Solve Enjoyable Enigmas with Mr.E's Challenge

[Enigman]

*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

 

[drizzle]

Spoiler

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?

 

[zoobyshoe]

Spoiler

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."

 

[drizzle]

Spoiler

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.

 

[Enigman]

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

 

[zoobyshoe]

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?

 

[Enigman]

Can't sleep

Spoiler

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

 

[collinsmark]

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.

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.

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.

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.]

 

[Office_Shredder]

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

 

[zoobyshoe]

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.

 

[Enigman]

Can't sleep

Spoiler

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?

 

[collinsmark]

Yoohooo! Anyone?

I have plenty of matches now. More than I know what to do with. Hmm.

 

[Enigman]

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

 

[consciousness]

I got one for matchstick-

Spoiler

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.

 

[Enigman]

Yep. And now the second method to get another square from the same arrangement by moving one matchstick.

 

[zoobyshoe]

Yep. And now the second method to get another square from the same arrangement by moving one matchstick.

Possibly what you're looking for:

Spoiler

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.

 

[zoobyshoe]

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.

 

[Enigman]

Possibly what you're looking for:

Spoiler

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...

Spoiler

that was a hint

 

[zoobyshoe]

Can't sleep

Spoiler

14 penny make 2 shilling, so 3 shilling and 3 penny

Penny wise, plural foolish.

 

[Enigman]

Penny wise, plural foolish.

sleep muddled, brain addled,
funny little Enigman

(was I hallucinating or did collinsmark too reply to this one?)

 

[zoobyshoe]

Nope its a perfect square and you will need all the matches for it...

Spoiler

that was a hint

Spoiler

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.

 

[Enigman]

Perfect.
Edit: Almost perfect.

Spoiler

place it diagonally rather than vertically (between horizontal left and upper vertical) and you get 4

 

[zoobyshoe]

Perfect.
Edit: Almost perfect.

Spoiler

place it diagonally rather than vertically (between horizontal left and upper vertical) and you get 4

Spoiler

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

 

[zoobyshoe]

A bottle of wine costs $10. The wine, itself, costs $9 more than the bottle. How much does the bottle cost?

 

[consciousness]

Spoiler

0.5$

I didn't like the solution of matchstick at all.

 

[Enigman]

I write it that way too but since we are typing its almost perfect.
And

Spoiler

with or without the wine?

 

[zoobyshoe]

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?"

 

[zoobyshoe]

I write it that way too but since we are typing its almost perfect.
And

Spoiler

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.

 

[consciousness]

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?"

Spoiler

2$ is the total profit

I insist my answer is correct!

 

[Enigman]

I didn't like the solution of matchstick at all.

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...

 

[zoobyshoe]

Spoiler

0.5$

Spoiler

This is correct, despite the fact it's unlikely you'd get this much for it at any recycling center.

 

[consciousness]

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...

Go away Schrodinger! Someone might observe YOU!

 

[consciousness]

Spoiler

This is correct, despite the fact it's unlikely you'd get this much for it at any recycling center.

What recycling center? *follows the examples of others around him and throws a bottle out the window*

 

[Enigman]

Go away Schrodinger! Someone might be observe YOU!

I am the ENIGMAN. I am unobservable. My mask indiscernible. My mystery (Mr.E) unsolvable. I am outside the box. I will do the observing...

 

[consciousness]

I am the ENIGMAN. I am unobservable. My mask indiscernible. My mystery (Mr.E) unsolvable. I am outside the box. I will do the observing...

So you cannot be observed? *Uses Occam's razor to cut Enigman out of reality!*

 

[zoobyshoe]

I am the ENIGMAN. I am unobservable. My mask indiscernible. My mystery (Mr.E) unsolvable. I am outside the box. I will do the observing...

I'm sorry to inform you you have been the victim of a misspelling all this time. You are actually the Eggman. No one's had the heart to tell you.

 

[zoobyshoe]

Spoiler

2$ is the total profit

I insist my answer is correct!

Well, I insist that it IS correct, and I'll keep on insisting it till you're blue in the face!

 

[Enigman]

Well, I insist that it IS correct, and I'll keep on insisting it till you're blue in the face!

So you pute his answer? Well, so do I!

 

[zoobyshoe]

"The instructive thing about this puzzle is that although it can easily be solved using elementary algebra, it can also be solved without any algebra at all-just by plain common sense. Moreover, the common sense solution is, in my judgement, for more interesting and informative - and certainly more creative - than the algebraic solution.

Fifty-six biscuits are to be fed to ten pets; each pet is either a cat or a dog. Each dog is to get six biscuits, and each cat is to get five. How many dogs and how many cats are there?

Any reader familiar with algebra can get this immediately. Also, the problem can be solved by trial and error: there are eleven possibilities for the number of cats (anywhere from zero to ten), so each possibility can be tried until the correct answer is found. But if you look at this problem in just the right light, there is a surprisingly simple solution that involves neither algebra nor trial and error."

What is this interesting and creative "common sense" answer?

 

[Enigman]

I'm sorry to inform you you have been the victim of a misspelling all this time. You are actually the Eggman. No one's had the heart to tell you.

Hushhh...that can't be known. I can neither confirm or deny that. If I tell you I will have to kill you...
EDIT: Possible News headline in the recent future:
the last zoobie killed by an eggman who Mr.E-ously disappears.

 

[collinsmark]

sleep muddled, brain addled,
funny little Enigman

(was I hallucinating or did collinsmark too reply to this one?)

Yes, I replied with the correct

Spoiler

3 Shillings and 3 pennies

answer. But then I saw that you had already provided the correct answer (I had missed it at first glance) so I deleted my post.

 

[Enigman]

Fifty-six biscuits are to be fed to ten pets; each pet is either a cat or a dog. Each dog is to get six biscuits, and each cat is to get five. How many dogs and how many cats are there?
What is this interesting and creative "common sense" answer?

Spoiler

The sum should be 56.
units place should be 6
A-units place of biscuits eaten by cats should 0 or 5
You can get 6 in units place by (adding 1 to A=5) or (adding 6 to A=0)
B- units place of biscuits eaten by dogs can't be 1
A=0 B=6
Multiple of 6 with units place 6--> 36,6
36 satisfies all conditions
6-->dogs 4-->cats

 

[Enigman]

Yes, I replied with the correct

Spoiler

3 Shillings and 3 pennies

answer. But then I saw that you had already provided the correct answer (I had missed it at first glance) so I deleted my post.

Don't do that... I have a paranoia of having paranoid schizophrenia...

 

[zoobyshoe]

Yes, I replied with the correct

Spoiler

3 Shillings and 3 pennies

answer. But then I saw that you had already provided the correct answer (I had missed it at first glance) so I deleted my post.

Spoiler

Your correct answer was much better than his correct answer. He seemed to have lost the ability to form plurals: "3 Shilling, 3 Penny."

 

[zoobyshoe]

Spoiler

The sum should be 56.
units place should be 6
A-units place of biscuits eaten by cats should 0 or 5
You can get 6 in units place by (adding 1 to A=5) or (adding 6 to A=0)
B- units place of biscuits eaten by dogs can't be 1
A=0 B=6
Multiple of 6 with units place 6--> 36,6
36 satisfies all conditions
6-->dogs 4-->cats

Spoiler

Absolutely not!

We're looking for an "interesting, creative, 'common-sense'" answer.

 

[Enigman]

Those are vague guidelines, I steered clear of the more set in ground ones- No algebra; No trial and error.

Common sense is the collection of prejudices acquired by age eighteen.
-A.E.

 

[zoobyshoe]

When you read the answer, you'll go "Doh!"

 

[Office_Shredder]

This is still math but maybe less enough that it counts:

Spoiler

If there were 10 dogs, we would need 60 biscuits. We need 4 fewer biscuits so there must be 4 cats.

 

[Enigman]

Spoiler

Maximum case 60 biscuits
Actual case 56 biscuits
Shortage 4
No. of cats 4
No. of dogs 6

O_S replied already...
won't delete it.

 

[consciousness]

Doh!

A man marries a woman, but is not her husband. How?