Showing posts with label play puzzle games. Show all posts

Friday, July 17, 2015

How Many Monks?

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how many monks riddle
There's a monastry of fifty monks that have taken a vow of silence. They go through the same routine every day. They wake up, pray. They go to lunch, their only meal of the day, were they all sit round a big circular table, and eat a simple meal. Then they go back to their rooms and pray. Then they fall asleep.

One day, the head monk of the area comes to visit. He's a bit different - he's allowed to talk and has a life outside the monastry. He tells them that there is currently a plague ravaging the land. People everywhere are dying. The disease manifests itself as red blotches on the forehead. The blotches are the only manifestation of the disease for three months, whereupon the next stage starts, a horribly painful death.

The head monk tells them that at least one of their midst will have the disease, probably more. Anybody with the disease should kill themselves, to save all of the pain and suffering. By killing themselves, they will restrict the movement of the disease, and will go to heaven. Anybody with the disease will show the first symptoms within a month. The chief monk then leaves. He returns two months later, to find that all of the infected monks have killed themselves, and they all did it on the same day.

Bear in mind the following:
  • They have no mirrors or any other way of seeing themselves
  • The blotches appear only on the forehead and cannot be seen by the monk
  • Infected monks feel no different - the only manifestation of the first part of the disease are the blotches.
  • The monks cannot talk to each other or in any other way communicate.
  • Any monks with the disease will display the blotches within a month of the head monk leaving.
  • At least one monk definitely has the disease.
  • The monks only see each other once per day, at lunch, when they are all sat round the round table.
  • These monks are brighter than the average monk....
How did they know whether or not they were infected and why did they all kill themselves on the same day?

How Many Monks? Puzzle Solution

The way they work this out and the reason it happens on the same day is as follows. It is important to note that not only are the monks intelligent but they all know all their fellow monks are as well.
Start with one monk:

He knows he is the only one who can have the disease so he kills himself on
day 1.

Then two monks:
They know that either one of them has the disease or both do. If monk1
doesn't see the mark of the disease on monk2 then he will realise that he
must have it so will kill himself. If he sees the mark then he knows that
monk2 will following the same reasoning and kill himself. If the next day
the other monk is still alive then he realises that the other monk must have
seen the mark on him and so they both have the disease and he must kill
himself. Both follow the same reasoning and kill themselves on day 2.

Three monks:
If only one of the monks has the disease he will see no mark on the other
two and so diagnose himself. He will kill himself on day one. If two monks
have the disease then they will each see one monk with the mark and one
without. When they see each other again the next day they will deduce that
the monk they see with the mark would only not have killed himself if he
could see someone else with the mark. They know it is not the third monk so
it must be them. Both diseased monks follow the same reasoning and kill
themselves on day two. If all three monks have the disease then they are all
in the same position as the healthy monk in looking on in the previous
example. Each of them can see two diseased men. When they haven't both
killed themselves on day two there can be only one reason - the viewing monk
must have the mark as well. All taking the same reasoning they all kill
themselves on day three.

And so on...
N monks all with the disease will all kill themselves on day N.

Saturday, June 27, 2015

Four Hats

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Four men have been buried all the way to the neck, only their heads stick out. They cannot turn their heads, so they can see only in front of them. A wall has been placed between A and B, so that A cannot see the other 3 (B, C, D), and viceversa. All of them know in which position the others have been buried. So, for example, B knows that C and D can see him, even though he can't see them.

A hat has been placed on top of each man's head. All of them know that there are two black hats and two white hats, but no one is told the colour of the hat he's wearing.

four hats
They will all be saved if at least one of them can safely say what colour is the hat he's wearing. Otherwise they'll all be decapitated.

Which one of them saved the day? And, most importantly, how?

Four Hats Puzzle Solution

C saves the day.

D clearly has the most information at his hands, but seeing one white and one black hat doesn't give him any certainty about his own hat's colour. Would B and C both have been wearing the same colour, D would have been able to provide the answer.

But C is one clever guy and he knows that if D doesn't answer, it means that B is wearing a different colour than him. Because B is wearing white, C knows he's wearing black.

Note that A is redundant: the puzzle could have included only B, C, D. That way, the hats would have been three, with two hats of the same unspecified colour, and one other hat of the opposite colour.

Monday, May 18, 2015

12 Balls

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12 balls riddle

You have twelve balls that all weigh the same except one, which is either slightly lighter or slightly heavier. The only tool you have is a balance scale that can only tell you which side is heavier.

Using only three weighings, how can you deduce, without a shadow of a doubt, which is the odd one out, and if it is heavier or lighter than the others?

12 Balls Puzzle Solution

We found a good solution to this puzzle on the Cut The Knot website. They show a solution taken from the book Mathematical Spectrum, by Brian D. Bundy. Bundy shows two solutions: the first one requires different courses of action depending on the outcome of previous weighings, so it is not particularly elegant or easy to remember. The second solution involves a fixed course of action in all circumstances, and is the one that follows.

In this method we weigh four specified balls against four other specified balls in each of the three weighings and note the result. If we observe say the left hand side of the balance, then for an individual weighing there are three possible alternatives: the left hand side is heavy (>), light (<) or equal (=) as compared with the right hand side of the balance.

Since three weighings are allowed, the number of different results that can be obtained is just the number of arrangements (with repetitions allowed) of the three symbols >, <, =, i.e. 3 3 = 27. If we use all twelve balls in the three weighings, and ensure that no particular ball appears on the same side of the balance in all three weighings, the outcomes >>>, <<<, === are not possible. We thus have only 24 possible outcomes and we shall show that it is possible to set up a one to one correspondence between these 24 outcomes and the conclusion that a particular ball among the twelve is heavy or light.

The 24 outcomes can be divided into two groups of twelve in each group. If we call the reverse of an outcome the outcome obtained by replacing > by <, < by > and leaving = unchanged, one group of twelve will be the reverses of the other group and vice versa. We can thus write the 24 outcomes in the form of two arrays, each array having three rows (the three weighings) and twelve columns (the twelve balls), so that each row contains four >'s, four <'s and four ='s. Thus we have, for example,
> > > >   < < < <   = = = =
< = = =   < < > >   > < > =
=
> < =   > = < >   > = < <
A
B C D   E F G H   I J K L
<
< < <   > > > >   = = = =
> = = =   > > < <   < > < =
= < > =   < = > <   < = > >
 
We consider just the top array, and for each weighing (row) place the balls corresponding to a > in the left, and the balls corresponding to a < in the right of the balance. Thus we would weigh A, B, C, D against E, F, G, H; then G, H, I, K against A, E, F, J; and finally B, E, H, I against C, G, K, L. The results of these three weighings as observed on the left of the balance are noted. If the outcome is >, <, = we conclude that ball A is heavy; if >, =, > ball B is heavy, etc; if =, =, < ball L is heavy. If we obtain an outcome that appears in the lower array, we conclude that the corresponding ball is light. Thus for <, >, = ball A is light, etc; =, =, > means ball L is light.

Thursday, April 23, 2015

Strawberry Ice Cream

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A man walks into a bar, orders a drink, and starts chatting with the bartender.

After a while, he learns that the bartender has three children. "How old are your children?" he asks.
"Well," replies the bartender, "The product of their ages is 72."

The man thinks for a moment and then says, "That's not enough information."

"All right," continues the bartender. "If you go outside and look at the building number posted over the door to the bar, you'll see the sum of the ages."

The man steps outside, and after a few moments he reenters and declares, "Still not enough!"

The bartender smiles and says, "My youngest just loves strawberry ice cream."

strawberry ice cream riddle

How old are the children?

Strawberry Ice Cream Puzzle Solution

First, determine all the ways that three ages can multiply together to get 72:
  • 72 1 1 (quite a feat for the bartender)
  • 36 2 1
  • 24 3 1
  • 18 4 1
  • 18 2 2
  • 12 6 1
  • 12 3 2
  • 9 4 2
  • 9 8 1
  • 8 3 3
  • 6 6 2
  • 6 4 3
As the man says, that's not enough information; there are many possibilities.

So the bartender tells him where to find the sum of the ages--the man now knows the sum even though we don't. Yet he still insists that there isn't enough info. This must mean that there are two permutations with the same sum; otherwise the man could have easily deduced the ages.

The only pair of permutations with the same sum are 8 3 3 and 6 6 2, which both add up to 14 (the bar's address). Now the bartender mentions his "youngest"--telling us that there is one child who is younger than the other two. This is impossible with 8 3 3--there are two 3 year olds. Therefore the ages of the children are 6, 6, and 2.

Pedants have objected that the problem is insoluble because there could be a youngest between two three year olds (even twins are not born exactly at the same time). However, the word "age" is frequently used to denote the number of years since birth. For example, I am the same age as my wife, even though technically she is a few months older than I am. And using the word "youngest" to mean "of lesser age" is also in keeping with common parlance. So I think the solution is fine as stated.

Friday, April 3, 2015

Trick Mules

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trick mules
Start with the three puzzle pieces arranged so that the two jockeys are correctly riding the two weary-looking mules.

Rearrange the pieces so that the mules miraculously break into a frenzied gallop!

 

Notes:
The mules do not overlap. There is a clever, but not deceptive solution to this puzzle. 

Trick Mules Puzzle Solution

This ambiguous "Trick Mules Puzzle" is solved by the realisation that the mule can have two different orientations. Here the same lines and contours have two interpretations, one horizontal and one vertical.
trick mules solution

Tuesday, January 6, 2015

Creative Lamps

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These unusual and very interesting lamps are developed by the Studio Cheha (Tel Aviv, Israel) and they all share one interesting secret which is not immediately visible. See the full post and find out!








The thing is that they are all flat, though it may seem otherwise.  Look at them once again and discover that these objects are  2-dimensiona!:)

Friday, October 10, 2014

Faulty Batch

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A little nation in Antarctica has its gold coins manufactured by eight different European companies. The Treasury Minister and his secretary were examining samples just delivered from the eight companies.

"How much should these coins weigh?" the Minister asked.

"Ten grams each, Sir."

"At least one of these coins - this one - is lighter than the others," said the Minister. "Let's check."

He put the coin on the scale, which showed that the coin weighed only nine grams. A bunch of coins, untidily placed on a tray, were frantically searched by the Minister and his secretary. Within the bunch, they found a handful of coins that also weighed one gram less than they should. The two men looked at each other; obviously, one of the manufacturing companies was producing coins with the wrong weight.

"Most of the coins are still packed in the plastic wrappers. It should be easy to tell which company is producing the faulty batch," said the secretary.

The two men placed eight packs of coins on the table, one pack from each company.

"How tedious," sighed the Minister. "Do we really have to use this scale eight more times, just to find the faulty batch of coins?"

"That won't be necessary, Sir," grinned the secretary. "We can find the lighter coins by using the scale only once."

How would they do it?



Notes:
By using the scale once, it means that only one reading can be taken after all the coins to be weighed are placed onto the scale. ie, you cannot read the values as you place the coins on -- that would make the puzzle too easy!

Faulty Batch Puzzle Solution

The secretary placed on the scale 1 coin from the first batch, 2 from the second, and so on until he put 8 from the eighth batch.

If all coins weighed 10 grams each, then the weight displayed on the scale should have been 360 grams ((1 + 2 + ... + 8) × 10). But, since one batch of coins weighs less, the difference between 360 grams and the weight displayed on the scale should point us to the faulty batch. For example, if the faulty batch was the fifth one, then the total weight displayed on the scale would be 355 grams. Or if it was the seventh batch, the weight would have been 353 grams, ie 7 grams less than the theoretical total weight of 360 grams.

An 'optimisation' on this solution is to omit the 8 coins from the eighth batch. In this case, the maximum weight of the coins would be 280 grams, and if it equals 280, then the eighth batch is the faulty one. Thanks to Denis Borris for this observation.

By using the same logic, one could omit the coins from any one of the other batches, instead of the eighth one. For example, if we omit the fourth batch, we'll be left with a theoretical 320 grams and, if it is indeed the total weight, then we will know that the fourth batch was the faulty one. Thanks to Glen Parnell for noticing this.

Tuesday, September 30, 2014

Orbiting Logic

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Colonel Tom Carpenter, during his fifth space mission, was being kept awake by the blabbering of the Cape Canaveral Control Centre operator, who offered him the following puzzle.

"Here's a deck of 52 cards, Tom. I'm taking the Aces and the Royals out of the deck. Do you copy that, Tom?"
cards riddle

"Roger," the yawning voice of the astronaut answered.

"Of the 36 remaining cards, I've drawn 5 of them. These 5 cards have the following properties:
(a) all four suits are represented here;
(b) there is no more than 2 consecutive ranks for each sequence (ie a 2 followed by a 3, or a 7 by an 8, or both, but not 2, 3, 4);
(c) the sum of the even ranks and the sum of the odd ranks produce two numbers: the difference between these two numbers is 9, but I won't specify whether it's the sum of odds being greater than the sum of evens, or viceversa.
(d) the sum of ranks of the red cards is exactly twice the sum of ranks of the black cards.
You awake, Tom?"

After a pause, Tom managed a faint "Roger."
"Ok, you should also know that:
(e) a hearts is a multiple of a clubs;
(f) the rank of a diamond is greater than that of a hearts;
(g) there are no 2 cards with the same rank.

Which cards did I draw? Tom, are you listening? Which cards have I got?"

Deduce which five cards he necessarily holds.

Orbiting Logic Puzzle Solution

During the next orbit, the sleepy but correct answer came forth from the astronaut: 2 of clubs, 5 of hearts, 7 of diamonds, 9 of spades, and 10 of hearts.

Following from (g) - no two cards are the same rank - and (b), the strongest combination (ie highest ranks) possible is 10, 9, 7, 6, 4, which adds up to 36. If we write down all combinations of 2 numbers that have a difference of 9, and a maximum sum of 36, we'll have:
  • 2, 11
  • 3, 12
  • 4, 13
  • 5, 14
  • 6, 15
  • 7, 16
  • 8, 17
  • 9, 18
  • 10,19
  • 11, 20
  • 12, 21
  • 13, 22
The number of odd ranks in the 5-card combination could be 1 or 3 (not 2 and not 4 because their sum would be an even number, and a difference of 9 between two even numbers is inexistent). If there was only 1 odd rank, then there would be 4 even ranks, and the weakest combination of evens would be 2, 4, 6, 8 which add up to 20: the difference, 9, would result in a single number, 11, which does not exist in the deck of cards (the Royals were excluded). Therefore we have 3 odd ranks. The sum of the weakest 3 odd ranks (excluding the Aces) is 15, ie 3 + 5 + 7. So, from the combinations above, we can exclude all combinations that contain an odd number less than 15. We are left with
  • 6, 15
  • 8, 17
  • 10, 19
  • 12, 21
The first one (6, 15) must be excluded because the 3 odd numbers (3, 5, 7) would be joined by the only 2 even numbers which would add up to 6 (2,4), and therefore we would have a sequence of 5 consecutive numbers, which doesn't match the constraint set by (b). We are left with 3 combinations:
  • 8, 17 (and the 5 ranks would be 2, 3, 5, 6, 9)
  • 10, 19 (and the 5 ranks would be 3, 4, 6, 7, 9)
  • 12, 21 (and the 5 ranks would be 2, 5, 7, 9, 10)
But (d) states that the sum of red cards is twice of the sum of black cards, so the first 2 combinations must be excluded, because it's not possible to find 2 sets of numbers, one of which is twice the other; therefore the 5 ranks can only be 2, 5, 7, 9, 10. The red cards must be 5, 7, 10.

Following (e), rank 2 must be clubs and 10 must be hearts, so rank 9 must be spades; following (f), rank 7 has got to be diamonds and rank 5 must be hearts.