Older riddles and logic problems

[Divinetrust]

I got an easy one to start with:
"How far can you walk into a forest?"

halfway, cuz then you're walking out

[Deleted]

The only thing that I can think of is that each prisoner promises only to toggle switch #1 until it's their third or fourth time in room 0. Then they flip switch #2 and the next person to enter to see #2 is in another position can state that everyone has been in the room at least once.

You're on track, I suppose, but what if you have two people one after each other who have to flip the second switch because as you said it's their 3rd or 4th time in room 0? Any person after those two will see that switch be in the position in which it was before even the first of those two other people entered, thus, losing count.

Since the warden can be as mean as he wants to, he might not send someone even once for a LOOOOOOOOOOONG time, so keep that in mind.

I'll give a hint though, only 1 person is (or should be) able to call out that all 23 of them have gone to room 0.

Also, Arnorei, nope, there is no way of communication, other than toggling the switches themselves. It's all about the switches, no gimmicks or tricks.

As for the daughter one, assuming that all the children are from the same mother, 3 daughters having 2 brothers each means there are 2 boys in total, since they (the brothers) aren't mutually exclusive to the daughters. So, 5. Woops, I missed Samin's post where he already answered it.

[Deleted]

Sorry for the spoiler tags, Going away for a day or so and would like people to at least know the answers if they choose to.

It takes 10 men 5 Hours to dig a hole 10ft Square, and 3ft Deep. How long does it take 5 Men to dig Half a hole ?
You cant Dig half a hole

I went to the woods and I found it, When I found it I could not see it, when I saw it I could not feel it, When I Felt it I took it home in my hand, What was it ?
A splinter

The Man who made it did Not want it, The Woman who bought it did not need it. The man that used it did not know he was using it. What was it ?
A coffin

You are the driver of an electric train, the train carry's a maximum 100 passangers. upon leaving the depot you come to the first stop where 10 people get on the train, the next stop 3 people get on and 2 people get off, the train then heads west for an hour and comes to another station where 12 people get on, at the next stop 5 people get on and 10 people get off. The last leg of the trains journey is south, there is a westerly wind blowing at 10mph. What way is the wind blowing the smoke from the trains engine and what is the train drivers name ?
There is no somke, its an electric train, You are the driver of the train

[Arcangnom]

Question about the warden riddle: Will the prisoners know when the first prisoner is taken to room 0 (so they can know that they're not the first)?

[Todgruppe]

There's a two mile track, the first mile you average 30mph. How fast do you have to go to average 60mph?

There are two circles, one is four inches in circumference the other is one inch. How many times does the one inch circle rotate while revolving around the four inch circle?

[Deleted]

Question about the warden riddle: Will the prisoners know when the first prisoner is taken to room 0 (so they can know that they're not the first)?

Nope, no-one has any idea who's the first to enter, and the original positions of the switches are unknown.

[Arcangnom]

Two mile track: You can't. You already spent 2 minutes driving the first mile. You'd have to drive the second mile in literally 0 time to average 60 mph.

[Divinetrust]

Two mile track: You can't. You already spent 2 minutes driving the first mile. You'd have to drive the second mile in literally 0 time to average 60 mph.

its not a timed race, just cuz they take 2 minutes to go the first mile doesnt mean they are out of time. It just means after 2 minutes they were half done. If they could floor it and go 90mph around the second mile, it would ave to 60.

[Braindwen]

its not a timed race, just cuz they take 2 minutes to go the first mile doesnt mean they are out of time. It just means after 2 minutes they were half done. If they could floor it and go 90mph around the second mile, it would ave to 60.

The flaw here is that they did 1 mile at 30mph. They would have to run the same amount of time at 90mph to average 60mph, which would be 3 miles. In order to average 60mph over a 2 mile run, after doing 1 mile at 30mph (spending 2 minutes), they'd have to do the second mile in 0 seconds, aka infinite speed.

Warden riddle: Prisoners know which cell # they're in, yes?

[Deleted]

warden riddle: just a few guesses. Obviously they would have to communicate before they go in their own cells and make one the "leader", the leader is the only one who can say that everyone was in it. Then they would have to agree that everyone switches one up, if switch one is up they switch the second switch. the leader can switch the first switch down. If that switch is already down he switches 2. So that would mean that 2 isn't bothering anyone and only switch one is important. If the leader switched the switch 1 23 times up, he can say that everyone was in it. Just a lucky guess

[Deleted]

"7. A mother has 3 daughters. Each of these 3 girls has 2 brothers, how many children does the mother have?"

5?

[Primernova]

this one is pure evil and I have never seen it be airtight solved, enjoy.

[Arcangnom]

warden riddle: just a few guesses. Obviously they would have to communicate before they go in their own cells and make one the "leader", the leader is the only one who can say that everyone was in it. Then they would have to agree that everyone switches one up, if switch one is up they switch the second switch. the leader can switch the first switch down. If that switch is already down he switches 2. So that would mean that 2 isn't bothering anyone and only switch one is important. If the leader switched the switch 1 23 times up, he can say that everyone was in it. Just a lucky guess

This is along the lines of what I've been thinking, but the system is flawed. If the first switch starts out being up, the leader would risk making a call that all 23 have been in the room when only 22 different people has been there for sure (assuming everyone will only switch the first switch once). Of course, the chances would be really really slim that the last person wouldn't have been in there while the switch was already up (assuming people are chosen at random), but it's not guaranteed. Not knowing the starting position of the switches AND the first person in not knowing he's the first person in makes this quite complicated.
I'm really squeezing my brain for this one Love it!

[Sykol]

this one is pure evil and I have never seen it be airtight solved, enjoy.

I'm thinking Phil should be 22, since an 11 year difference would put Paul at 22 when Phil was 11. 11x3=33. But I may be way off and he's probably a dead twin or something.

[Braindwen]

warden riddle: just a few guesses. Obviously they would have to communicate before they go in their own cells and make one the "leader", the leader is the only one who can say that everyone was in it. Then they would have to agree that everyone switches one up, if switch one is up they switch the second switch. the leader can switch the first switch down. If that switch is already down he switches 2. So that would mean that 2 isn't bothering anyone and only switch one is important. If the leader switched the switch 1 23 times up, he can say that everyone was in it. Just a lucky guess

Expanding on this (just going to assume prisoners know their own cell #). edit: I guess this part actually doesn't matter. Let's assume all the prisoners name themselves #1 through #23, so I don't have to change anything else. They can just decide #1 does the calling, cell # doesn't matter.
Assign #1 as the leader. Whenever the leader enters cell 0:
If switch 1 is UP, switch #1 to down.
If switch 1 is DOWN, switch #2 to whatever it isn't.

Every other prisoner:
If switch 1 is DOWN and that prisoner hasn't changed switch #1 before, switch #1 to UP
If switch 1 is UP or that prisoner has already switched #1 before, switch #2 to whatever it isn't.

So, every prisoner switches switch #1 once except for prisoner 1, so he'll know when everyone's switched it...

EXCEPT!!!!!

The prisoners don't know whether switch 1 begins up or down. If switch 1 starts up, all 23 prisoners will have entered the room when #1 enters the room and switches switch 1 down the 23rd time. If switch 1 begins down, prisoner #1 will switch it down 22 times by the time the last prisoner has entered, but no one will ever switch it up again for him to switch it down, and he'll never be certain that all 23 have entered (he can GUESS, but it's possible only 22 have entered and the last prisoner has just never gone in to switch it up again so they all die).

SO!

Instead, switch prisoner 2-23's logic just a little bit:
If switch 1 is DOWN and that prisoner hasn't changed switch #1 twice before, switch #1 to UP
If switch 1 is UP or that prisoner has already switched #1 twice before, switch #2 to whatever it isn't.

Prisoner 1 remains the same.

After Prisoner 1 has switched switch #1 down 44 times, declare everyone has been in the room:

at 43 times, if switch #1 started up, Prisoner 1 would have switched it down, and then it's possible that 21 of the other prisoners all came in twice leading to 43 total switches, with 1 prisoner never entering.
at 45 times, if switch #1 started down, 22 prisoners all entering twice would lead to 44 total switches, and prisoner 1 would never call it.

at 44 times, if switch #1 started up, 21 prisoners would enter twice and the last would enter once. If switch #1 started down, 22 prisoners would all enter twice. In both cases, the scenario ends eventually, and prisoner #1 knows that everyone's entered at least once.

[Deleted]

I think I'll post a riddle I found a year ago or so. It's not really hard but it's kinda long:


A warden meets with 23 prisoners. He tells them the following:

Each prisoner will be placed into a room numbered 1-23. Each will be alone in the room, which will be soundproof, lightproof, etc. In other words, they will NOT be able to communicate with each other.

They will be allowed one planning session before they are taken to their rooms.

There is a special room, room 0. In this room are 2 switches, which can each be either UP or DOWN. They cannot be left in between, they are not linked in any way (so there are 4 possible states), and they are numbered 1 and 2. Their current positions are unknown.

One at a time, a prisoner will be brought into room 0. The prisoner MUST change one and only one switch. The prisoner is then returned to his cell.

At any time t, given some N>0, there exists a finite t_0 by which time every prisoner will have visited room 0 at least N times. (In other words, there is no fixed pattern to the order or frequency with which prisoners visit room 0, but at any given time, every prisoner is guaranteed to visit room 0 again. If you’re still confused by this statement, ignore it, and you should be ok).

At any time, any prisoner may declare that all 23 of them have been in room 0. If right, the prisoners go free. If wrong, they are all executed.

What initial strategy is 100% guaranteed to let all go free?

Ended up googling it. Not that I was on the right track - had the right idea, but I missed out on something said in the beginning and that threw me off completely. Also, the way you worded it confused me (not to say you didn't say it right, I just got it better after reading it somewhere else). All in all, good one.

Edit: Oh, it was already answered. I thought of some way of counting and signaling through the switches, but I couldn't figure out how they communicate the code to each other and I started focusing on that (forgetting they are allowed to hold a meeting at the start). /facepalm

[Ardrienne]

The person who makes them doesn't want them himself. The person who wants one obviously doesn't need it or he wouldn't be there getting one. And the one who actually needs it isn't really in a position to notice much of anything...

Is the answer a coffin?

[Deleted]

I really enjoyed this riddle, please let me know if I'm on the right track.

The prisoners nominate a control speaker to make the call (thanks for the hint). They devise a system of communication using the switches. On entering the prisoner does the following:

• The prisoners aim is to flip Switch 1 down. If it is already down he arbitrary flips Switch 2. If it is up and he has not yet flipped Switch 1 down, they do so flip it.
• The assigned leader is the control. Upon entering he flips Switch 1 up and takes mental note (1 count), resetting for the next new prisoner. If it is still up he arbitrary flips Switch 2. He keeps doing this until he has counted 22.

However this is where it gets tricky, what if the switch started down? They would have to flip the control switch down twice, totaling to 44 flips. There will just be one prisoner who only flips it down once and rest twice.

Freedom?

Edit: Someone beat me by 20 mins. Damn you sir.

[Deleted]

Expanding on this (just going to assume prisoners know their own cell #). edit: I guess this part actually doesn't matter. Let's assume all the prisoners name themselves #1 through #23, so I don't have to change anything else. They can just decide #1 does the calling, cell # doesn't matter.
Assign #1 as the leader. Whenever the leader enters cell 0:
If switch 1 is UP, switch #1 to down.
If switch 1 is DOWN, switch #2 to whatever it isn't.

Every other prisoner:
If switch 1 is DOWN and that prisoner hasn't changed switch #1 before, switch #1 to UP
If switch 1 is UP or that prisoner has already switched #1 before, switch #2 to whatever it isn't.

So, every prisoner switches switch #1 once except for prisoner 1, so he'll know when everyone's switched it...

EXCEPT!!!!!

The prisoners don't know whether switch 1 begins up or down. If switch 1 starts up, all 23 prisoners will have entered the room when #1 enters the room and switches switch 1 down the 23rd time. If switch 1 begins down, prisoner #1 will switch it down 22 times by the time the last prisoner has entered, but no one will ever switch it up again for him to switch it down, and he'll never be certain that all 23 have entered (he can GUESS, but it's possible only 22 have entered and the last prisoner has just never gone in to switch it up again so they all die).

SO!

Instead, switch prisoner 2-23's logic just a little bit:
If switch 1 is DOWN and that prisoner hasn't changed switch #1 twice before, switch #1 to UP
If switch 1 is UP or that prisoner has already switched #1 twice before, switch #2 to whatever it isn't.

Prisoner 1 remains the same.

After Prisoner 1 has switched switch #1 down 44 times, declare everyone has been in the room:

at 43 times, if switch #1 started up, Prisoner 1 would have switched it down, and then it's possible that 21 of the other prisoners all came in twice leading to 43 total switches, with 1 prisoner never entering.
at 45 times, if switch #1 started down, 22 prisoners all entering twice would lead to 44 total switches, and prisoner 1 would never call it.

at 44 times, if switch #1 started up, 21 prisoners would enter twice and the last would enter once. If switch #1 started down, 22 prisoners would all enter twice. In both cases, the scenario ends eventually, and prisoner #1 knows that everyone's entered at least once.

DING DING! You are correct sir, good job !

---------- Post added 2012-06-08 at 11:08 AM ----------

Here's another one for you, our English teacher asked us this one in 3rd grade of junior high, so it's not that hard, definitely easier than the warden riddle:

3 men were caught stealing, they were taken to the king. The king being a forgiving person said that he has 4 hats in total, 2 white and 2 black.

He then puts them in a line where the furthest back can see the other two, the one in the middle can see only the first person and the first person couldn't see anyone else. No communication whatsoever is allowed. The king then proceeded to put the hats on them and said "If any of you finds out what colour hat he's wearing in 10 minutes then I'll let all of you go, if not, I'll throw you in prison."

5 minutes passed and no-one has come up with an answer. Then suddenly, the middle man yells out his hat's colour correctly ! How did he do this ?

[Reluctant]

this one is pure evil and I have never seen it be airtight solved, enjoy.

Im thinking its 22? couse the difference between themm is 11 years.