Dudeney’s Christmas puzzles


For years on end the famous British puzzle maker Henry Ernest Dudeney published puzzles in his weekly and monthly columns. Several of these have a Christmas theme, most of which I brought here together.

I know that some people claim other puzzles as Dudeneys Christmas puzzles, and I may or may not elaborate on that later. These have been selected because they have the word “Christmas” in them.

In his book “The Canterbury Puzzles” we find:
1) The Christmas Geese**/*****
Squire Hembrow, from Weston Zoyland—wherever that may be—proposed the following little arithmetical puzzle, from which it is probable that several somewhat similar modern ones have been derived: Farmer Rouse sent his man to market with a flock of geese, telling him that he might sell all or any of them, as he considered best, for he was sure the man knew how to make a good bargain. This is the report that Jabez made, though I have taken it out of the old Somerset dialect, which might puzzle some readers in a way not desired. “Well, first of all I sold Mr. Jasper Tyler half of the flock and half a goose over; then I sold Farmer Avent a third of what remained and a third of a goose over; then I sold Widow Foster a quarter of what remained and three-quarters of a goose over; and as I was coming home, whom should I meet but Ned Collier: so we had a mug of cider together at the Barley Mow, where I sold him exactly a fifth of what I had left, and gave him a fifth of a goose over for the missus. These nineteen that I have brought back I couldn’t get rid of at any price.” Now, how many geese did Farmer Rouse send to market? My humane readers may be relieved to know that no goose was divided or put to any inconvenience whatever by the sales.

You can check your solution here

2) Tasting the Plum Puddings**/*****

“Everybody, as I suppose, knows well that the number of different Christmas plum puddings that you taste will bring you the same number of lucky days in the new year. One of the guests (and his name has escaped my memory) brought with him a sheet of paper on which were drawn sixty-four puddings, and he said the puzzle was an allegory of a sort, and he intended to show how we might manage our pudding-tasting with as much dispatch as possible.” I fail to fully understand this fanciful and rather overstrained view of the puzzle. But it would appear that the puddings were arranged regularly, as I have shown them in the illustration, and that to strike out a pudding was to indicate that it had been duly tasted. You have simply to put the point of your pencil on the pudding in the top corner, bearing a sprig of holly, and strike out all the sixty-four puddings through their centres in twenty-one straight strokes. You can go up or down or horizontally, but not diagonally or obliquely; and you must never strike out a pudding twice, as that would imply a second and unnecessary tasting of those indigestible dainties. But the peculiar part of the thing is that you are required to taste the pudding that is seen steaming hot at the end of your tenth stroke, and to taste the one decked with holly in the bottom row the very last of all.

You can check your solution here

3) Under the Mistletoe Bough***/*****



“At the party was a widower who has but lately come into these parts,” says the record; “and, to be sure, he was an exceedingly melancholy man, for he did sit away from the company during the most part of the evening. We afterwards heard that he had been keeping a secret account of all the kisses that were given and received under the mistletoe bough. Truly, I would not have suffered any one to kiss me in that manner had I known that so unfair a watch was being kept. Other maids beside were in a like way shocked, as Betty Marchant has since told me.” But it seems that the melancholy widower was merely collecting material for the following little osculatory problem.

The company consisted of the Squire and his wife and six other married couples, one widower and three widows, twelve bachelors[Pg 92] and boys, and ten maidens and little girls. Now, everybody was found to have kissed everybody else, with the following exceptions and additions: No male, of course, kissed a male. No married man kissed a married woman, except his own wife. All the bachelors and boys kissed all the maidens and girls twice. The widower did not kiss anybody, and the widows did not kiss each other. The puzzle was to ascertain just how many kisses had been thus given under the mistletoe bough, assuming, as it is charitable to do, that every kiss was returned—the double act being counted as one kiss.

You can check your solution here

Dudeney lists another puzzle, “Buying presents”, but this involves now outdated British coins, for which reason I do not include it in this collection.

New puzzles are published at least twice a month on Fridays. Solutions are published after one or more weeks. You are welcome to remark on the difficulty level of the puzzles, discuss alternate solutions, and so on. Puzzles are rated on a scale of 1 to 5 stars.

Eleusis


There are few good puzzle games. Puzzles rarely make good games, and good games rarely contain puzzles.

The first classical exception is mastermind. In recent years, Escape room shave become popular. A game which I learned as a student is Eleusis. This post concentrates on Eleusis. I wrote about Eleusis before in December 2013, you can find that post here.

The game of Eleusis was invented by Robert Abbott in 1956, and is totally different from such games as bridge or poker. Eleusis is played with a standard card deck of 52 cards. One player thinks of a secret rule and preferably writes this down. He plays two cards which obey the secret rule. All other players receive a number of cards, for example each player receives 5 cards.

The two cards are the beginning of a line of cards. The other players now take turns in playing a card to the end of the line. When a player plays a card, the Rule Inventor indicates whether the card obeys the rule. If it does, it is added to the end of the line. If it does not, the card is placed below the line and the player draws two extra cards from the deck. In both cases, the turn passes to the next player. The player who first gets rid of all his cards wins.

In the image above, the Rule Inventor started the row with 10 of clubs and jack of spades. The first player played 3 of spades, which was wrong. The next two cards, 3 of diamonds and 6 of spades, were also wrong. The fourth player tried 9 of hearts, which was correct.

The question is of course: With your hand depicted at the bottom, which of the 5 cards labeled A-E do you play?

You can check your solution here

Playing card puzzles by Henry Dudeney


For the past couple of months, I have been publishing puzzles with playing cards. Henry Dudeney was British formost puzzle master of the late 19th / early 20th century. In this series his puzzles, as published in “Amusement in mathematics”, may not be omitted.

1) 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.


You can check your solution here

2) 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.

You can check your solution here

3) 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?

You can check your solution here

4) 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 Pg 116is 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.

You can check your solution here

Bongard problem 40


Which rule satisfies the 6 figures on the left but is obeyed by none of the 6 figures on the right?
1)Bongard problem 40***/*****


In 1967 the Russian scientist M.M. Bongard published a book containing 100 problems. Each problem consists of 12 small boxes: six boxes on the left and six on the right. Each of the six boxes on the left conform to a certain rule. Each and every box on the right contradicts this rule. Your task, of course, is to figure out the rule.

You can check your solution here

You can find more Bongard problems here on this site and at Harry Foundalis’ site.

Pradeesh Mutalik can be credited for taking Bongard problems from the realm of geometry to the realm of numbers and language.

New puzzles are published at least twice a month on Fridays. Solutions are published after one or more weeks. You are welcome to remark on the difficulty level of the puzzles, discuss alternate solutions, and so on. Puzzles are rated on a scale of 1 to 5 stars.

Playing card puzzles (4)


1) A 5×5 grid***/*****
Which of the 5 cards on the right should take the place of the card turned down?

You can check your solution here

New puzzles are published at least twice a month on Fridays. Solutions are published after one or more weeks. You are welcome to remark on the difficulty level of the puzzles, discuss alternate solutions, and so on. Puzzles are rated on a scale of 1 to 5 stars.

Bongard problem 39


Which rule satisfies the 6 figures on the left but is obeyed by none of the 6 figures on the right?
1)Bongard problem 10**/*****


In 1967 the Russian scientist M.M. Bongard published a book containing 100 problems. Each problem consists of 12 small boxes: six boxes on the left and six on the right. Each of the six boxes on the left conform to a certain rule. Each and every box on the right contradicts this rule. Your task, of course, is to figure out the rule.

You can check your solution here

You can find more Bongard problems here on this site and at Harry Foundalis’ site.

New puzzles are published at least twice a month on Fridays. Solutions are published after one or more weeks. You are welcome to remark on the difficulty level of the puzzles, discuss alternate solutions, and so on. Puzzles are rated on a scale of 1 to 5 stars.

Dice puzzles (3)


We had two previous posts on dice problems, which you can find here and here.

This is the third post in a small series on dice puzzles. The first one was about the polar bear puzzle and its variations, the second one posed some alternate patterns.
In this post I want to explore some dice puzzles by the British grand master of puzzles, Henry Dudeney.

1) The dice numbers.

I have a set of four dice, not marked with spots in the ordinary way, but with Arabic figures, as shown in the illustration. Each die, of course, bears the numbers 1 to 6. When put together they will form a good many, different numbers. As represented they make the number 1246. Now, if I make all the different four-figure numbers that are possible with these dice (never putting the same figure more than once in any number), what will they all add up to? You are allowed to turn the 6 upside down, so as to represent a 9. I do not ask, or expect, the reader to go to all the labour of writing out the full list of numbers and then adding them up. Life is not long enough for such wasted energy. Can you get at the answer in any other way?
This puzzles was published as puzzle 96 in “Amusement in Mathematics”

You can check your solutions here

2) A trick with dice

The first problem I want to have a look at was published by British puzzler Henry Dudeney in his book “Amusement in Mathematics” (and before that probably in one of the magazines in which he had a monthly column).
I ask you to throw three dice without me seeing them. Then I tell you to multiply the points of the first die by 2 and add 5. Multiply the result by 5 and add the point of the second die. Multiply the result by 10 and add the points on the third die. Mention me the result and I will immediately tell you the points on your three dice.
For example, if you throw 1, 3 and 6, the result will be 386, from which I could at once say what you had thrown. How do I do that?
This puzzles was published as puzzle 386 in “Amusement in Mathematics”

You can check your solutions here

3) The Montenegrin dice game
It is said that the inhabitants of Montenegro have a little dice game that is both ingenious and well worth investigation. The two players first select two different pairs of odd numbers (always higher than 3) and then alternately toss three dice. Whichever first throws the dice so that they add up to one of his selected numbers wins. If they are both successful in two successive throws it is a draw and they try again. For example, one player may select 7 and 15 and the other 5 and 13. Then if the first player throws so that the three dice add up 7 or 15 he wins, unless the second man gets either 5 or 13 on his throw.

The puzzle is to discover which two pairs of numbers should be selected in order to give both players an exactly even chance.

You can check your solutions here

Bongard problem 38


Which rule satisfies the 6 figures on the left but is obeyed by none of the 6 figures on the right?
1)Bongard problem 38***/*****

In 1967 the Russian scientist M.M. Bongard published a book containing 100 problems. Each problem consists of 12 small boxes: six boxes on the left and six on the right. Each of the six boxes on the left conform to a certain rule. Each and every box on the right contradicts this rule. Your task, of course, is to figure out the rule.

You can check your solution here

You can find more Bongard problems here on this site and at Harry Foundalis’ site.

New puzzles are published at least twice a month on Fridays. Solutions are published after one or more weeks. You are welcome to remark on the difficulty level of the puzzles, discuss alternate solutions, and so on. Puzzles are rated on a scale of 1 to 5 stars.

Playing card puzzles (3)


This is the third post with puzzles about playing cards.

1) Which hands are valid?****/*****
In a game, the following six hands are valid plays:

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Which of the following three hands is / are valid?

You can check your solution here

New puzzles are published at least twice a month on Fridays. Solutions are published after one or more weeks. You are welcome to remark on the difficulty level of the puzzles, discuss alternate solutions, and so on. Puzzles are rated on a scale of 1 to 5 stars.

Playing card puzzles (2)


Last month I published two puzzles with playing cards. Here are two more in the same line:

For those who missed the puzzles, look at the figure below. It shows 16 cards, one of which is hidden. At the bottom you find 4 cards. Which of those 4 cards should replace the hidden card?

3) Playing cards puzzle 3***/*****

You can check your solution here

4) Playing cards puzzle 4***/*****

You can check your solution here

New puzzles are published at least twice a month on Fridays. Solutions are published after one or more weeks. You are welcome to remark on the difficulty level of the puzzles, discuss alternate solutions, and so on. Puzzles are rated on a scale of 1 to 5 stars.