Solve Enjoyable Enigmas with Mr.E's Challenge

[zoobyshoe]

Spoiler

b) There are no monks with red eyes, but they are not 100% sure if the tourist is telling the truth or lying. The original wording said that something dramatic happens so we can assume that at least one monk will attempt to gouge one of his eyes out (that's the dramatic part). The rest of the monks, seeing a brown eyed monk gouging one of his eyes out, realize the tourist is lying. After examining the eye, after washing the blood off, even the now one-eyed monk realizes the tourist is lying.

Spoiler

That's pretty dramatic. That audience won't believe it though, unless we write in a stipulation that they take their vow of silence so seriously, they'd rather lose an eye than break it.

 

[collinsmark]

Spoiler

I think Enigman is on the right track. But I'll put my own spin on it.
One of two things happens:
a) The one monk with the red eyes gouges one of his eyes out. The then washes the blood off, examines it with his remaining eye, realizes he has red eyes, and proceeds to kill himself that night at midnight.
b) There are no monks with red eyes, but they are not 100% sure if the tourist is telling the truth or lying. The original wording said that something dramatic happens so we can assume that at least one monk will attempt to gouge one of his eyes out (that's the dramatic part). The rest of the monks, seeing a brown eyed monk gouging one of his eyes out, realize the tourist is lying. After examining the eye, after washing the blood off, even the now one-eyed monk realizes the tourist is lying.

[Edit: I intentionally didn't include the situation where there are several red eyed monks because then nothing dramatic would happen in that case.]

Spoiler

Hmmm. Thinking about this a little more, things might actually get a little interesting if there are several red-eyed monks.

Suppose that there are exactly two red-eyed monks. Nothing would happen on the first day (and first midnight) because even the red-eyed monks know that there is at least one red-eyed monk.

However, each red-eyed monk would expect the other to gouge his eyes out and commit suicide if there were only 1 red eyed monk. But on the following day, neither monk has killed himself. Because the other red eyed monk is still alive, each red eyed monk can logically deduce to himself, "that red eyed monk is able to see the red eyes of another monk (otherwise he would have killed himself by now), and since I only see the red eyes of the one monk, the other monk must be me!" So on the second day, each of the red eyed monks gouge their own respective eyes out* and commit suicide at the second midnight.

Similarly if there are exactly three red eyed monks they would gouge an eye out on the third day and kill themselves on the third midnight and so on.

So in conclusion, if there are n red eyed monks they would gouge their eye out on the nth day, and kill themselves on the nth midnight.

*(They gouge their own eye out for good measure.)

 

[consciousness]

Collinsmark, you have pretty much got it. The way I thought about this one was that the monks believe the tourist, so there is no eye gouging.

There is a nice way to organise your solution using

Spoiler

Mathematical Induction

.

 

[collinsmark]

Here's sort of a mathy one that doesn't involve gouging out sensory organs or committing suicide. Instead, it's only about trains. Nice, relaxing trains.

A man retires from his job and moves to the countryside. He's sick of the hustle-and-bustle of the city and decides never to even wear a wristwatch any more or even keep clocks in his house.

He's always enjoyed trains (as in railroad) though, and happens to live next to a railroad track. Each day, at a completely random time, he walks out to tracks and waits for a train. After watching a train go by he goes back home for the rest of the day and records whether the train was a yellow train or a red train.

After months and months of data, he notices that for every red train he has seen, he has seen about 5 yellow trains. (i.e. 5 yellow trains to every 1 red train.)

One day he goes into the town to get groceries and mentions this to the shopkeeper, who knows quite a lot about local trains and train schedules. The shopkeeper informs him that the trains are on a very tight schedule and the red train passes on the tracks near his house on the hour every hour (12:00, 1:00, 2:00, 3:00, etc.). And to the man's surprise, the shopkeeper also informs him that the red and yellow trains alternate, one after the other, also at a set schedule, and there are an equal number of red trains as yellow trains that pass on the tracks near his house. And those are the only trains that ever use those tracks. [Edit: by all that I mean the yellow trains are also on a fixed schedule. And for any given hour of the day, two trains will pass: one red and one yellow.]

How can this be? Why did he see so many more yellow trains than red trains?

Stipulations:
(a) The red trains and the yellow trains are all of equal size/length, and that size is rather short: just a couple of cars or so.
(b) Although trains are involved, this enigma has nothing to do with special relativity or the Doppler effect. They are just normal trains (albeit short ones) moving at normal train speeds.

 

[zoobyshoe]

You say the trains will "pass". Does this mean red trains only go in one direction and yellow trains in the opposite direction?

 

[collinsmark]

You say the trains will "pass". Does this mean red trains only go in one direction and yellow trains in the opposite direction?

Hmm. Either way will work.

But when I said "pass" I mean pass by his house. The trains do not pass by his house and each other at the same time (if they even pass each other at all -- all trains can all be going in the same direction at the same speed for this). [They can also travel in different directions but it's not necessary.]

So whichever direction, from the location of his house, there is nor more than one train going by at any point in time.

 

[zoobyshoe]

Hmm. Either way will work.

But when I said "pass" I mean pass by his house. The trains do not pass by his house and each other at the same time (if they even pass each other at all -- all trains can all be going in the same direction at the same speed for this). [They can also travel in different directions but it's not necessary.]

So whichever direction, from the location of his house, there is nor more than one train going by at any point in time.

Thanks.

The red trains are on the hour every hour. The yellow trains are exactly one per hour, but we don't know if it's a quarter past the hour, a quarter before, on the half hour, or what, right?

 

[collinsmark]

Thanks.

The red trains are on the hour every hour. The yellow trains are exactly one per hour, but we don't know if it's a quarter past the hour, a quarter before, on the half hour, or what, right?

Yes, that's the right idea! Think about this one some more, the answer will arrive.

Spoiler

[Hint: Remember, the man comes to wait for a train (any train) at a completely random time, once per day. ]

 

[zoobyshoe]

Spoiler

Well, I would suppose he's actually being prompted by the passing of a red train without realizing it. The yellow trains probably follow the red trains by a small amount of time. He hears a train and subliminally is prompted to go out and wait.

Once in a while he doesn't make it out in time to see a yellow train and happens to catch the next red one, then goes home.

 

[collinsmark]

Spoiler

Well, I would suppose he's actually being prompted by the passing of a red train without realizing it. The yellow trains probably follow the red trains by a small amount of time. He hears a train and subliminally is prompted to go out and wait.

Once in a while he doesn't make it out in time to see a yellow train and happens to catch the next red one, then goes home.

Spoiler

Not quite. I thought you almost had it there for a second, but then you mentioned the prompting part. He's not prompted by anything (subliminally or otherwise). He really goes out to wait at a random time each day.

 

[collinsmark]

Spoiler

Well, I would suppose he's actually being prompted by the passing of a red train without realizing it. The yellow trains probably follow the red trains by a small amount of time. He hears a train and subliminally is prompted to go out and wait.

Once in a while he doesn't make it out in time to see a yellow train and happens to catch the next red one, then goes home.

Spoiler

Hmmm. I might just give it to you. But just to solidify everything, let me ask this first: at what time of the hour do the yellow trains pass by his house?

 

[zoobyshoe]

Spoiler

Hmmm. I might just give it to you. But just to solidify everything, let me ask this first: at what time of the hour do the yellow trains pass by his house?

Spoiler

I don't see there's any way to determine the exact time, but it would be more likely to be shortly after the red trains pass, assuming it doesn't take him long to get to the tracks.

 

[lisab]

Spoiler

this is the quietest thread on the forum, i think

 

[zoobyshoe]

Spoiler

I suddenly realized where you might be going: if the yellow train always precedes the red by a few minutes or so, then it is more likely that's the one he'll see on authentically random excursions. If they both go past within 10 minutes of each other, with the yellow always first, then there is a 50 minute period during which his decision to go out will feel "random," but will, 5 times out of 6, be concluded by seeing a yellow train. He'll see it, then leave before the red one goes by. Once out of 6 times, though, he'll get out there just after the yellow train has passed by and he'll see the red train.

 

[consciousness]

Spoiler

The yellow trains arrive 10 minutes before the red trains.

 

[collinsmark]

Spoiler

I suddenly realized where you might be going: if the yellow train always precedes the red by a few minutes or so, then it is more likely that's the one he'll see on authentically random excursions. If they both go past within 10 minutes of each other, with the yellow always first, then there is a 50 minute period during which his decision to go out will feel "random," but will, 5 times out of 6, be concluded by seeing a yellow train. He'll see it, then leave before the red one goes by. Once out of 6 times, though, he'll get out there just after the yellow train has passed by and he'll see the red train.

Spoiler

The yellow trains arrive 10 minutes before the red trains.

Yes, these are both correct answers.

 

[zoobyshoe]

Yes, these are both correct answers.

That was an interesting situation, where two equal things looked to be quite unequal.

 

[zoobyshoe]

It was time to send the kids to camp, and Sally and Jim were shopping for supplies. They spent half of the money they had plus $4 on socks for the kids; half of what was then left plus $3 on name tapes; and half of what was then left plus $2 on a small wallet for each child. They found themselves with $3 left over. How much did they start with?

FWIW: The book says only 70% of Mensa members who tackled this one got it right. I don't know why. It didn't seem to be that tricky to me.

 

[zoobyshoe]

Here's the kind Gad and Enigman like:

My first is in sugar but not in tea
My second in swim but not in sea
My third in apple and also pear
My fourth in ring and also hare
My last in ten but not in herd
My whole a very complimentary word.

 

[drizzle]

is there 'o' in the word?

 

[zoobyshoe]

is there 'o' in the word?

I don't think that's something I should reveal.

 

[drizzle]

 

[Office_Shredder]

For the Gad/Enigman enjoying puzzle:

Spoiler

I think it's "smart"

 

[zoobyshoe]

For the Gad/Enigman enjoying puzzle:

Spoiler

I think it's "smart"

Intelligent answer.

 

[drizzle]

Hmm, I wasn't thinking this way at all.

 

[consciousness]

It was time to send the kids to camp, and Sally and Jim were shopping for supplies. They spent half of the money they had plus $4 on socks for the kids; half of what was then left plus $3 on name tapes; and half of what was then left plus $2 on a small wallet for each child. They found themselves with $3 left over. How much did they start with?

FWIW: The book says only 70% of Mensa members who tackled this one got it right. I don't know why. It didn't seem to be that tricky to me.

Spoiler

60$

The challenge is to solve this one without a paper and without hit and trail. I suppose one could form an equation in their mind but there is a nicer alternative.

 

[zoobyshoe]

The challenge is to solve this one without a paper and without hit and trail. I suppose one could form an equation in their mind but there is a nicer alternative.

I had to write out and solve 3 equations. What's "hit and trail"?

 

[zoobyshoe]

My first is in fish but not in snail
My second in rabbit but not in tail
My third in up but not in down
My fourth in tiara not in crown
My fifth in tree you plainly see
My whole a food for you and me.

 

[drizzle]

Spoiler

fruit

 

[collinsmark]

My first is in fish but not in snail
My second in rabbit but not in tail
My third in up but not in down
My fourth in tiara not in crown
My fifth in tree you plainly see
My whole a food for you and me.

Spoiler

I think Gad might like this one!

 

[zoobyshoe]

Spoiler

fruit

All you want, yes!

 

[Travis_King]

It was time to send the kids to camp, and Sally and Jim were shopping for supplies. They spent half of the money they had plus $4 on socks for the kids; half of what was then left plus $3 on name tapes; and half of what was then left plus $2 on a small wallet for each child. They found themselves with $3 left over. How much did they start with?

FWIW: The book says only 70% of Mensa members who tackled this one got it right. I don't know why. It didn't seem to be that tricky to me.

Spoiler

It's much easier backwards, like 60 times easier

 

[zoobyshoe]

Spoiler

It's much easier backwards, like 60 times easier

Spoiler

That is correct. I'm not sure how you'd do it the other way round.

 

[Travis_King]

Someone sent me this a while back, I'm sure it's searchable, but don't cheat! It's a tough-ish one but not so bad.

You are given eight cards with numbers written on them: 4,4,3,3,2,2,1,1. Your task is to arrange the cards in such a way that the ones are separated by one digit, the twos are separated by two digits, the threes are separated by three digits, and the fours are separated by four digits. You must use all the cards. What is the number you create?

 

[drizzle]

Spoiler

23421314...