The Brain Teaser Thread

[ShadowOfMyself]

Originally posted by: coldmeat

Originally posted by: TuxDave

Originally posted by: brikis98

Originally posted by: brikis98

Originally posted by: TuxDave
Here's a puzzle that was told to me that I kinda liked (and probably retold numerous times in this forum)

100 pirates stumble across a box with 5050 coins in it. Someone came up with a game with these rules:

They would write the numbers 1,2,3...98,99,100 on separate pieces of paper and toss them in a hat. They would randomly line up and each pick out a number without replacing and secretly look at it. That number would represent the number of coins they would take out of the box.

After the round completed, everyone who was unhappy with their number could toss their number back in the hat and they would randomly line up and pick a new one from the ones tossed in. Those who were happy would keep the number and the corresponding number of coins.

What is the best strategy of the game assuming everyone there was a logic mastermind?

interesting question. just to make sure i understand it correctly, is this what's happening?

1. In random order, all 100 pirates pick out a number from the hat.
2. AFTER everyone has picked, any of the pirates not happy with their numbers return JUST their numbers to the hat, get in line (in random order?), and pick again.
3. This continues until all numbers are used up.
4. Can a pirate choose to sit out one round of picking and jump in at a later one?

actually, if they are all logic masterminds and assuming they are maximally greedy, wouldn't the following happen:

1. They all pick their numbers.
2. The pirate who got 100 is happy, he won't put his number back in the hat.
3. The pirate who got 99 knows that someone else got 100 and that guy won't put it back in the hat. Therefore, the pirate who got 99 can't do any better and he holds on to his number.
4. The pirate who got 98 knows #2 and #3 above, and for the same reasoning holds on to his number.
5. This continues on down the line. No one would put their number back in the hat, what you got on the first try is what you keep.

And that would be the right answer.

So the answer is they would do nothing? Not much of a puzzle.

LOL it was kinda funny though... Didnt think about it at first, but it makes total sense... Its just ridiculous when you think about a guy getting 5 coins and not taking them back because he knows anyone above 6 coins wont either, etc

 

[TuxDave]

Originally posted by: coldmeat

Originally posted by: TuxDave

Originally posted by: brikis98

Originally posted by: brikis98

Originally posted by: TuxDave
Here's a puzzle that was told to me that I kinda liked (and probably retold numerous times in this forum)

100 pirates stumble across a box with 5050 coins in it. Someone came up with a game with these rules:

They would write the numbers 1,2,3...98,99,100 on separate pieces of paper and toss them in a hat. They would randomly line up and each pick out a number without replacing and secretly look at it. That number would represent the number of coins they would take out of the box.

After the round completed, everyone who was unhappy with their number could toss their number back in the hat and they would randomly line up and pick a new one from the ones tossed in. Those who were happy would keep the number and the corresponding number of coins.

What is the best strategy of the game assuming everyone there was a logic mastermind?

interesting question. just to make sure i understand it correctly, is this what's happening?

1. In random order, all 100 pirates pick out a number from the hat.
2. AFTER everyone has picked, any of the pirates not happy with their numbers return JUST their numbers to the hat, get in line (in random order?), and pick again.
3. This continues until all numbers are used up.
4. Can a pirate choose to sit out one round of picking and jump in at a later one?

actually, if they are all logic masterminds and assuming they are maximally greedy, wouldn't the following happen:

1. They all pick their numbers.
2. The pirate who got 100 is happy, he won't put his number back in the hat.
3. The pirate who got 99 knows that someone else got 100 and that guy won't put it back in the hat. Therefore, the pirate who got 99 can't do any better and he holds on to his number.
4. The pirate who got 98 knows #2 and #3 above, and for the same reasoning holds on to his number.
5. This continues on down the line. No one would put their number back in the hat, what you got on the first try is what you keep.

And that would be the right answer.

So the answer is they would do nothing? Not much of a puzzle.

I didn't see youuuuuu posting an answer.

 

[coldmeat]

Originally posted by: TuxDave
I didn't see youuuuuu posting an answer.

 

[Chaotic42]

Originally posted by: Killmenow
Can't you just follow the same logic and break the bars into 10 equal parts... should require less effort...

But the grinder is due back to the rental place in four hours and we've barely used it.

 

[her209]

You have a plane. You have a conveyor belt.

 

[Atomic Playboy]

Originally posted by: her209
You have a plane. You have a conveyor belt.

The plane doesn't fit.

 

[hellokeith]

Use a RAID 5 or RAID 6 setup, starting with the first engineer:

"I am #1. #2 has a black hat, #3 has a black hat, and #4 has a red hat. My hat color is xxx". - 50/50 chance

Next engineer
"I am #2. #3 has a black hat, #4 has a red hat, and #5 has a black hat. My hat color is black." - 100% assuming #1 was honest and accurate.

Etc. Each engineer will hear his # and hat color 3 times, and then he can decide if A) the color matches 3 times and B) does he trust the previous 3 engineers.

Assuming all engineers are honest and accurate, only the first engineer may die.

Next puzzle:

A man is placed in a steel cage. The door is welded shut. 8 steel walls are placed around the cage and all welded together, forming an air-tight cube around the cage. The man's air supply is now time limited.

He has:
* his clothing with no accessories
* a wooden chair
* a wooden table
* a hand saw (no, the saw is not capable of cutting through a steel cage let alone a steel wall
* a small led flashlight (no, you can't use the battery acid in the led light battery to burn through cage bars let alone a steel wall)

How does he escape?

 

[coldmeat]

Originally posted by: hellokeith
Use a RAID 5 or RAID 6 setup, starting with the first engineer:

"I am #1. #2 has a black hat, #3 has a black hat, and #4 has a red hat. My hat color is xxx". - 50/50 chance

Next engineer
"I am #2. #3 has a black hat, #4 has a red hat, and #5 has a black hat. My hat color is black." - 100% assuming #1 was honest and accurate.

Etc. Each engineer will hear his # and hat color 3 times, and then he can decide if A) the color matches 3 times and B) does he trust the previous 3 engineers.

Assuming all engineers are honest and accurate, only the first engineer may die.

It has already been explained what the best way is, and in the original puzzle the engineers aren't allowed to say anything other than "red" or "black".

edit: and 8 walls don't make a cube.

 

[CycloWizard]

Originally posted by: brikis98
can you explain this i don't even get it

try out some examples.

say initially you had 25 coins, 10 of them tails up. when the lights go off, pick up any 10 and make one stack. take the rest and make a second stack. flip the first stack (of 10) over. What could have happened:

1. All 10 coins that were tails ended up in the first stack. When you flip them, they are all heads up, so now both stacks have 0 coins tails up.
2. All 10 coins that were tails up ended up in the second stack. When you flip the first stack, all 10 of those coins are now tails up as well. You now have 10 tails up coins in each stack.
3. Some combination in between. Say you got 3 coins in the first stack tails up (7 heads up) and the other 7 are in the second stack. When you flip the first stack over, you now have 7 tails up, so both stacks are again equal.

If you work it out, you'll find that MotionMan's strategy works for any starting position where there are exactly N coins tails up.[/quote]
This approach works only in the case where the total number of coins is >=2N. If I have 11 coins and 10 are tails up, then it all falls apart. Thus, "many coins on the table" is sufficient if it implies that the number of coins is >=2N.

 

[her209]

Stolen from another thread:

You're sitting in the middle of a circular lake in a canoe. There's a goblin waiting for you on the shore; He'll always run to the closest point on the shore to your canoe, and he can run four times faster than you can row. How do you escape the lake without being eaten once you hit the shore?

 

[Basilisk6]

Originally posted by: CycloWizard

Originally posted by: brikis98
can you explain this i don't even get it

try out some examples.

say initially you had 25 coins, 10 of them tails up. when the lights go off, pick up any 10 and make one stack. take the rest and make a second stack. flip the first stack (of 10) over. What could have happened:

1. All 10 coins that were tails ended up in the first stack. When you flip them, they are all heads up, so now both stacks have 0 coins tails up.
2. All 10 coins that were tails up ended up in the second stack. When you flip the first stack, all 10 of those coins are now tails up as well. You now have 10 tails up coins in each stack.
3. Some combination in between. Say you got 3 coins in the first stack tails up (7 heads up) and the other 7 are in the second stack. When you flip the first stack over, you now have 7 tails up, so both stacks are again equal.

If you work it out, you'll find that MotionMan's strategy works for any starting position where there are exactly N coins tails up.
This approach works only in the case where the total number of coins is >=2N. If I have 11 coins and 10 are tails up, then it all falls apart. Thus, "many coins on the table" is sufficient if it implies that the number of coins is >=2N.

Maybe I'm doing the logic wrong in my head, but why exactly does his approach not work for number of coins < 2N? In your example, for instance (11 coins and 10 tails):

- 2 stacks, 1 with 10 coins, one with 1
- Possible results:
1. All 10 tails in bigger stack, none in smaller stack. Flip coins in bigger stack, 0 tails in either one
2. 9 tails in bigger stack, 1 in smaller stack. Flip coins in bigger stack, 1 tail in each stack.

 

[Killmenow]

Originally posted by: her209
Stolen from another thread:

You're sitting in the middle of a circular lake in a canoe. There's a goblin waiting for you on the shore; He'll always run to the closest point on the shore to your canoe, and he can run four times faster than you can row. How do you escape the lake without being eaten once you hit the shore?

You leave your canoe and swim?

 

[thecrecarc]

Originally posted by: Killmenow

Originally posted by: her209
Stolen from another thread:

You're sitting in the middle of a circular lake in a canoe. There's a goblin waiting for you on the shore; He'll always run to the closest point on the shore to your canoe, and he can run four times faster than you can row. How do you escape the lake without being eaten once you hit the shore?

You leave your canoe and swim?

go to one side of the island, he will wait for you on the shore. then quickly row to the opposite side?

That's all I got.

 

[TuxDave]

Originally posted by: her209
Stolen from another thread:

You're sitting in the middle of a circular lake in a canoe. There's a goblin waiting for you on the shore; He'll always run to the closest point on the shore to your canoe, and he can run four times faster than you can row. How do you escape the lake without being eaten once you hit the shore?

I like math riddles. .

If the lake had a radius r, put your canoe r/4-some infinitely small amount from the center and start paddling in circles.

The circumference you paddle is smaller than 1/4 the circumference that he runs on the outside so you should outpace him. Keep going till he's exactly on the other side of the canoe and then paddle straight to shore.

The distance you travel = 3/4*r
The distance he needs to travel = pi * r > 4x your distance so you should get to shore before he gets to you.

Then you die.

 

[Killmenow]

Originally posted by: thecrecarc

Originally posted by: Killmenow

Originally posted by: her209
Stolen from another thread:

You're sitting in the middle of a circular lake in a canoe. There's a goblin waiting for you on the shore; He'll always run to the closest point on the shore to your canoe, and he can run four times faster than you can row. How do you escape the lake without being eaten once you hit the shore?

You leave your canoe and swim?

go to one side of the island, he will wait for you on the shore. then quickly row to the opposite side?

That's all I got.

Well, according to the post, he's only after your canoe =] Just hop out and swim it.

 

[brikis98]

Originally posted by: Basilisk6

Originally posted by: CycloWizard

Originally posted by: brikis98
can you explain this i don't even get it

try out some examples.

say initially you had 25 coins, 10 of them tails up. when the lights go off, pick up any 10 and make one stack. take the rest and make a second stack. flip the first stack (of 10) over. What could have happened:

1. All 10 coins that were tails ended up in the first stack. When you flip them, they are all heads up, so now both stacks have 0 coins tails up.
2. All 10 coins that were tails up ended up in the second stack. When you flip the first stack, all 10 of those coins are now tails up as well. You now have 10 tails up coins in each stack.
3. Some combination in between. Say you got 3 coins in the first stack tails up (7 heads up) and the other 7 are in the second stack. When you flip the first stack over, you now have 7 tails up, so both stacks are again equal.

If you work it out, you'll find that MotionMan's strategy works for any starting position where there are exactly N coins tails up.
This approach works only in the case where the total number of coins is >=2N. If I have 11 coins and 10 are tails up, then it all falls apart. Thus, "many coins on the table" is sufficient if it implies that the number of coins is >=2N.

Maybe I'm doing the logic wrong in my head, but why exactly does his approach not work for number of coins < 2N? In your example, for instance (11 coins and 10 tails):

- 2 stacks, 1 with 10 coins, one with 1
- Possible results:
1. All 10 tails in bigger stack, none in smaller stack. Flip coins in bigger stack, 0 tails in either one
2. 9 tails in bigger stack, 1 in smaller stack. Flip coins in bigger stack, 1 tail in each stack.

edit: woops, misread your post, as you're saying, everything works fine... not sure what issue Cyclo is seeing...

 

[brikis98]

Originally posted by: hellokeith
Use a RAID 5 or RAID 6 setup, starting with the first engineer:

"I am #1. #2 has a black hat, #3 has a black hat, and #4 has a red hat. My hat color is xxx". - 50/50 chance

Next engineer
"I am #2. #3 has a black hat, #4 has a red hat, and #5 has a black hat. My hat color is black." - 100% assuming #1 was honest and accurate.

Etc. Each engineer will hear his # and hat color 3 times, and then he can decide if A) the color matches 3 times and B) does he trust the previous 3 engineers.

Assuming all engineers are honest and accurate, only the first engineer may die.

Next puzzle:

A man is placed in a steel cage. The door is welded shut. 8 steel walls are placed around the cage and all welded together, forming an air-tight cube around the cage. The man's air supply is now time limited.

He has:
* his clothing with no accessories
* a wooden chair
* a wooden table
* a hand saw (no, the saw is not capable of cutting through a steel cage let alone a steel wall
* a small led flashlight (no, you can't use the battery acid in the led light battery to burn through cage bars let alone a steel wall)

How does he escape?

as coldmeat explained, the proper solution was already posted and you can't say anything other than "red" or "black", or all engineers are executed.

as for your puzzle:

1. why 8 steel walls?
2. i hope this isn't like the silly kids puzzle where you cut the table in half and then put the two halves together to make a "hole" [sic] that you can climb out of.

 

[thecrecarc]

Originally posted by: brikis98

Originally posted by: hellokeith
Use a RAID 5 or RAID 6 setup, starting with the first engineer:

"I am #1. #2 has a black hat, #3 has a black hat, and #4 has a red hat. My hat color is xxx". - 50/50 chance

Next engineer
"I am #2. #3 has a black hat, #4 has a red hat, and #5 has a black hat. My hat color is black." - 100% assuming #1 was honest and accurate.

Etc. Each engineer will hear his # and hat color 3 times, and then he can decide if A) the color matches 3 times and B) does he trust the previous 3 engineers.

Assuming all engineers are honest and accurate, only the first engineer may die.

Next puzzle:

A man is placed in a steel cage. The door is welded shut. 8 steel walls are placed around the cage and all welded together, forming an air-tight cube around the cage. The man's air supply is now time limited.

He has:
* his clothing with no accessories
* a wooden chair
* a wooden table
* a hand saw (no, the saw is not capable of cutting through a steel cage let alone a steel wall
* a small led flashlight (no, you can't use the battery acid in the led light battery to burn through cage bars let alone a steel wall)

How does he escape?

as coldmeat explained, the proper solution was already posted and you can't say anything other than "red" or "black", or all engineers are executed.

as for your puzzle:

1. why 8 steel walls?
2. i hope this isn't like the silly kids puzzle where you cut the table in half and then put the two halves together to make a "hole" [sic] that you can climb out of.

I think that question needed a mirror. It was something like "you look in a mirror and see what you saw. Then you take the saw and cut the table then you jump through the whole/hole of the table to get out"

Yes, it was lame.

 

[brikis98]

bump for more puzzles.

 

[QED]

Here's a good one:

Ann and Bob's safe combination is comprised of two whole numbers, each greater than 1 and less than 100. For security purposes,
neither one are told what the two numbers are. Ann is only told the product of the two numbers, while Bob is only told of their sum.

Ann says "I know their product, but I I don't know what the two numbers are"
Bob says "I knew that you didn't know what the two numbers are. I don't know them either, but I know their sum."
Ann says "Based on what you told me, I now know what the two numbers are."
Bob says "And now based on what you told me, I now know what the two numbers are."

What are the two numbers?

 

[jonks]

Originally posted by: QED
Here's a good one:

Ann and Bob's safe combination is comprised of two whole numbers, each greater than 1 and less than 100. For security purposes,
neither one are told what the two numbers are. Ann is only told the product of the two numbers, while Bob is only told of their sum.

Ann says "I know their product, but I I don't know what the two numbers are"
Bob says "I knew that you didn't know what the two numbers are. I don't know them either, but I know their sum."
Ann says "Based on what you told me, I now know what the two numbers are."
Bob says "And now based on what you told me, I now know what the two numbers are."

What are the two numbers?

After googling the answer, I have determined this riddle is not for regular humans to solve.

 

[Atomic Playboy]

Originally posted by: sirjonk

Originally posted by: QED
Here's a good one:

Ann and Bob's safe combination is comprised of two whole numbers, each greater than 1 and less than 100. For security purposes,
neither one are told what the two numbers are. Ann is only told the product of the two numbers, while Bob is only told of their sum.

Ann says "I know their product, but I I don't know what the two numbers are"
Bob says "I knew that you didn't know what the two numbers are. I don't know them either, but I know their sum."
Ann says "Based on what you told me, I now know what the two numbers are."
Bob says "And now based on what you told me, I now know what the two numbers are."

What are the two numbers?

After googling the answer, I have determined this riddle is not for regular humans to solve.

This is possibly the stupidest riddle I've ever heard. It can be literally any combination of integers between 1 and 100 given the information provided. $20 says that if Ann and Bob wrote down what they thought the numbers were on slips of paper, they'd come up with different answers.

 

[jonks]

Originally posted by: Atomic Playboy

Originally posted by: sirjonk

Originally posted by: QED
Here's a good one:

Ann and Bob's safe combination is comprised of two whole numbers, each greater than 1 and less than 100. For security purposes,
neither one are told what the two numbers are. Ann is only told the product of the two numbers, while Bob is only told of their sum.

Ann says "I know their product, but I I don't know what the two numbers are"
Bob says "I knew that you didn't know what the two numbers are. I don't know them either, but I know their sum."
Ann says "Based on what you told me, I now know what the two numbers are."
Bob says "And now based on what you told me, I now know what the two numbers are."

What are the two numbers?

After googling the answer, I have determined this riddle is not for regular humans to solve.

This is possibly the stupidest riddle I've ever heard. It can be literally any combination of integers between 1 and 100 given the information provided. $20 says that if Ann and Bob wrote down what they thought the numbers were on slips of paper, they'd come up with different answers.

Some of the versions of this riddle I found limit the sun to <100, if that helps. Having read the answer, it wouldn't help me. Unless phrases like "of course, all even numbers can be expressed as the sum of two primes" roll off your tongue with ease, I suggest skipping.

 

[brikis98]

Originally posted by: sirjonk

Originally posted by: Atomic Playboy

Originally posted by: sirjonk

Originally posted by: QED
Here's a good one:

Ann and Bob's safe combination is comprised of two whole numbers, each greater than 1 and less than 100. For security purposes,
neither one are told what the two numbers are. Ann is only told the product of the two numbers, while Bob is only told of their sum.

Ann says "I know their product, but I I don't know what the two numbers are"
Bob says "I knew that you didn't know what the two numbers are. I don't know them either, but I know their sum."
Ann says "Based on what you told me, I now know what the two numbers are."
Bob says "And now based on what you told me, I now know what the two numbers are."

What are the two numbers?

After googling the answer, I have determined this riddle is not for regular humans to solve.

This is possibly the stupidest riddle I've ever heard. It can be literally any combination of integers between 1 and 100 given the information provided. $20 says that if Ann and Bob wrote down what they thought the numbers were on slips of paper, they'd come up with different answers.

Some of the versions of this riddle I found limit the sun to <100, if that helps. Having read the answer, it wouldn't help me. Unless phrases like "of course, all even numbers can be expressed as the sum of two primes" roll off your tongue with ease, I suggest skipping.

It's an interesting riddle, because you definitely CAN see patterns and eliminate lots of possibilities... but it is WAY too much paper work to track it down to the final answer. I could prob write a computer program to simplify it... but, while I'm bored at work, I don't think i'm THAT bored

any1 else got interesting puzzles to post?

 

[TuxDave]

Originally posted by: sirjonk

Originally posted by: QED
Here's a good one:

Ann and Bob's safe combination is comprised of two whole numbers, each greater than 1 and less than 100. For security purposes,
neither one are told what the two numbers are. Ann is only told the product of the two numbers, while Bob is only told of their sum.

Ann says "I know their product, but I I don't know what the two numbers are"
Bob says "I knew that you didn't know what the two numbers are. I don't know them either, but I know their sum."
Ann says "Based on what you told me, I now know what the two numbers are."
Bob says "And now based on what you told me, I now know what the two numbers are."

What are the two numbers?

After googling the answer, I have determined this riddle is not for regular humans to solve.

I like this puzzle but the exhaustive search of finding what numbers qualify after each piece of information is pretty difficult.