Chestnuts (Puzzles most of us have already seen)

[Zag]

A friend told me a puzzle recently, and I realized quickly that it was a chestnut. After I explained the concept, he asked what other chestnuts I knew, and I was chagrined only to be able to think of 6 or so. (Yes, I'm getting old and senile.) Anyway, as an effort to remember more, and hopefully to stave off the newcomer who insists on posting several of them, I thought I would start a chestnuts thread. It should be fun to peruse, if only for nostalgia.

So post your chestnuts here! A couple of rules:

1. Please follow the puzzle with the answer in spoiler tags. I'm sure that there are people around who haven't seen them all and would like to try to solve them. After all, these puzzles became chestnuts because they were good puzzles to begin with. so let people who haven't already seen them enjoy them fully.

2. One puzzle per post.

3. Keep it to a few puzzles per person per week. I don't want the thread to explode and then be left forgotten. Save some for next week.

4. Plz rite n English. Sum1 else shud b able 2 read it (even old farts like me).

5. Try to include the obvious words that people would search on if they were looking for this one. The corollary, if this thread gets longish, is to search a bit to see if it has already been posted before you post one.

6. Please don't debate the puzzles in this thread. If you think the person writing it is wrong, then PM that person. I am assuming that no one is going to bother to post a puzzle unless he is quite sure he fully understands the answer. If you're right, that person will probably get 40 PM's in the first day, and he can edit to have the right answer.

7. Please put a title in bold at the top of your puzzle.

Thanks!
Zag



Monty Hall Puzzle
Find the odd coin with a balance scale
Pirates
Three Hats
gry
Bridge Crossing with Flashlight
Choosing a puppy
Hole in a Sphere
River Crossing with Untrustworthy Animals
A Knight and a Knave (Liar and Truth-teller)
The Hardest Logic Puzzle
The Fly and the Bicycles
The Fallen Signpost
21 from 1, 5, 6, 7
River Crossing with Three Missionaries and Three Cannibals
Truel (Three-way Duel)
How many Fs?
Dropping Glasses
62 Squares
100 Prisoners Wearing Hats
The Snail in the Well
9 dots with 4 Straight Lines
The Bookworm
What Color is the Bear?
Forming a Line Based on Hat Color
Hat Puzzle with Simultaneous Guessing
River Crossing With Convict and Angry Family
The Missing Dollar
Perfect Square Matchstick Problem
East and West Coast Same Time
What's greater than God ...
Planet Census
Rotating Square Table
100 Prisoners and a Light Switch
100 Prisoners and Two Light Switches
Identifying Light Switches
Slamming Locker Problem
incompetant Firing Squad
Carried in a wheelbarrow
The Plate Game
1, 11, 21, 1211
O, T, T, F, F, S, S
How old are my children?
How many horse races?
100 ants on a stick
Lilies in a pond
Ant on an Elastic
First Alphabetic Odd Number
Combinations of Patio Tiles in a Path
Mark of Death
Four Ants on a Square
4 Triangles Using 6 Matches
English Words with Special Properties
Constructing a square with 1/5 area of a given square
Breaking a Hersey Bar

[Zag]

Monty Hall Puzzle

You are a contestant on "Let's Make a Deal," the old game show hosted by Monty Hall. He shows you three doors and explains that one of them hides a new Lexus while the other two hide a rusty roller skate each.

After you choose a door (let's say door #1), he opens a door OTHER than the one you chose (door #3) and you see it has a roller skate behind it. Now he invites you either to stick with your original choice or to switch to the remaining closed door. Does it improve your odds any to switch?

For the sake of this puzzle, we'll assume that Monty knows which door has the real prize, and, whenever he plays this game, he always opens one non-winning door and makes this offer.

The answer is that you absolutely should switch -- it gives you a 2/3 chance of winning, whereas your original choice only had a 1/3 chance of winning. The critical factor is that Monty added information because he knows which door NOT to open.

Consider that, instead of 3 doors, there were 10,000 doors. You choose door #5491, and then Monty opens every door except your choice and door #7033. Do you switch now?

[ralphmerridew]

You are given 12 coins. 11 are the same weight, but one's different.

You have three weighings on a balance in which to determine which of the coins is different, and whether it's lighter or heavier than the others.
(Variant: or maybe they're all the same weight.)

Label the coins.
MADO vs LIKE
METO vs FIND
FAKE vs COIN
After that, it's easy to see which coin is off.

[Duke Gnome]

Pirates

After the Captain has taken his share and the crew has been paid there are 100 gold coins remaining from a recent haul. The Captain delcares that his 5 officers should determine the allocation as follows.

The highest ranked officer suggests a distribution
The officers, including the proposer, all vote on whether to accept it.
If a majority accept then it's settled. If not the proposer is thrown to the sharks and the process begins again with the remaining officers.

All 5 officers are widely known to be perfect logicians, and all have the same priorities. Ultimately they want to survive, then they want as much gold as possible and finally if it doesn't affect their gold, they'd like to see someone thrown to the sharks. (Promotion )

You're the first officer. How do you propose the split?


[
You offer yourself 97 gold coins, the third officer 1 gold coin and the fourth or fifth officer 2 gold coins.

Imagine you're the fifth officer and all others are dead. Naturally you'd propose to keep the lot. So imagine what, knowing this, the fourth officer would do. There's nothing he can possibly give the fifth officer so it doesn't matter what he does. He's a dead man. So, the third officer knowing that he only needs 1 vote doesn't need to bribe the 4th officer as he will accept any deal to save his skin. The third officer will therefore offer himself the entire 100 coins.

The second officer, realising he needs 2 more votes will offer the 4th and 5th offers one coin each. After all it's better than the 0 that the 3rd officer would give them. So now you, as the 1st officer, know you need 2 more votes. You can give the third officer 1 gold, but in order to get the other vote you need to beat the 2nd officer's potential bribe of 1 gold to the 4th and 5th guys. You only need one, so pick the guy you like best and offer him 2.]

[Jack_Ian]

Three Hats
You are one of the three wisest sages in the land brought before the king to see which of you is worthy to become the king's advisor. After passing many tests of cunning and invention, you are pitted against each other in a final battle of the wits.

You are all placed seated around a small table and blind-folded as the king describes the test.

"Upon each of your heads I will place a hat. Now you are either wearing a blue hat or a white hat. All I will tell you is this- at least one of you is wearing a blue hat. The first to correctly announce the color of his hat shall be my advisor. Be warned however, he who guesses wrongly shall be beheaded. If not one of you answers within the hour, you will be sent home and I will seek elsewhere for wisdom."

Your blindfolds are removed and you see before you that the two other sages are wearing a blue hat.

For what seemed like hours no one spoke. Finally you stand up and say, "The color of the hat I am wearing is..."
"Blue!"
Consider the scenario of 1 blue hat and 2 white hats. Immediately the person wearing the blue hat would announce the correct answer.
Now consider the scenario of 2 blue hats and 1 white hat. Imagine you are wearing one of the blue hats. You would see 1 white and 1 blue. Since the person that they can see with the blue hat does not announce immediately that they are wearing a blue hat, they must not be seeing 2 white hats so you can confidently announce that you must also be wearing a blue hat.
Now in the given scenario you can see 2 blue hats. If your hat was white, then it would describe the second scenario above and since the other 2 are wise, one of them would eventually realise that they are wearing a blue hat. However, after a very long wait, neither of them has dared make such a claim. This would only occur if all 3 were wearing blue hats.

[Scurra]

gry

This is one of those language ones that is horribly contentious because all the answers are really tricks, based entirely upon how the question is worded.

One traditional form goes like this:
Angry and Hungry are two common words that end -gry. There are three words in the English language; what's the third?

[One possible answer would be language, as it is the third word in "the English language". There are many other, equally dubious answers.

What's interesting, of course, is that there really are only those two common words that end -gry. There are a few very obscure or obsolete words to be found in various dictionaries, but I don't think anyone uses any of them in daily conversation unless discussing this question!
And there is one slightly uncommon word that begins Gry-. ]
_________________
_{still Quiz Olympiad champion. Must get a life.
New definitions: COFFEE - someone who is coughed upon}

[/dev/joe]

Bridge Crossing with Flashlight

Four people have to cross a narrow, weak bridge at night. The bridge can only support two people at a time and the person/people crossing have to use a flashlight to avoid falling off. They have only one flashlight and the gap is too far to throw the flashlight back across. The fastest person can cross the bridge in one minute, the next-fastest can cross in two minutes, the third can cross in five minutes, and the slowest person requires ten minutes to cross the bridge. When two people cross together, the faster person has to slow down to the slower one's speed. What is the shortest time required for all four to get across the bridge and how do they do it?

The people are often given names, which vary from one telling of the puzzle to another. One popular version has the members of the band U2, with Bono crossing in 1 minute, Edge in 2 minutes, Adam in 5 minutes, and Larry in 10 minutes.

First the two fast people cross in two minutes. The fastest one goes back with the flashlight in one minute. Then the two slowest people cross in ten minutes. The second-fastest person goes back across in two minutes, and finally the two fast people cross again in two minutes. Total time is 2 + 1 + 10 + 2 + 2 = 17 minutes.

[Zag]

Choosing a puppy

This is poignant to me because we just had puppies in my house one week ago. Zag's puppies This puzzle is usually told with a farmer who has the pups, but I'll play that part.

You and your spouse have just heard about some adorable Pugle-Welsh Terrier puppies just born at Zag's house, and you decide to get one. Your spouse wants a female, so you call up Zag and have the following conversation.

You: Do you still have any of those adorable puppies left?
Zag: Yes, there are two still unclaimed
You: Is there at least one female available?
Zag: Yes.
You: Thanks! We'll be over in 15 minutes! Bye.

So you get in the car with your spouse, who, being a fickle sort says, "I've changed my mind. I want to get a male puppy."

Knowing what you know, what are the chances that Zag has a male puppy available? (Assume that each puppy has an independent 50% chance for either gender.)

The answer, perhaps surprisingly, is 2/3! When you learned that there were two puppies available, then you knew that there were four possible combinations for those puppies (assuming we always state the larger puppy first): F&F, F&M, M&F, M&M. When your second question was answered, you only eliminated one of the combinations (M&M) and you gained no information that would re-weight the others -- they are still equal possibility. Of those three (equally probable) combinations, two of them include a male.

Suppose that, instead, you had asked "Is the larger one a female?" and you were told that it is. Then you would have eliminated two of the three combinations, and you would be left with a 50% chance on the sex of the smaller dog.

[bonanova]

Hole in a sphere

You drill a hole that is 6" long through the center of a sphere.
What is the volume of the portion of the sphere that remains?

[36 pi in ³

Either believe the answer does not depend on sphere's diameter [it wasn't specified] and choose 6" for diameter, or go through the calculus.]
_________________
_{Vidi, vici, veni.}

[Antrax]

River Crossing with Untrustworthy Animals

You want to get a wolf, a sheep and a cabbage to the other side of the river. Alas, your boat only has room for two of these at a time, and as the title suggests, these animals are sneaky - if allowed, the wolf will eat the sheep, and the sheep will eat the cabbage (though for some reason this won't happen if all three are present at one side at the same time). How do you get all three across the river without any eating taking place?

Go across with W and C, come back with only C, leave C behind and take S across, come back with W leaving S behind, load C and shuttle both C and W across.
_________________
After years of disappointment with get rich quick schemes, I know I'm gonna get rich with this scheme. And quick!

[groza528]

A Knight and a Knave
As you're trying to find your way out of the Puzzlanian haunted forest you reach a fork in the path. You know that one of these paths will lead you out of the forest and the other will send you to your doom.
At this fork stand two men: One is a knight who is honorbound to tell the truth in all cases; the other is a knave who will lie with every statement. However, you do not know which man is which.
What one question can you ask to one of the men in order to find your way out of the woods?

Ask either man "Which way would your companion tell me leads out of the woods?" The knave would falsely claim that the knight would point down the wrong path and the knight would accurately state that the knave would point down the wrong path. In either case you want to take the opposite path from the one indicated.

And just for fun, here's a comedic twist on that chestnut: http://www.giantitp.com/comics/oots0327.html

[Jack_Ian]

In a similar vein..

The Hardest Logic Puzzle by George Boolos

Three gods A, B, and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are 'da' and 'ja', in some order. You do not know which word means which.

Clarifications:

• It could be that some god gets asked more than one question (and hence that some god is not asked any question at all).
• What the second question is, and to which god it is put, may depend on the answer to the first question. (And of course similarly for the third question.)
• Whether Random speaks truly or not should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he speaks truly; if tails, falsely.
• Random will answer 'da' or 'ja' when asked any yes-no question.

----------------------------
The answer becomes a little involved, so rather than post it here, I'll provide this link.

[/dev/joe]

The Fly and the Bicycles

Two men ride bicycles toward each other at constant speeds of 10 km/hr, starting 10 km apart. As they set out, a fly takes off from the front of one of the bicycles, headed toward the other bicycle at 20 km/hr. When it gets there, it turns around, maintaining its speed. Each subsequent time it reaches one of the bicycles, it turns around, until eventually the bikes collide, crushing the fly. How far did the fly fly?

The easy way to solve this is to notice that the bikes approach each other at a net speed of 20 km/hr, so it takes them half an hour to make the trip. In this time, the fly flies 10 km.

It's possible to solve this puzzle a harder way by calculating the distance the fly travels in each segment and summing the infinite series. When this puzzle was posed to him, John von Neumann is said to have done it the hard way in his head in a few seconds. I cannot verify the accuracy of this final claim, but it is included for the purposes of including his name for the search.

[groza528]

The Fallen Signpost
You're walking out of Puzzlania to visit a friend in beautiful, sunny Enigmatopolis. You reach an intersection with roads leading out in 8 directions, evenly-spaced. Unfortunately, the signpost telling you which road leads to which city has fallen. Also unfortunately, you have no idea how to reach Enigmatopolis.

How do you determine the correct route?

Orient the signpost such that the sign to Puzzlania points back in the direction you've just come from.

[Antrax]

21 from 1, 5, 6, 7

Using the numbers 1, 5, 6 and 7 and the four basic arithmetic operations, can you get to 21?

Yep. 21 = 6 / (1 - (5/7))
_________________
After years of disappointment with get rich quick schemes, I know I'm gonna get rich with this scheme. And quick!

[Ctuchik]

River Crossing with Three Missionaries and Three Cannibals

Three Missionaries are travelling through the jungle together with three Cannibals from a jungle tribe. One day they reach a river with only a small boat which can only carry up to two people at once. The Cannibals are normally not a threat, but should they ever outnumber the missionaries on either side of the river, they will attack and eat them.
How do they all get across the river without the missionaries getting eaten?

Solution:
Send two cannibals over and one cannibal back. Then do the same thing again. Now send two missionaries over and a missionary + a cannibal back. Send two missionaries over and a cannibal back.
The missionaries are now all on the other side while the cannibals are back where they started.
Send two cannibals over and one comes back and picks up the last one.
Now they've all reached the other side, without the missionaries ever being outnumbered.

[lostdummy]

Truel

A, B, and C are to fight a three-cornered pistol duel. All know that A's chance of hitting his target is 0.3, C's is 0.5, and B never misses. They are to fire at their choice of target in succession in the order A, B, C, cyclically (but a hit man loses further turns and is no longer shot at) until only one man is left. What should A's strategy be?


While solution is simple, reasoning behind it is fairly complex, so instead of posting here, I'll provide link: Truel Solution

For those interested in more general (also harder) Multi-duel problem, link to my GL post: Multi-Duel

[Duke Gnome]

How many Fs?

How many Fs are there in the following sentence?

[/dev/joe]

Dropping Glasses

You have two identical drinking glasses and a building 100 floors tall. You know that each of them will break if dropped from floor N or higher of a tall building, but not when dropped from any lower floor. N is an unknown integer from 1 to 101. Unbroken dropped glasses are unharmed and can be retrieved to be reused, but once a glass breaks it is useless.

What method should you use to determine N such that you're guaranteed to find it exactly, and the worst case for the number of drops required is minimized? And what is the maximum number of drops needed under that strategy?

Suppose you have only one glass. Then the only way to ensure you get the answer is to try dropping it from every floor, starting from the bottom, until it breaks. After your first glass breaks, this is the situation you are in. The strategy with the first glass is then to drop it from certain floors, going upward, and when it breaks, try the second glass on the floors between the last two drops.

To minimize the worst case number of drops, pick a number of drops and maximize the number of floors that can be covered with that number of drops. So with a maximum of K drops, you begin by dropping the first glass from floor K. If it breaks, you use up to K-1 drops with the second glass on floors 1 to K-1 to find the exact floor. If the first glass doesn't break, the next drop for the first glass is on floor K+(K-1). This is the second drop, so you have K-2 drops with the second glass on floors between K and K+(K-1) if the first glass breaks this time.

This continues, and you end up on floor K+(K-1)+(K-2)+...+1 = K(K+1)/2 on the Kth drop with the first glass, at which point you will either know N is one of the floors up to that one, or that it is some indeterminate higher floor. We need K(K+1)/2 to be at least 100 (if it is exactly 100, and N is 101, we'll know it is 101 because we are told that is the limit for N). This occurs for K=14 where the value is 105. Thus, the strategy is as above for K=14 and the maximum number of drops required is 14.

[groza528]

62 Squares
Is it possible to tile a standard 8x8 chessboard with 31 2x1 dominoes such that the only two squares remaining exposed are opposite corners?

No, it isn't-- each domino will cover one light square and one dark square so it's impossible to leave only two squares of the same color exposed.

[Amb]

Re the Gry puzzle that Scurra posted.

The wording of the puzzle was something like "There are three english words ending in Gry. Hungry, Angry and one other. If you listened carefully I have already told you what the third word was. What is it?".

The answer is Gry itself, which once appeared in the Webster dictionary but has since been obsoleted.

[Duke Gnome]

100 Prisoners Wearing Hats

100 people have been kidnapped by an evil psycopath. He explains that in 20 minutes he will be releasing knockout gas. Whilst unconcious he will line them all up at random facing the same direction, so that the guy at the back can see all 99 of his friends but the person at the front can't see anyone. He will place a hat on everybody's head that is either black or white, each independantly determined by a fair coin toss (At least he's an honest psycopath).

Starting with the guy at the back and working forward he will ask each person in turn what colour their hat is. If they are correct then they live, otherwise he shoots them immediately. The room is small enough that everybody will hear the response and the gunshot. Any attempts at conveying information via pausing, tone of voice, saying anything other than "White" or "Black" or any similar loopholes will put the captor into a rage and he will kill everybody.

Given that everybody is anxious to help as many of their friends survive as possible what strategy should they use, and how many can they save?

Bonus: What if the captor used a six sided die to choose between Red, Orange. Yellow, Green, Blue and Violet instead of a coin?

[You can save 99 people with certainty. The person at the back says "White" if he sees an odd number of white hats and says "Black" if he sees an odd number of black hats. Each subsequent person will be able to work out their colour based on what they can see and what colours they know have already been said.

Bonus: Again you can save 99 with certainty. Assign each hat colur a value R=0, O=1, Y=2 etc. The person at the back adds up the value of all hats, then subtracts the highest multiple of 6 that he can without going negative. (For example if he had a total of 345, he would subtract 342). He then says the colour of the hat with the same value. All further prisoners can again calculate their colour based on what they can see.]

[Victoria Silverwolf]

I might as well add a very simple one.

The Snail in the Well

A snail is trapped at the bottom of a 10 meter deep well. She can crawl up 3 meters each day, but she slips back 2 meters each night. How long will it take her to escape from the well?

At the end of 7 days, she has climbed up 7 meters. The next day she will climb up 3 meters, reach the top of the well, and escape, so the answer is 8 days and not the "obvious" 10 days.
_________________
Reality is a crutch for people who can't face up to science fiction

[Antrax]

9 Dots with 4 Straight Lines

Place 9 dots in a 3x3 matrix. Can you connect all nine dots with a continous line segment comprised of four straight lines?

yep, but you need to think "outside the box". So, if the dots are in [0,2]x[0,2], your four lines are as follows:
(0, 0) to (3, 0) - covering the top row.
(3, 0) to (0, 3) - covering the rightmost in the middle row and the middle in the bottom row.
(0, 3) to (0, 0) - covering the leftmost column
(0, 0) to (2, 2) - covering the diagonal
_________________
After years of disappointment with get rich quick schemes, I know I'm gonna get rich with this scheme. And quick!

[Scurra]

[Victoria Silverwolf]

The Bookworm

You have a ten-volume encyclopaedia on your bookshelf. A hungry bookworm starts at page one of the first volume, and eats through the set all the way to the last page of volume ten. If each volume is one inch thick, how far has she gone when she is finished? (Disregard the thickness of the covers.)

The way that a set of books is normally arranged on a bookshelf, the first page of volume one is next to the last page of volume two, if we disregard the covers. Similarly, the last page of volume ten is next to the first page of volume nine. Thus, the bookworm only goes through volumes two through nine inclusive. Therefore she has gone only eight inches, and not the "obvious" ten.
_________________
Reality is a crutch for people who can't face up to science fiction

[/dev/joe]

The Bear

A hunter walked 50 miles south, 50 miles west, and 50 miles north, and ended up where he started. Along the way he shot a bear. What color was the bear?

The traditional answer has the man start at the north pole. This is one of the places where the directions work, and any bear within 50 miles of the pole would be a polar bear, and thus white.

It's also possible to make this trip starting at certain locations near the south pole, where the 50 mile west trip would result in one or more complete loops around the pole, and the 50 mile north trip retraces his path south. But there are no bears near the south pole, so this answer does not work if the bear is needed.

[Zag]

Forming a Line Based on Hat Color.

100 of you are locked in a room by the king of Puzzleania and hear the following: "In 15 minutes, knockout gas will be released. When you all wake up, you will each have a red or black hat on, such that you can see everyone else's color but not your own. With NO communication, you must form yourselves in a queue with all the red hats on one side and all the black hats on the other. Any communication after you awaken or any deviation in the line and you will all be killed. You have 13 minutes remaining in which to agree on a strategy." What is the strategy you propose (and, hopefully, the idiots in the room listen to)?


The solution is that two people start the line by standing next to each other. The next person steps between them if they are different or to one edge if they are the same. Once both colors are in the line, each person steps into the split between the colors and the line stretches back to accommodate him.


Another, better, strategy by BraveHat (because stepping into the split in the line could potentially be construed as communicating):

Before getting knocked out, tell everyone that when they first come to, they must count the number of people they see wearing black hats. In fact tell them to count three times and make sure they get the same number each time before doing anything. Designate one side of the room for people who count an odd number, and the other side of the room for people who count an even number (zero being even). They will automatically be congregating with like colors.

[Antrax]

Hat puzzle with Simultaneous Guessing
100 of you are locked in a room by the king of Puzzleania and hear the following: "In 15 minutes, knockout gas will be released. When you all wake up, you will each have a red or black hat on, such that you can see everyone else's color but not your own. With NO communication [edit: specifically, the prisoners are prohibited from moving - say they're all in cells, able to watch each other but not move, and any signal gets everyone killed. Thanks to mtc32], you must each simultaneously guess the color of the hat on your head. I have 100 lobes and 200 ears, so I can make out what everyone said. If you hesitate or get it wrong, you die. You have 11 minutes remaining in which to agree on a strategy." What strategy ensures the survival of the maximal amount of people, and how many people is that?

You can assure the survival of 50 people - have the 100 people pair up. The "first" person of every pair (say, the taller one or whatever) guesses the color of hat he sees on the head of the other. The "second" person of each such pair guesses the opposite of the color the other has on his head. That way, if the hats are of the same color, the first survives, and in the other case, the second survives, so exactly one of each pair survives.

This was proposed by a friend of mine when I told him the puzzle. It turns out to be the same solution, and it might seem more comprehensible to some: Designate half the people who will assume that the final parity of the red hats will be parity odd, and the other half of the people will assume that it will be parity even. That is, 50 people assume that the complete count, including their own, will be an odd number, and 50 assume that the complete count will be an even number. Everyone guesses according to make their assumption correct. So if you're a "parity even" person and you count an even number of red hats, you guess black.

The real number will either be parity even or parity odd. All the people who assumed the right one will have guessed correctly.
_________________
After years of disappointment with get rich quick schemes, I know I'm gonna get rich with this scheme. And quick!

[mtc32]

River Crossing With Convict and Angry Family

Once again you have a river to cross and only two people can be on at a time. This time you need to get eight people to cross which are as follows.

Son
Son
Daughter
Daughter
Angry Mom (Can't be left with one or both sons without Angry Dad)
Angry Dad (Can't be left with one or both daughters without Angry Mom)
Cop
Convict (Can't be left with any family member without cop. Can be left alone.)

With these restrictions, get all eight people across the river.

Solution:
Bring cop and convict over.
Send cop back.
Bring cop and son over.
Send cop and convict back.
Bring angry dad and son over.
Send angry dad back.
Bring angry dad and mom over.
Send angry mom back.
Bring cop and convict over.
Send angry dad back.
Bring angry dad and mom over.
Send angry mom back.
Bring angry mom and daughter over.
Send cop and convict back.
Bring cop and daughter over.
Send cop back.
Bring cop and convict over.

[jja]

The Missing Dollar

This is usually told about a bellhop and a hotel room; I give a more modern formulation below.

Three friends have just finished lunch at a restaurant. The server brings a bill for $30 and each of them chips in ten dollars. The server brings the $30 back to the register, where the manager says that the bill should only have been $25, and tells the server to refund $5 to the three friends.

On the way back to their table, the server reasons that it's impossible to split $5 easily among three people, and they didn't give a tip anyway... So the server quietly pockets two dollars, and gives each of the friends a one dollar refund. Everyone is happy!

But wait a minute. The three friends initially paid out $30, and each got a dollar back, so the total amount they paid out is 3x9 = $27...

But when you add the $2 from the server, that only makes $29!

What happened to the missing dollar?

This puzzle depends on misdirection; there is no missing dollar. The $2 that the server took is part of the $27 that the friends paid out, and should not be added again. Of the original $30, the restaurant got $25, the server got $2 and the friends got $3.
_________________
JJA
"Sancte," inquit, "quare, quaeso, rapis arborem festam?"

[mtc32]

Perfect Square Matchstick Problem

This matchstick problem uses four matchsticks. They are arranged in a cross pattern as follows.
(0,0)-(1,0)
(0,0)-(0,1)
(0,0)-(-1,0)
(0,0)-(0,-1)

The object is to move one matchstick to make a perfect square.

Solution: Move matchstick (0,0)-(1,0) to (-1,0)-(-1,1). This will have the matchsticks shows the number "4" which is a "perfect square".

[/dev/joe]

East and West Coast Same Time

Sorry for the U.S.-centric puzzle, but they can't all be for everyone.
Two friends are talking on a phone call. One of them lives in a U.S. state that borders the Pacific ocean, the other in a U.S. state that borders the Atlantic. During the call, they realize that it is the same time in both locations. How is this possible?

One of them lives in southeastern Oregon, near Boise, Idaho. This area is in the mountain time zone. The other lives in western Florida, south of Alabama. This area is in the central time zone. Normally these areas are one hour apart. However, when Daylight Saving Time ends, it does say at the same time of day in each time zone, not all at once across the country. Thus, it ends for the Florida friend first, and the hour "fall back" makes it the same time there as for the Oregon friend, for one hour.

[Courk]

Oooh! I'm adding that to my list of weird, random things to do, right after "Give birth while crossing International Date Line."

[Chuck]

Back when the drinking age was raised from 18 to 21 in the U.S., people already 18 or older were still allowed to continue drinking. That could be used for some kind of puzzle. Identical twins could have been born on opposite sides of midnight 18 years earlier, so the firstborn could go out drinking with his friends every night while his identical twin brother would be too immature to even transport alcoholic beverages for another three years. If they were born on a train changing time zones then it could be the firstborn twin who's not allowed to drink while his younger brother can because the train crossed a time zone boundary into the previous day just before he was born.

[StupidVisitor]

What is it?

What's greater than God and worse than the devil? The rich want it, the poor have it, and if you eat it, you die.

Nothing.
_________________
"Oh, you can't help that," said the Cat: "we're all mad here. I'm mad. You're mad."
"How do you know I'm mad?" said Alice.
"You must be," said the Cat, "or you wouldn't have come here."

[mtc32]

Planet Census
You are an astronaut that has landed on a small planet that has only one species. Your job is to count how many of them are on the planet. Some have stars on their bellies and some have stars on their backs. You make a count of 54 having starred bellies and 78 with starred backs. You then discover that you may have counted some twice as 23 have a star on both the belly and back. You then also discover that some were not counted at all as 33 have neither star.

With all the information, determine how many aliens are on the planet.

The answer is 1. You are the only alien on that planet while the 142 creatures are native to that planet.

[/dev/joe]

Rotating Square Table

There is a square table with a well in each corner for holding a glass, with a glass in each well. The glasses are identical, but some of them are right-side up and some are upside down. The goal is to get all the glasses turned the same way. You are blindfolded and only get to touch two glasses per round, and the table rotates after each round so that you cannot tell which glass is which except by whether they are right-side up or upside down.

You'll be told when you have won the game. Devise a method for ensuring you win the game in no more than 5 rounds of checking and possibly turning glasses.

1. Turn two adjacent glasses right-side up. If they were already that way, leave them.

2. Turn two opposite glasses right-side up. Since one of these must be a new glass you haven't touched yet, now at least three glasses are right-side up. If you have not won the game yet, you know there is exactly one upside-down glass.

3. Touch two opposite glasses. If one is upside down, flip it over and win. Otherwise, turn one glass upside down; this will make there be two adjacent upside-down glasses and two adjacent right-side-up glasses.

4. Touch two adjacent glasses. If they are turned the same way, flip both over to win the game. If they are turned different ways, flip them over anyway; this preserves the number of upside-down glasses but makes them be opposite each other.

5. Flip over two opposite glasses. This wins if you have not won yet.

[Duke Gnome]

100 Prisoners and a Light Switch

The King of Puzzlania has gathered 100 of his prisoners in the prison exercise yard. He explains that there is a room in the palace that is completely bare except for a light switch. The light swich is currently set for Off. The prisoners will have 20 minutes to discuss strategy after which they will be taken to solitary confinement. Every day thereafter at exactly noon one prisoner, selected at random without regard for the past, will be taken to the palace where she may switch the light or not at her discretion.

At any time any prisoner may declare that everyone has been to the palace at least once. If she is correct then everyone will be set free, otherwise they will all be subjected to mathgrant's music. Since nobody wants even the slightest chance of having to listen to mathgrant's music they all agree that they won't make any declaration unless they are 100% percent sure. Given that they are all in good health and can be expected to survive indefinitely, give a method that can ensure release eventually.


[The simplest method is to elect one person as counter. She will turn the light off when she sees it on, and do nothing when it's off. Everybody else will turn the light on if they have never turned it on before and do nothing if the light is on. The counter may declare after she has turned the light off 99 times. There are better methods that can cut average time down but this is the simplest.]

[lostdummy]

100 Prisoners and Two Light Switches

After seeing above post, I remembered this one too. It has exactly same problem setup as one above, except for:
- there are two light swithces in room
- prisoners do not know initial position of those switches


Since this post ended on new page, here is full text of altered problem:

The King of Puzzlania has gathered 100 of his prisoners in the prison exercise yard. He explains that there is a room in the palace that is completely bare except for two light switches. The light swiches are currently in unknown position. The prisoners will have 20 minutes to discuss strategy after which they will be taken to solitary confinement. Every day thereafter at exactly noon one prisoner, selected at random without regard for the past, will be taken to the palace where she may switch the light or not at her discretion.

At any time any prisoner may declare that everyone has been to the palace at least once. If she is correct then everyone will be set free, otherwise they will all be subjected to mathgrant's music. Since nobody wants even the slightest chance of having to listen to mathgrant's music they all agree that they won't make any declaration unless they are 100% percent sure. Given that they are all in good health and can be expected to survive indefinitely, give a method that can ensure release eventually.