Solve Enjoyable Enigmas with Mr.E's Challenge

  • Thread starter Enigman
  • Start date
In summary: Four princes approach the king vying for the hand of the princess. In order to choose the best among the four suitors the king and the princess arrange a test: the suitors are brought to a large rectangular hall. The floor is covered with a carpet all over except at the corners- where there are four squares of bare floor and the suitors are told to stand at these corner. Each suitor takes a corner and stands there while the princess stands at the center of the hall. The king then proclaims the prince who without leaving their respective squares shall put a ring on the princess's hand will be declared to be the bridegroom of his daughter and the heir to Enigmania. No ropes or rods are
  • #421
zoobyshoe said:
I don't understand. Isn't the goal to end up with one ball at each separate end?

Yes, this is the first step towards that goal. Second step is same as in Enigman's solution.
 
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  • #422
Possible solution to mask enigma-

The running man is a thief. He stole a diamond ring from his wife's sister and was planning to sell it for cash. He has hidden the ring under the bed etc. away from his wife. He was working peacefully when he remembered that a pest control man would be coming to his home that day. Realising that the probability of his wife discovering the theft would rise exponentially if he started shifting stuff, he runs towards home to hide the ring somewhere else. He is afraid because the discovery might already have been made.
Notes-
The pest control man is masked.
His wife will recognise the the ring if she sees it.
The man is not too smart to hide the ring there.
His wife and children are the spectators.
 
  • #423
Too complicated conciousness...Enigma doesn't give enough data to conclude all that. I think we will all be face-palming when we get the solution and go "Doh!"...
- though you may be still right but the enigma says he is afraid of going in there not afraid of what might happen if he did not get there...
 
  • #424
His wife might be waiting for him with a rolling pin though!
BTW I don't think we can solve problems of this nature without making assumptions...the trick is to make reasonable ones.
 
  • #425
Enigman said:
You don't have to spoiler everything...
No, conciousness' method would not work if
the axis passes through the end of tube but if if the axis is somewhere between the center of mass of the spheres it would get the job done.
We don't know the diameter of the cylinder. It could be large enough that both balls can rest against an end circle side by side in a line perpendicular to the length of the pipe.
Edit:
More specifically, we don't know the inside diameter of the cylinder. That makes knowing the ball dimensions moot.
 
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  • #426
collinsmark said:
A man is running home, but he's afraid to get there, because there is another man already there who is wearing a mask."

What is the masked man's occupation?
I reworded the question slightly removing the emphasis on "job." suffice it to say the masked man is at his occupation at the time.

zoobyshoe said:
People are watching the masked man?

Yes.

Office_Shredder said:
On the masked man:
The two men are married (hence the diamond). The job the man at home is doing is that he does webcam shows for money and wears a mask while doing them - the man returning home was surfing online for pornography while at work and found his husband doing these shows, and is running home to confront him in the act.

It's a reach but I figured I'd post it.

:bugeye: I'm going to have to go with "no" on that.

----------------------
Here is another clue:

The riddle is culturally biased insofar that exposure to such professions is most probable if one lives in the United States. That said, it's not completely limited to the United States. It's also possible that the situation is happening in Cuba, Japan, Australia, and I think even the United Kingdom, among other possible countries.
 
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  • #427
zoobyshoe said:
We don't know the diameter of the cylinder. It could be large enough that both balls can rest against an end circle side by side in a line perpendicular to the length of the pipe.
Edit:
More specifically, we don't know the inside diameter of the cylinder. That makes knowing the ball dimensions moot.

er... just shake it till you feel balls are in a line...
Any ideas about collinsmark's puzzle? Last time I was this lost was the chinese forensics enigma...darned baseball...
:grumpy:
Where is that occam's razor when you need it?
 
  • #428
Enigman said:
Any ideas about collinsmark's puzzle? Last time I was this lost was the chinese forensics enigma...darned baseball...
:grumpy:
Where is that occam's razor when you need it?

I think you just got it.
 
  • #429
Mask:
"Home" is home plate in the game of baseball. The man with the mask he fears is the catcher: is someone throws the catcher the ball before the man get's "home" the man will be "out". Baseball is played on a "diamond".
 
  • #430
zoobyshoe said:
Mask:
"Home" is home plate in the game of baseball. The man with the mask he fears is the catcher: is someone throws the catcher the ball before the man get's "home" the man will be "out". Baseball is played on a "diamond".

Yes, that's right.

The man in the mask is a professional baseball player, specifically the catcher on a professional baseball team. The running man is also a professional baseball player, although his specific position on the team is arbitrary.

[Edit: I also would have accepted "umpire."]
 
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  • #431
That awkward moment when you realize that you are an idiot...
Next time I don't get one- my answer is baseball.
Anyway the answer to this is definitely not baseball:
Enigman said:
A monk climbs to the top of a certain mountain with unequal speeds and random stops of random durations, he reaches the top at the sunset of the 13th day from the start. After meditating there for a week, he starts climbing down the mountain at the sunrise with unequal speeds and random stops. The speed while climbing down is obviously greater than speed climbing up. Assuming that he follows the exact same path for both journeys prove that there exists a time of day where the monk was at the same position on the path for both journeys.
 
  • #432
Monk question:

You are saying that there is some specific time of day, like, say, 2:40 P.M. when it is guaranteed the monk will be at the very same point on the path going down the mountain as he was going up the mountain?
 
  • #433
I am asking for a proof that there will exist a time of day when the monk is at the same point on the path on both journeys. The time of day can be any thing so can be the point on the path - depending on the monk and his various whimsical speeds.
Hint: The speed while climbing down is obviously greater than speed climbing up- Just given to confuse you.
 
  • #434
Enigman said:
The time of day can be any thing...
So, "time of day" could be as vague as "morning," "afternoon," or "night"?
 
  • #435
No- 2:14:35 pm something like that.
Time of day could belong to the whole range of 6:00 am to 6:00 pm* and would be determined how the monk decides to carry out his journeys.**
*assuming sunrise and sunset at those times.
EDIT-**the proof should hold for all cases- all possible speeds and combined with all possible breaks of all possible durations.
Its really not a mathematical proof. Think out of the box...you do remember what happens to the cats in the box...
 
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  • #436
Enigman said:
Time of day could belong to the whole range of 6:00 am to 6:00 pm* and would be determined how the monk decides to carry out his journeys.**
*assuming sunrise and sunset at those times.
The monk only travels during daylight hours, then?
 
  • #437
Yes, but it doesn't really matter all that is of significance is both the trips start and end at the same time of day/night...
 
  • #438
Enigman said:
...all that is of significance is both the trips start and end at the same time of day/night...
Both trips have to start at sunrise and end at sunset?
 
  • #439
The only essential thing is that both of the journeys start at the same time of the day and the monk does complete both trips; anything else is extraneous.
There is no math involved, just some elegant reasoning.
 
  • #440
You can see for yourself you left this "essential" fact out of your statement of the enigma:
A monk climbs to the top of a certain mountain with unequal speeds and random stops of random durations, he reaches the top at the sunset of the 13th day from the start. After meditating there for a week, he starts climbing down the mountain at the sunrise with unequal speeds and random stops. The speed while climbing down is obviously greater than speed climbing up. Assuming that he follows the exact same path for both journeys prove that there exists a time of day where the monk was at the same position on the path for both journeys.
There's nothing to indicate the first journey was started at the same time of day the second was.
 
  • #441
oops...sorry.
Okay rephrasing the question:
A monk climbs to the top of a certain mountain starting at sun-rise with unequal speeds and random stops of random durations, he reaches the top at the sunset of the same day. After meditating there for a week, he starts climbing down the mountain at the sun-rise with unequal speeds and random stops. Assuming that he follows the exact same path for both journeys prove that there exists a time of day where the monk was at the same position on the path for both journeys.
(Sorry again...should never trust the net to give accurate statements...I should have phrased it myself but I just couldn't remember the wordings...)
EDIT:Although the way the enigma was previously stated is solvable and is similar to what I had in mind...I can't decide which phrasing is easier...but 13 day thing may make things more complicated so let's stick to this one
 
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  • #442
Just to be clear the previous version won't need the " same time of the day" constraint to be solved.
I think the one in the last post is easier compared to that...(I just assumed that the statement from the web was the same thing as what I had in mind...:redface:) But these are just technicalities...
As a compensation for the confusion a hint:
Think of the two trips simultaneously...
 
  • #443
On about 1/3 of the path on the way up?
 
  • #444
Enigman said:
oops...sorry.
Okay rephrasing the question:
A monk climbs to the top of a certain mountain starting at sun-rise with unequal speeds and random stops of random durations, he reaches the top at the sunset of the same day. After meditating there for a week, he starts climbing down the mountain at the sun-rise with unequal speeds and random stops. Assuming that he follows the exact same path for both journeys prove that there exists a time of day where the monk was at the same position on the path for both journeys.
You specify that he starts the first trip at sunrise and reaches the top at sunset. You also specify that he starts the second trip at sunrise, but there is no destination time given for this return trip. Is the time he finishes the second trip immaterial?
 
  • #445
Nope. It doesn't matter.
EDIT: and the fact that he reaches at sunset is pretty useless too...
 
  • #446
Gad said:
On about 1/3 of the path on the way up?

Depends on his whimsy...it might be anywhere...What I ask for is a proof that it will happen.
 
  • #447
I have a feeling that it will never happen, as the path he follows starts from one end of the mountain and ends on the other side of it. :biggrin:
 
  • #448
He follows the exact same path i.e. he retraces his path from the opposite direction. And it will happen.
 
  • #449
Enigman said:
Think of the two trips simultaneously...

Well? What happens?
 
  • #450
Are we talking about a single monk here?
 
  • #451
*no comment*...Er, the monk is the same*...can't say anything more than that...not after #449
Ah, well- *how you want to think of it is another matter
 
  • #452
Well, assuming 2 monks start the journey from both sides of the path, they would definitely meet/collide at some point, so, is this is it?
 
  • #453
Imagine a straight line segment with a dot at both ends. The line represents the mountain path, the dots represent the monk on his two different trips. The dots simultaneously start to move to the other end from where they started. Their respective motions can be smooth or erratic, fast or slow, but there must, inevitably, be a point where they meet and pass each other. That point is "the same time of day for both journeys."
 
  • #454
zoobyshoe said:
Imagine a straight line segment with a dot at both ends. The line represents the mountain path, the dots represent the monk on his two different trips. The dots simultaneously start to move to the other end from where they started. Their respective motions can be smooth or erratic, fast or slow, but there must, inevitably, be a point where they meet and pass each other. That point is "the same time of day for both journeys."

I always miss the point. :grumpy:
 
  • #455
YES!
(The thirteen day thing would also work because of the long time span given.)
And now I can sleep, really tired.
:zzz:
 

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