving five for the additions. They are: —
| (1—0) |
(0—0) |
(0—2) |
(2—1) |
(1—3) |
| (4—0) |
(0—0) |
(0—2) |
(2—1) |
(1—0) |
| (2—0) |
(0—0) |
(0—1) |
(1—3) |
(3—0) |
Now, how many similar arrangements are there of five dominoes that shall give
six instead of five in the two additions?
Solution 380.—THE
DOMINO FRAME PUZZLE.
It will be seen in the illustration that the full set of twenty-eight
dominoes is arranged in the form of a square frame, with 6 against 6, 2 against
2, blank against blank, and so on, as in the game. It will be found that the
pips in the top row and left-hand column both add up 44. The pips in the other
two sides sum to 59 and 32 respectively. The puzzle is to rearrange the dominoes
in the same form so that all of the four sides shall sum to 44. Remember that
the dominoes must be correctly placed one against another as in the game.
Solution 381.—THE
CARD FRAME PUZZLE.
In the illustration we have a frame constructed from the ten playing cards,
ace to ten of diamonds. The children who made it wanted the pips on all four
sides to add up alike, but they failed in their attempt and gave it up as
impossible. It will be seen that the pips in the top row, the bottom row, and the left-hand
side all add up 14, but the right-hand side sums to 23. Now, what they were
trying to do is quite possible. Can you rearrange the ten cards in the same
formation so that all four sides shall add up alike? Of course they need not add
up 14, but any number you choose to select.
Solution 382.—THE
CROSS OF CARDS.
In this case we use only nine cards—the ace to nine of diamonds. The puzzle
is to arrange them in the form of a cross, exactly in the way shown in the
illustration, so that the pips in the vertical bar and in the horizontal bar add
up alike. In the example given it will be found that both directions add up 23.
What I want to know is, how many different ways are there of rearranging the
cards in order to bring about this result? It will be seen that, without
affecting the solution, we may exchange the 5 with the 6, the 5 with the 7, the
8 with the 3, and so on. Also we may make the horizontal and the vertical bars
change places. But such obvious manipulations as these are not to be regarded as
different solutions. They are all mere variations of one fundamental solution.
Now, how many of these fundamentally different solutions are there? The pips
need not, of course, always add up 23.
Solution 383.—THE
"T" CARD PUZZLE.
An entertaining little puzzle with cards is to take the nine cards of a suit,
from ace to nine inclusive, and arrange them in the form of the letter "T," as
shown in the illustration, so that the pips in the horizontal line shall count
the same as those in the column. In the example given they add up twenty-three
both ways. Now, it is quite easy to get a single correct arrangement. The puzzle
is to discover in just how many different ways it may be done. Though the number
is high, the solution is not really difficult if we attack the puzzle in the
right manner. The reverse way obtained by reflecting the illustration in a
mirror we will not count as different, but all other changes in the relative
positions of the cards will here count. How many different ways are there?
Solution 384.—CARD
TRIANGLES.
Here you pick out the nine cards, ace to nine of diamonds, and arrange them
in the form of a triangle, exactly as shown in the illustration, so that the
pips add up the same on the three sides. In the example given it will be seen
that they sum to 20 on each side, but the particular number is of no importance
so long as it is the same on all three sides. The puzzle is to find out in just how many
different ways this can be done.
If you simply turn the cards round so that one of the other two sides is
nearest to you this will not count as different, for the order will be the same.
Also, if you make the 4, 9, 5 change places with the 7, 3, 8, and at the same
time exchange the 1 and the 6, it will not be different. But if you only change
the 1 and the 6 it will be different, because the order round the triangle is
not the same. This explanation will prevent any doubt arising as to the
conditions.
Solution 385.—"STRAND"
PATIENCE.
The idea for this came to me when considering the game of Patience that I
gave in the Strand Magazine for December, 1910, which has been reprinted
in Ernest Bergholt's Second Book of Patience Games, under the new name of
"King Albert."
Make two piles of cards as follows: 9 D, 8 S, 7 D, 6 S, 5 D, 4 S, 3 D, 2 S, 1
D, and 9 H, 8 C, 7 H, 6 C, 5 H, 4 C, 3 H, 2 C, 1 H, with the 9 of diamonds at
the bottom of one pile and the 9 of hearts at the bottom of the other. The point
is to exchange the spades with the clubs, so that the diamonds and clubs are
still in numerical order in one pile and the hearts and spades in the other.
There are four vacant spaces in addition to the two spaces occupied by the
piles, and any card may be laid on a space, but a card can only be laid on
another of the next higher value—an ace on a two, a two on a three, and so on.
Patience is required to discover the shortest way of doing this. When there are
four vacant spaces you can pile four cards in seven moves, with only three
spaces you can pile them in nine moves, and with two spaces you cannot pile more
than two cards. When you have a grasp of these and similar facts you will be
able to remove a number of cards bodily and write down 7, 9, or whatever the
number of moves may be. The gradual shortening of play is fascinating, and first
attempts are surprisingly lengthy.
Solution 386.—A
TRICK WITH DICE.
Here is a neat little trick with three dice. I ask you to throw the dice
without my seeing them. Then I tell you to multiply the points of the first die
by 2 and add 5; then multiply the result by 5 and add the points of the second
die; then multiply the result by 10 and add the points of the third die. You
then give me the total, and I can at once tell you the points thrown with the
three dice. How do I do it? As an example, if you threw 1, 3, and 6, as in the
illustration, the result you would give me would be 386, from which I could at
once say what you had thrown.
Solution 387.—THE
VILLAGE CRICKET MATCH.
In a cricket match, Dingley Dell v. All Muggleton, the latter had the
first innings. Mr. Dumkins and Mr. Podder were at the wickets, when the wary
Dumkins made a splendid late cut, and Mr. Podder called on him to run. Four runs
were apparently completed, but the vigilant umpires at each end called, "three
short," making six short runs in all. What number did Mr. Dumkins score? When
Dingley Dell took their turn at the wickets their champions were Mr. Luffey and
Mr. Struggles. The latter made a magnificent off-drive, and invited his
colleague to "come along," with the result that the observant spectators
applauded them for what was supposed to have been three sharp runs. But the
umpires declared that there had been two short runs at each end—four in all. To
what extent, if any, did this manœuvre increase Mr. Struggles's total?
Solution 388.—SLOW
CRICKET.
In the recent county match between Wessex and Nincomshire the former team
were at the wickets all day, the last man being put out a few minutes before the
time for drawing stumps. The play was so slow that most of the spectators were
fast asleep, and, on being awakened by one of the officials clearing the ground,
we learnt that two men had been put out leg-before-wicket for a combined score
of 19 runs; four men were caught for a combined score or 17 runs; one man was
run out for a duck's egg; and the others were all bowled for 3 runs each. There
were no extras. We were not told which of the men was the captain, but he made
exactly 15 more than the average of his team. What was the captain's score?
Solution 389.—THE
FOOTBALL PLAYERS.
"It is a glorious game!" an enthusiast was heard to exclaim. "At the close of
last season, of the footballers of my acquaintance four had broken
their left arm, five had broken their right arm, two had the right arm sound,
and three had sound left arms." Can you discover from that statement what is the
smallest number of players that the speaker could be acquainted with?
It does not at all follow that there were as many as fourteen men, because,
for example, two of the men who had broken the left arm might also be the two
who had sound right arms.
Solution 390.—THE
HORSE-RACE PUZZLE.
There are no morals in puzzles. When we are solving the old puzzle of the
captain who, having to throw half his crew overboard in a storm, arranged to
draw lots, but so placed the men that only the Turks were sacrificed, and all
the Christians left on board, we do not stop to discuss the questionable
morality of the proceeding. And when we are dealing with a measuring problem, in
which certain thirsty pilgrims are to make an equitable division of a barrel of
beer, we do not object that, as total abstainers, it is against our conscience
to have anything to do with intoxicating liquor. Therefore I make no apology for
introducing a puzzle that deals with betting.
Three horses—Acorn, Bluebottle, and Capsule—start in a race. The odds are 4
to 1, Acorn; 3 to 1, Bluebottle; 2 to 1, Capsule. Now, how much must I invest on
each horse in order to win £13, no matter which horse comes in first? Supposing,
as an example, that I betted £5 on each horse. Then, if Acorn won, I should
receive £20 (four times £5), and have to pay £5 each for the other two horses;
thereby winning £10. But it will be found that if Bluebottle was first I should
only win £5, and if Capsule won I should gain nothing and lose nothing. This
will make the question perfectly clear to the novice, who, like myself, is not
interested in the calling of the fraternity who profess to be engaged in the
noble task of "improving the breed of horses."
Solution 391.—THE
MOTOR-CAR RACE.
Sometimes a quite simple statement of fact, if worded in an unfamiliar
manner, will cause considerable perplexity. Here is an example, and it will
doubtless puzzle some of my more youthful readers just a little. I happened to
be at a motor-car race at Brooklands, when one spectator said to another, while
a number of cars were whirling round and round the circular track:—
"There's Gogglesmith—that man in the white car!"
"Yes, I see," was the reply; "but how many cars are running in this
race?"
Then came this curious rejoinder:—
"One-third of the cars in front of Gogglesmith added to three-quarters of
those behind him will give you the answer."
Now, can you tell how many cars were running in the race?
Solution
Previous Next
|
