1. have you ever heard of professer layton for ds. all it is is brain-teasers.
here are some math related ones.

1.
the distance 3 racehorses can around the racetrack in one minute is listed below

HORSE A: 2 laps
HORSE B: 3 laps
C: 4 laps

The horses line up at the starting line and start running in the same direction. how many minutes until all 3 horses line up at the starting line again

all post more 8) [/b][/i]

2.

3. Is the answer 1 minute?
because the speed of the horses are 2 laps/min, 3 l/m, and 4 l/m so that means in one minute the horses have ran 2, 3, and 4 laps (respectively). And that should also mean that they are all at the finish/start line, right?

4. Originally Posted by DivideByZero
because the speed of the horses are 2 laps/min, 3 l/m, and 4 l/m so that means in one minute the horses have ran 2, 3, and 4 laps (respectively). And that should also mean that they are all at the finish/start line, right?

right!

most people think it is 12, the lowest common multible of the 3 numbers :wink:

5. now HERE'S a hard one

When asked about her birthday , a young woman gives the following information:

"the day after tomorrow, i turn 22, but i was still 19 on new year's day last year"

when is her birthday?

6. Originally Posted by mathformonkeys
Originally Posted by DivideByZero
because the speed of the horses are 2 laps/min, 3 l/m, and 4 l/m so that means in one minute the horses have ran 2, 3, and 4 laps (respectively). And that should also mean that they are all at the finish/start line, right?

right!

most people think it is 12, the lowest common multible of the 3 numbers :wink:
12 would be for if the “laps” and the “minutes” were interchanged in the question – i.e. if the horses took 2 minutes, 3 minutes and 4 minutes to complete one lap, then they would meet up after 12 minutes. Most people fall into the trap of thinking that’s what the question should be, not what the question actually is.

Originally Posted by mathformonkeys
now HERE'S a hard one

When asked about her birthday , a young woman gives the following information:

"the day after tomorrow, i turn 22, but i was still 19 on new year's day last year"

when is her birthday?
2nd of January.

7. The warden meets with the 23 prisoners when they arrive. He tells them:

You may meet together today and plan a strategy, but after today you will be in isolated cells and have no communication with one another.

There is in this prison a "switch room" which contains two light switches, labelled "A" and "B", each of which can be in the "on" or "off" position. I am not telling you their present positions. The switches are not connected to any appliance. After today, from time to time, whenever I feel so inclined, I will select one prisoner at random and escort him to the "switch room", and this prisoner will select one of the two switches and reverse its position (e.g. if it was "on", he will turn it "off"); the prisoner will then be led back to his cell. Nobody else will ever enter the "switch room".

Each prisoner will visit the switch room aribtrarily often. That is, for any N it is true that eventually each of you will visit the switch room at least N times.)

At any time, any of you may declare to me: "We have all visited the switch room." If it is true (each of the 23 prisoners has visited the switch room at least once), then you will all be set free. If it is false (someone has not yet visited the switch room), you will all remain here forever, with no chance of parole.

Devise for the prisoners a strategy which will guarantee their release.

8. Originally Posted by Harold14370
The warden meets with the 23 prisoners when they arrive. He tells them:

You may meet together today and plan a strategy, but after today you will be in isolated cells and have no communication with one another.

There is in this prison a "switch room" which contains two light switches, labelled "A" and "B", each of which can be in the "on" or "off" position. I am not telling you their present positions. The switches are not connected to any appliance. After today, from time to time, whenever I feel so inclined, I will select one prisoner at random and escort him to the "switch room", and this prisoner will select one of the two switches and reverse its position (e.g. if it was "on", he will turn it "off"); the prisoner will then be led back to his cell. Nobody else will ever enter the "switch room".

Each prisoner will visit the switch room aribtrarily often. That is, for any N it is true that eventually each of you will visit the switch room at least N times.)

At any time, any of you may declare to me: "We have all visited the switch room." If it is true (each of the 23 prisoners has visited the switch room at least once), then you will all be set free. If it is false (someone has not yet visited the switch room), you will all remain here forever, with no chance of parole.

Devise for the prisoners a strategy which will guarantee their release.

So is each prisoner picked randomly until all have gone, then they get picked randomly again? In that case on a prisoner's second attempt he will know that everyone has gone before him already so he can say "We have all visited the switch room". right?

9. Originally Posted by DivideByZero
because the speed of the horses are 2 laps/min, 3 l/m, and 4 l/m so that means in one minute the horses have ran 2, 3, and 4 laps (respectively). And that should also mean that they are all at the finish/start line, right?
Why not half a minute? It wasn't said that the answer had to be an integer.

10. Originally Posted by thyristor
Originally Posted by DivideByZero
because the speed of the horses are 2 laps/min, 3 l/m, and 4 l/m so that means in one minute the horses have ran 2, 3, and 4 laps (respectively). And that should also mean that they are all at the finish/start line, right?
Why not half a minute? It wasn't said that the answer had to be an integer.
Not half a minute. After half a minute, Horse A and Horse C are at the starting line, but Horse B isn’t.

11. Originally Posted by DivideByZero
So is each prisoner picked randomly until all have gone, then they get picked randomly again? In that case on a prisoner's second attempt he will know that everyone has gone before him already so he can say "We have all visited the switch room". right?
No, the same prisoner could be picked twice in a row or more.

12. Originally Posted by mathformonkeys
now HERE'S a hard one

When asked about her birthday , a young woman gives the following information:

"the day after tomorrow, i turn 22, but i was still 19 on new year's day last year"

when is her birthday?
2nd of January.[/quote]

right again 8)

13. i know a well-known one, but I'll post it anyway

A glass jar holds a single germ. After 1 minute, the germ splits into two germs.after another minute, the 2 germs split forming 4 germs.continuing
at this rate, a single germ can fill the jar in 1 hour.

knowing this, how long in minutes would it take to fill the jar if you started with 2 germs?

14. That’s easy. 59 minutes.

As for Harold’s puzzle, I’ve seen a version of it before, but I can’t remember the complete solution.

15. Originally Posted by JaneBennet
That’s easy. 59 minutes.

As for Harold’s puzzle, I’ve seen a version of it before, but I can’t remember the complete solution.
right

lol, i showed my dad this one and he said "its unsolvable, you don't know how big the jar is" or something like that :wink:

anyway here's one more,

You've scattered a deck of 52 cards and 1 joker facedown on a table so that you don't know which card is where.

Next you start turning the cards over 1 by 1. Assuming that you cant flip the same card twice, what are the percentage odds that you will turn over all 4 aces before the joker

16. geomtry puzzle

a room is perfectly square and completely empty.later, 4 normal chhickn eggs were placd on the floor of the room. later again, a man came in with a giant steel cylinder and rolled it all over the floor. not a single egg was broken. where were the eggs?

17. Originally Posted by mathformonkeys
geomtry puzzle

a room is perfectly square and completely empty.later, 4 normal chhickn eggs were placd on the floor of the room. later again, a man came in with a giant steel cylinder and rolled it all over the floor. not a single egg was broken. where were the eggs?
I think I solved this one without attempting to:
"4 normal chhickn eggs were placd on the floor"
therefore the answer is: The eggs were placed on the floor.

Is that cheating? :wink:

on a more serious note, were the eggs inside the (hollow) cylinder?

18. Originally Posted by DivideByZero
Originally Posted by mathformonkeys
geomtry puzzle

a room is perfectly square and completely empty.later, 4 normal chhickn eggs were placd on the floor of the room. later again, a man came in with a giant steel cylinder and rolled it all over the floor. not a single egg was broken. where were the eggs?
I think I solved this one without attempting to:
"4 normal chhickn eggs were placd on the floor"
therefore the answer is: The eggs were placed on the floor.

Is that cheating? :wink:

on a more serious note, were the eggs inside the (hollow) cylinder?

BINGO!

not
try again

19. more

this one has no trick to it.(sort of)

a cube has a height of 5 inches. will a rod that has a infinitesimal thickness and a height of 8 inchs fit in the box?

2.

a teacher was yelling at a lazy student and told him the following:

"at the very least, you need to study once a day for an entire week before a test! Don't skimp on time either! Each time you study i want you to study for i minimum of 2 hours!"

the boy wanted to spend as little time doing so, but follow the teachers orders exactly.
how many hours did he end up studying

just for little hint i got an expression

x<14

20. Originally Posted by mathformonkeys
You've scattered a deck of 52 cards and 1 joker facedown on a table so that you don't know which card is where.

Next you start turning the cards over 1 by 1. Assuming that you cant flip the same card twice, what are the percentage odds that you will turn over all 4 aces before the joker
One chance in 5. Each of the 5 cards, 4 aces and joker, have an equal chance of being last turned over.

21. Originally Posted by Harold14370
Originally Posted by mathformonkeys
You've scattered a deck of 52 cards and 1 joker facedown on a table so that you don't know which card is where.

Next you start turning the cards over 1 by 1. Assuming that you cant flip the same card twice, what are the percentage odds that you will turn over all 4 aces before the joker
One chance in 5. Each of the 5 cards, 4 aces and joker, have an equal chance of being last turned over.
right!

22. Originally Posted by mathformonkeys
more
2.

a teacher was yelling at a lazy student and told him the following:

"at the very least, you need to study once a day for an entire week before a test! Don't skimp on time either! Each time you study i want you to study for i minimum of 2 hours!"

the boy wanted to spend as little time doing so, but follow the teachers orders exactly.
how many hours did he end up studying

just for little hint i got an expression

x<14
12 hours. He studies from 11pm to 1am six times.

23. Originally Posted by mathformonkeys
more

this one has no trick to it.(sort of)

a cube has a height of 5 inches. will a rod that has a infinitesimal thickness and a height of 8 inchs fit in the box?
Yes. The diagonal of a cube is the length of a side multiplied by the square root of 3.

a teacher was yelling at a lazy student and told him the following:

"at the very least, you need to study once a day for an entire week before a test! Don't skimp on time either! Each time you study i want you to study for i minimum of 2 hours!"

the boy wanted to spend as little time doing so, but follow the teachers orders exactly.
how many hours did he end up studying
8 hours?

24. Originally Posted by Harold14370
Originally Posted by mathformonkeys
more

this one has no trick to it.(sort of)

a cube has a height of 5 inches. will a rod that has a infinitesimal thickness and a height of 8 inchs fit in the box?
Yes. The diagonal of a cube is the length of a side multiplied by the square root of 3.

right, but all i did was use the pythagorean thingy twice :wink:

a teacher was yelling at a lazy student and told him the following:

"at the very least, you need to study once a day for an entire week before a test! Don't skimp on time either! Each time you study i want you to study for i minimum of 2 hours!"

the boy wanted to spend as little time doing so, but follow the teachers orders exactly.
how many hours did he end up studying
8 hours?
good, if you start studying at 11 pm you'll study for 8 hours. 12 hours is wrong because the 2 hours he studies goes into 2 days, still following his orders

25. Originally Posted by mathformonkeys
Originally Posted by Harold14370
Originally Posted by mathformonkeys
more

this one has no trick to it.(sort of)

a cube has a height of 5 inches. will a rod that has a infinitesimal thickness and a height of 8 inchs fit in the box?
Yes. The diagonal of a cube is the length of a side multiplied by the square root of 3.

right, but all i did was use the pythagorean thingy twice :wink:

a teacher was yelling at a lazy student and told him the following:

"at the very least, you need to study once a day for an entire week before a test! Don't skimp on time either! Each time you study i want you to study for i minimum of 2 hours!"

the boy wanted to spend as little time doing so, but follow the teachers orders exactly.
how many hours did he end up studying
8 hours?
good, if you start studying at 11 pm you'll study for 8 hours. 12 hours is wrong because the 2 hours he studies goes into 2 days, still following his orders
oooo

26. Originally Posted by DivideByZero
Originally Posted by mathformonkeys
Originally Posted by Harold14370
Originally Posted by mathformonkeys
more

this one has no trick to it.(sort of)

a cube has a height of 5 inches. will a rod that has a infinitesimal thickness and a height of 8 inchs fit in the box?
Yes. The diagonal of a cube is the length of a side multiplied by the square root of 3.

right, but all i did was use the pythagorean thingy twice :wink:

a teacher was yelling at a lazy student and told him the following:

"at the very least, you need to study once a day for an entire week before a test! Don't skimp on time either! Each time you study i want you to study for i minimum of 2 hours!"

the boy wanted to spend as little time doing so, but follow the teachers orders exactly.
how many hours did he end up studying
8 hours?
good, if you start studying at 11 pm you'll study for 8 hours. 12 hours is wrong because the 2 hours he studies goes into 2 days, still following his orders
oooo

umm..... do you get it or not?

27. Originally Posted by mathformonkeys
geomtry puzzle

a room is perfectly square and completely empty.later, 4 normal chhickn eggs were placd on the floor of the room. later again, a man came in with a giant steel cylinder and rolled it all over the floor. not a single egg was broken. where were the eggs?
The roller can't get into the corners.

28. this puzzle has 2 different answers, mathematical and logical,but im looking for the logical one.

THere are 10 2-seater cars attached to the fairs ferris wheel.the ferris wheel turns so that one car rotates through the exit platform every minute. so it taks 10 minutes for a full ride.

The wheel began operating at 10 in the morning and shut down 30 minutes later. what is the maximum number of popl that could've rode during that 30 minutes?

29. Originally Posted by Harold14370
Originally Posted by mathformonkeys
geomtry puzzle

a room is perfectly square and completely empty.later, 4 normal chhickn eggs were placd on the floor of the room. later again, a man came in with a giant steel cylinder and rolled it all over the floor. not a single egg was broken. where were the eggs?
The roller can't get into the corners.

d=(^_^)=b
right

30. Originally Posted by mathformonkeys
Originally Posted by JaneBennet
That’s easy. 59 minutes.

As for Harold’s puzzle, I’ve seen a version of it before, but I can’t remember the complete solution.
right

lol, i showed my dad this one and he said "its unsolvable, you don't know how big the jar is" or something like that :wink:

anyway here's one more,

You've scattered a deck of 52 cards and 1 joker facedown on a table so that you don't know which card is where.

Next you start turning the cards over 1 by 1. Assuming that you cant flip the same card twice, what are the percentage odds that you will turn over all 4 aces before the joker
Can solve by using the arithmetic progress anyway right?

31. Originally Posted by Harold14370
The warden meets with the 23 prisoners when they arrive. He tells them:

You may meet together today and plan a strategy, but after today you will be in isolated cells and have no communication with one another.

There is in this prison a "switch room" which contains two light switches, labelled "A" and "B", each of which can be in the "on" or "off" position. I am not telling you their present positions. The switches are not connected to any appliance. After today, from time to time, whenever I feel so inclined, I will select one prisoner at random and escort him to the "switch room", and this prisoner will select one of the two switches and reverse its position (e.g. if it was "on", he will turn it "off"); the prisoner will then be led back to his cell. Nobody else will ever enter the "switch room".

Each prisoner will visit the switch room aribtrarily often. That is, for any N it is true that eventually each of you will visit the switch room at least N times.)

At any time, any of you may declare to me: "We have all visited the switch room." If it is true (each of the 23 prisoners has visited the switch room at least once), then you will all be set free. If it is false (someone has not yet visited the switch room), you will all remain here forever, with no chance of parole.

Devise for the prisoners a strategy which will guarantee their release.
What is the solution for this?

32. Originally Posted by ChrisLee
Originally Posted by Harold14370
The warden meets with the 23 prisoners when they arrive. He tells them:

You may meet together today and plan a strategy, but after today you will be in isolated cells and have no communication with one another.

There is in this prison a "switch room" which contains two light switches, labelled "A" and "B", each of which can be in the "on" or "off" position. I am not telling you their present positions. The switches are not connected to any appliance. After today, from time to time, whenever I feel so inclined, I will select one prisoner at random and escort him to the "switch room", and this prisoner will select one of the two switches and reverse its position (e.g. if it was "on", he will turn it "off"); the prisoner will then be led back to his cell. Nobody else will ever enter the "switch room".

Each prisoner will visit the switch room aribtrarily often. That is, for any N it is true that eventually each of you will visit the switch room at least N times.)

At any time, any of you may declare to me: "We have all visited the switch room." If it is true (each of the 23 prisoners has visited the switch room at least once), then you will all be set free. If it is false (someone has not yet visited the switch room), you will all remain here forever, with no chance of parole.

Devise for the prisoners a strategy which will guarantee their release.
What is the solution for this?
First a hint. then if nobody gets it, I'll give the solution. Switch B is not used to record any information. It is only operated when a prisoner has to change one of the two switches, by rule, and does not want to change switch A.

33. Originally Posted by ChrisLee
Can solve by using the arithmetic progress anyway right?
Are you talking about the cards one, or the germs one? The germs one would be a geometric progression, and the cards one is just using probability I think.

First a hint. then if nobody gets it, I'll give the solution. Switch B is not used to record any information. It is only operated when a prisoner has to change one of the two switches, by rule, and does not want to change switch A.
That hasn't helped much

For two switches, each with two possible staes there is four possible states for the switches to appear in. I thought about how the prisoners could cycle through the four binary states, until the last prisoner reads it off, and knows...but that doesn't work at all with 23 prisoners who are being randomly picked, cause the same states appear more than once.

I can't see any possible way to store the information in the switches, so maybe there is some other means of storing the information?

34. Okay, here’s what happens with the prisoners and the switches.

One of the prisoners will be in charge of counting how many prisoners besides him have visited the switch room. To do this, he always flips the switches so that they are both off or both on. Each of the other prisoners flips the switches so that one of them is on and the other is off, doing so exactly once. (If the swtiches are already one off and one on when he first enters, he leaves them alone and waits for a subsequent revisit when the switches are both on or off.) When the first prisoner sees that the switches have been changed from the both-off or both-on position 22 times, he will know that every one of his fellow inmates has already been in the switch room.

There is probably a simpler solution than the one I’ve just given. The puzzle I originally encountered uses one lightbulb (which can be turned on or off) instead of two switches and so was a bit harder than this one. You should be able to “encode” more information using two switches than one lightbulb, so it should be possible to make the solution above more efficient – but I’m too lazy to work it out right now.

35. Originally Posted by JaneBennet
Okay, here’s what happens with the prisoners and the switches.

One of the prisoners will be in charge of counting how many prisoners besides him have visited the switch room. To do this, he always flips the switches so that they are both off or both on. Each of the other prisoners flips the switches so that one of them is on and the other is off, doing so exactly once. (If the swtiches are already one off and one on when he first enters, he leaves them alone and waits for a subsequent revisit when the switches are both on or off.) When the first prisoner sees that the switches have been changed from the both-off or both-on position 22 times, he will know that every one of his fellow inmates has already been in the switch room.

There is probably a simpler solution than the one I’ve just given. The puzzle I originally encountered uses one lightbulb (which can be turned on or off) instead of two switches and so was a bit harder than this one. You should be able to “encode” more information using two switches than one lightbulb, so it should be possible to make the solution above more efficient – but I’m too lazy to work it out right now.
Jane, there are a couple of problems with your solution. One, the prisoners are not allowed to leave the switches alone. They have to flip one. (See my hint). Secondly the counter/spokesman prisoner will not know on his first visit if the switches were changed or in their original positions, so his count could be off by one.

36. Well, it seems I still don’t fully remember the solution to the other similar problem I came across.

37. Regaring the eggs on floor issue, I'd assumed they'd simply hatched.

Harold - the number 23 appears slightly familiar. Is it not the same number as in the birthdays problem: 23 people randomly collected will have a more than 50% chance that at least two of them will share a birthday?

Is this relevant to this problem?

38. Originally Posted by sunshinewarrior
Harold - the number 23 appears slightly familiar. Is it not the same number as in the birthdays problem: 23 people randomly collected will have a more than 50% chance that at least two of them will share a birthday?

Is this relevant to this problem?
I don’t think so. The problem I came across before had 100 prisoners.

39. Jane, you were close to the solution. The problem with 2 switches, where you have to switch one, reduces to the same as 1 switch where you don't have to change it if you don't want to.
Shanks - the number 23 has no special significance. It's just the number they used in the problem I copied and pasted.

40. Originally Posted by JaneBennet
Originally Posted by sunshinewarrior
Harold - the number 23 appears slightly familiar. Is it not the same number as in the birthdays problem: 23 people randomly collected will have a more than 50% chance that at least two of them will share a birthday?

Is this relevant to this problem?
I don’t think so. The problem I came across before had 100 prisoners.
Well, you see...

I began thinking along th lines of:

Suppose the prisoners decide that the switch on the left will always be left alone, and the switch on the right is the one they will flick as a toggle - ignoring its current position.

hen they need only have one or two nominated prisoners, whose job is, for A, to ensure that he flicks the left hand switch to the 1 position, and for B, that he flicks the left hand switch to the 0 position.

Or some such...

Is there a way, probabilistically, to decide over the course of a year, if both A or B have seen a switch flipped at least once by the other, whether or not the chance of any of them not having visited is effectively zero?

41. Ok. I don't see any way that the prisoners could get out in a relatively short time. :?

They choose one guy that has to assess the position of the switches and then to alert the warden, but then that means that he would have to visit the room at least 22 times, which could mean that the room was visited 22*22 times in total. The other prisoners have to flip the right twitch on if it is off and if they have done it before they have to leave it and flip the first switch in any direction. When the counter guy comes in he flips the right switch off if it is down and flip the left one any way if not. Then every time he comes in and sees the right switch on he counts one extra. So when the counter guy sees the switch stay under for more than 22 times he can tell the warden. Oh, but he can't know if he is the first or last guy in line :? , so he has to assume that he is last. That means he can only alert the warden after the right switch stays put for 44 times. So they have to wait 44*44=1936 (like 5 years) times max!

42. So the prisoners elect a leader. He is the guy who will keep count of how many prisoners have been to the room. The leader follows these rules: if switch A is on, turn it off; otherwise, flip switch B. The nonleaders follow these rules: if switch A is off, turn it on the first two times you encounter this; otherwise, flip switch B. The leader counts the number of times he has to turn A off, and once that number reaches 44, he makes his announcement, and all prisoners are released forthright.

I'll let people think about why this works.

43. Originally Posted by serpicojr
So the prisoners elect a leader. He is the guy who will keep count of how many prisoners have been to the room. The leader follows these rules: if switch A is on, turn it off; otherwise, flip switch B. The nonleaders follow these rules: if switch A is off, turn it on the first two times you encounter this; otherwise, flip switch B. The leader counts the number of times he has to turn A off, and once that number reaches 44, he makes his announcement, and all prisoners are released forthright.

I'll let people think about why this works.
Good thinking, thatr man. I like.

Harold, I presume this is good?

Actually - I've thought of a possible glitch... what if the switch is on to begin with and the leader, unbeknownst to him, is the first person into the switch room? He can;t, in any case, wait for 45 flips, because after a while the switch will be stuck in the off position and he wouldn't know for sure for which reason. No?

44. That's the reason the nonleaders turn it off twice and why the leader counts to 44. The only way he can switch A off without it having been turned on by another prisoner is if he is the first guy in the room and A is on. So the nonleaders must have turned A on at least 43 times by the time he counts 44. Is it possible for one of the 22 nonleaders to not have turned A on if A has been turned on 43 times?

45. Originally Posted by serpicojr
That's the reason the nonleaders turn it off twice and why the leader counts to 44. The only way he can switch A off without it having been turned on by another prisoner is if he is the first guy in the room and A is on. So the nonleaders must have turned A on at least 43 times by the time he counts 44. Is it possible for one of the 22 nonleaders to not have turned A on if A has been turned on 43 times?
No. Got it.

46. Originally Posted by KALSTER
Ok. I don't see any way that the prisoners could get out in a relatively short time. :?

They choose one guy that has to assess the position of the switches and then to alert the warden, but then that means that he would have to visit the room at least 22 times, which could mean that the room was visited 22*22 times in total. The other prisoners have to flip the right twitch on if it is off and if they have done it before they have to leave it and flip the first switch in any direction. When the counter guy comes in he flips the right switch off if it is down and flip the left one any way if not. Then every time he comes in and sees the right switch on he counts one extra. So when the counter guy sees the switch stay under for more than 22 times he can tell the warden. Oh, but he can't know if he is the first or last guy in line :? , so he has to assume that he is last. That means he can only alert the warden after the right switch stays put for 44 times. So they have to wait 44*44=1936 (like 5 years) times max!
You got it, but it wasn't stated how often the prisoners visited the switch room. Could be daily or could be more or less.

Nevertheless the strategy is correct. Except maybe if the visits were so infrequent that the prison terms could exceed a prisoner's natural life, in which case it might pay to take an educated guess.

47. I found the site where I first saw the problem. It appears that the solution given there is wrong.

http://www.mathisfunforum.com/viewtopic.php?id=4384

That problem uses 100 rather than 23 prisoners – in which case, by Kalster’s and Serpico’s reasoning, the correct number of times should be 198, not 99.

48. Jane, that one's a little different in that the prisoners choose the initial position. That fact is buried in the middle of the thread.

49. new one

2 us coins make up 55 cents. 1 of them is not a nickel

what are they?

just think

50. Originally Posted by mathformonkeys
new one

2 us coins make up 55 cents. 1 of them is not a nickel

what are they?

just think
A 50 cent piece and a nickel.

51. A mother is 21 years older than her son. In 6 years this son is 5 times younger than his mother.

Question: Where is the father? 8)

52. Originally Posted by accountabled
A mother is 21 years older than her son. In 6 years this son is 5 times younger than his mother.

Question: Where is the father? 8)
huh?......

53. It’s just a joke. :P

54. Originally Posted by accountabled
A mother is 21 years older than her son. In 6 years this son is 5 times younger than his mother.

Question: Where is the father? 8)
Inside of the mommy, I believe. Kid's age is -3/4 when he is 21 years younger than mommy, which is -9 months, i.e. the baby's conception.

55. I originally posted this in the “Logic puzzles” but I’m moving it here because the puzzle involves maths as well.

Five girls read five books by different authors, taking different numbers of days to finish reading. From the clues below, work out which girl read which author in how many days.

(1) The average number of days taken by the girls to read their books was 7, but no girl finished her book in exactly 7 days. No two girls took the same number of days to finish reading, and the number of days each girl took to finish her book is an integer.

(2) Elizabeth was not the one who read Margaret Atwood; the latter took an odd number of days to finish reading while the number of days taken by the former to complete her book was a prime number.

(3) The girl who read D.H. Lawrence could not put her book down; not surprisingly she completed her reading in the shortest time, zipping through her steamy novel in only 4 days. On the other hand, the girl who chose to read Stephen King took over a week to get over her horror novel – only one other girl took even longer to finish her book.

(4) If l is the number of days Lydia took to finish her book, m is the number of days Mary took to finish hers, and n is the number of days the girl who read Michael Crichton took to complete her reading, then both l+m and m+n are odd numbers.

(5) Jane was the slowest reader, while Kitty, who did not read the Charles Dickens book, was not the fastest.

56. This is one isn't really a math puzzle, but it's one of my favorites. This thread has tickled my brain quite a bit though.

You're in a totally enclosed room with 3 switches. You're told they operate 3 lights in a room on the other side of the building. There is no possible way that you could operate the switches while viewing the lights (just a statement to close that loophole).

You're told you can operate the switches one time, in whatever manner you choose. Then, you can walk to other side of the building, and check on the three lights, one time, and one time only. How can you tell which switch operates which light, with 100% accuracy.

Hint: Peacocks and Bananas. (......)

57. Originally Posted by the great Gizmo
This is one isn't really a math puzzle, but it's one of my favorites. This thread has tickled my brain quite a bit though.

You're in a totally enclosed room with 3 switches. You're told they operate 3 lights in a room on the other side of the building. There is no possible way that you could operate the switches while viewing the lights (just a statement to close that loophole).

You're told you can operate the switches one time, in whatever manner you choose. Then, you can walk to other side of the building, and check on the three lights, one time, and one time only. How can you tell which switch operates which light, with 100% accuracy.

Hint: Peacocks and Bananas. (......)

58. Originally Posted by mathformonkeys
geomtry puzzle

a room is perfectly square and completely empty.later, 4 normal chhickn eggs were placd on the floor of the room. later again, a man came in with a giant steel cylinder and rolled it all over the floor. not a single egg was broken. where were the eggs?
In the corners, I guess.

59. Originally Posted by accountabled
A mother is 21 years older than her son. In 6 years this son is 5 times younger than his mother.
I'm not even sure what "5 times younger" would mean...

60. Originally Posted by JaneBennet
Five girls read five books by different authors, taking different numbers of days to finish reading. From the clues below, work out which girl read which author in how many days.

(1) The average number of days taken by the girls to read their books was 7, but no girl finished her book in exactly 7 days. No two girls took the same number of days to finish reading, and the number of days each girl took to finish her book is an integer.

(2) Elizabeth was not the one who read Margaret Atwood; the latter took an odd number of days to finish reading while the number of days taken by the former to complete her book was a prime number.

(3) The girl who read D.H. Lawrence could not put her book down; not surprisingly she completed her reading in the shortest time, zipping through her steamy novel in only 4 days. On the other hand, the girl who chose to read Stephen King took over a week to get over her horror novel – only one other girl took even longer to finish her book.

(4) If l is the number of days Lydia took to finish her book, m is the number of days Mary took to finish hers, and n is the number of days the girl who read Michael Crichton took to complete her reading, then both l+m and m+n are odd numbers.

(5) Jane was the slowest reader, while Kitty, who did not read the Charles Dickens book, was not the fastest.
First find out how many days each girl took to read their book. 4 is mentioned in the clues, and that's also the smallest number. So , where . None of a, b, c, d are equal to 7 and at least two of them are greater than 7. If three of them are greater than 7, they must be 8, 9, and 10. They would take 27 days between them. This would leave a=4, a contradiction. Therefore there must be exactly two of them greater than 7. This has to leave a=5 and b=6. And so c + d = 20, with . The only possibilities are 8 + 12 = 20 and 9 + 11 = 20. From clue (2), there must be at least two odd numbers, so it can't be 8+12, hence it must be 9+11.

So the number of days taken to read the books are (from fastest to slowest): 4, 5, 6, 9, 11. And that's as far as I've gotten.

61. Good work, Faldo! :-D

Now I hope you go on and solve the puzzle as well. It’s been such a long time since I posted it that, would you believe it, I’ve lost the solution to my own puzzle. :?

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