Solve Enjoyable Enigmas with Mr.E's Challenge

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The forum thread invites puzzle enthusiasts to share various types of puzzles, including cryptograms and whodunnits, while emphasizing that participants should know the answers without resorting to online searches. A code message is presented, which participants attempt to decode, leading to discussions about its meaning and possible interpretations. Participants also engage in solving additional puzzles, such as cutting a cake into pieces with minimal cuts and a physics challenge involving water and matchsticks. The conversation highlights the enjoyment of problem-solving and the creative thinking required to tackle these enigmas. Overall, the thread fosters a collaborative atmosphere for sharing and solving intriguing puzzles.
  • #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
 
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  • #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.
 
  • #455
YES!
(The thirteen day thing would also work because of the long time span given.)
And now I can sleep, really tired.
:zzz:
 
  • #456
Pete came back from a trip around the world with some souvenirs for his mother. On the remote Island of Andranda he'd purchased 3 handcrafted silver dishes. He told her they cost one Andrandan Shilling and five pennies each. The total for all three had, therefore, been five Andrandan Shillings and one penny.

His mother was confused. "Tell me," she said, "what would you have paid if you had only bought two of them?"

How much would he have paid had he only bought two silver dishes?
 
  • #457
Can't sleep :frown:
14 penny make 2 shilling, so 3 shilling and 3 penny
Shift one matchstick and make a square.
(two methods to do it)
https://www.physicsforums.com/attachment.php?attachmentid=62376&stc=1&d=1380671069
 

Attachments

  • #458
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.
(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

Actually, in order for it to be guaranteed that the monk will be at the same place, the same time of day, I'm pretty sure that at least one of the trips (either ascending or descending) must be greater than or equal to one day.

For example, it wouldn't hold if he makes it to the top of the mountain by sunset the same day he started, then started down at sunrise by first taking a long break (at the top of the mountain) until after sunset then made it all the way down the mountain before dawn.

Enigman said:
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.

I won't use any mathematical rigor. Instead I will prove it graphically.

Here is essentially a proof (albeit a very non-rigorous one). In order for this proof to always work, no matter what, at least one of the trips must be greater than or equal to a day.

We plot the distance to the mountain top during the monk's ascent.

attachment.php?attachmentid=62373&d=1380670249.jpg


I made it so he took about three days to reach the top. But all that is necessary is that it takes a day or more. Actually, it can be less than a day as long as the decent takes over a day, but since the original enigma specified that he descends faster, his ascent must take more than a day.

We can also plot his descent referencing the time at which he started his decent. We can overlay the plots on one another. In other words, we just line up 6:30 AM with one day when he was ascending with 6:30 AM on a day he was descending.

attachment.php?attachmentid=62375&d=1380670882.jpg


And since the span of the y-axis is the same for both curves, it is guaranteed that there be at least one time of day where the monk is in the same place.

[Edit: 'looks like Zooby beat me to the answer.]
 

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  • #459
collinsmark, I think in your counterexample proposal the monk would count as being at the top of the mountain at the same time on both days. He's officially started his journey at sun-rise even if he chooses to sit there for 10 hours
 
  • #460
collinsmark said:
Actually, in order for it to be guaranteed that the monk will be at the same place, the same time of day, I'm pretty sure that at least one of the trips (either ascending or descending) must be greater than or equal to one day.

For example, it wouldn't hold if he makes it to the top of the mountain by sunset the same day he started, then started down at sunrise by first taking a long break (at the top of the mountain) until after sunset then made it all the way down the mountain before dawn.

All that matters is that he starts both trips at the same time of day. This has to be specific, like 6:14:02 A.M. kind of specific.

He could do the journey up in an hour if it was possible, then, on the return trip take one step and sit down for a 5 hour break, but there would still be an inevitable overlap of the same spot, same time of day.

He can't delay the trip in either direction. It has to start at exactly the same time.
 
  • #461
Enigman said:
Can't sleep :frown:
14 penny make 2 shilling, so 3 shilling and 3 penny
Shift one matchstick and make a square.
(two methods to do it)
https://www.physicsforums.com/attachment.php?attachmentid=62376&stc=1&d=1380671069

Yoohooo! Anyone?
 
  • #462
Enigman said:
Yoohooo! Anyone?
I have plenty of matches now. More than I know what to do with. Hmm.
 
  • #463
collinsmark said:
I have plenty of matches now. More than I know what to do with. Hmm.
Try these; couldn't make any viable enigmas out of them...
https://www.youtube.com/watch?v=q0yfrZkqPxQ

https://www.youtube.com/watch?v=LWCQul0SbjY

https://www.youtube.com/watch?v=keGhyXwtdUc
 
  • #464
I got one for matchstick-
Shift the upper matchstick by the width of one matchstick (keeping orientation same). A small square with side equal to the width of the matchsticks will form.
 
  • #465
Yep. And now the second method to get another square from the same arrangement by moving one matchstick.
 
  • #466
Enigman said:
Yep. And now the second method to get another square from the same arrangement by moving one matchstick.
Possibly what you're looking for:

Lay a match on top of another match leaving a length of the bottom match equal to the width of a match visible. That visible portion of the bottom match will be a square. Kinda lame, but the only second solution I can think of.
 
  • #467
Office_Shredder said:
collinsmark, I think in your counterexample proposal the monk would count as being at the top of the mountain at the same time on both days. He's officially started his journey at sun-rise even if he chooses to sit there for 10 hours
I didn't get this at first, but you're right. So long as we define him as having started the trip he doesn't even need to move on the return leg. The same time, same place coincidence will happen, in this case, to take place at the first journey's end point and the second journey's start point.
 
  • #468
zoobyshoe said:
Possibly what you're looking for:

Lay a match on top of another match leaving a length of the bottom match equal to the width of a match visible. That visible portion of the bottom match will be a square. Kinda lame, but the only second solution I can think of.
Nope its a perfect square and you will need all the matches for it...
:biggrin:
that was a hint
 
  • #469
Enigman said:
Can't sleep :frown:
14 penny make 2 shilling, so 3 shilling and 3 penny
Penny wise, plural foolish.
 
  • #470
zoobyshoe said:
Penny wise, plural foolish.
sleep muddled, brain addled,
funny little Enigman
:blushing:
(was I hallucinating or did collinsmark too reply to this one?)
 
  • #471
Enigman said:
Nope its a perfect square and you will need all the matches for it...
:biggrin:
that was a hint
Take the horizontal match on the right, place it vertically on the upper left end of the horizontal match on the left. This will create the numeral 4. 4 is the square of 2.
 
  • #472
Perfect.
Edit: Almost perfect.
place it diagonally rather than vertically (between horizontal left and upper vertical) and you get 4
 
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  • #473
Enigman said:
Perfect.
Edit: Almost perfect.
place it diagonally rather than vertically (between horizontal left and upper vertical) and you get 4
That's not the way I learned to write 4. There are different systems out there.
http://www.montessori-spirit.com/4218-large_default/printed-numerals-print.jpg
 
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  • #474
A bottle of wine costs $10. The wine, itself, costs $9 more than the bottle. How much does the bottle cost?
 
  • #475
0.5$

I didn't like the solution of matchstick at all.:cry:
 
  • #476
I write it that way too but since we are typing its almost perfect.
And
with or without the wine?
 
  • #477
From a puzzle book:

"The amazing thing about this puzzle is that people always seem to fight over the answer! Yes, different people work it out in different ways and come up with different answers, and each insists his answer is correct. The puzzle is this:

A dealer bought an article for $7, sold it for $8, bought it back for $9, and sold it for $10. How much profit did he make?"
 
  • #478
Enigman said:
I write it that way too but since we are typing its almost perfect.
And
with or without the wine?
The empty bottle. How much does it cost by itself when you subtract the cost of the wine from the total cost of wine + bottle.
 
  • #479
zoobyshoe said:
From a puzzle book:

"The amazing thing about this puzzle is that people always seem to fight over the answer! Yes, different people work it out in different ways and come up with different answers, and each insists his answer is correct. The puzzle is this:

A dealer bought an article for $7, sold it for $8, bought it back for $9, and sold it for $10. How much profit did he make?"

2$ is the total profit

I insist my answer is correct! :-p
 
  • #480
consciousness said:
I didn't like the solution of matchstick at all.:cry:

Don't stay inside the box or you are just like the cat...I am a curious guy, I like opening boxes... <psychopathic grin>
What did they say? Ah, yes. Curiosity killed the cat...
 

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