Solve Enjoyable Enigmas with Mr.E's Challenge

  • Thread starter Thread starter Enigman
  • Start date Start date
AI Thread Summary
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.
  • #251
Enigman said:
Yep. EnigmaX's spell has got some serious limitations but nothing a bit of logic* can't overcome.

*which I didn't have apparently as I failed this one...
I'm not sure how yet, but assuming they can use one weigh-in to locate the underweight bars, they have some way of connecting those bars to the dwarf who cast them?
 
Physics news on Phys.org
  • #252
The dwarves will bring only the gold bars they themselves have cast when the weighing is done. So, yes.
EDIT: but that won't help much, if more than one dwarf's gold is brought you can't tell the bars apart.
 
  • #253
Huge Hint: think in terms of number sequences and their sums...
 
  • #254
Enigman said:
Huge Hint: think in terms of number sequences and their sums...
I think this is it (the clue was vital) :

Arbitrarily give each dwarf a number from 1-1000 and record which dwarf got which number. Then make each dwarf put a number of gold bars on the scale equal to the arbitrary number that has been assigned him. In this way you create an arithmetical sequence that can be summed. Calculate the sum and then calculate what the sum should weigh. Then weigh all the bars in the series.

The actual weight will be something less than the expected weight because some number of bars weighing only 999grams will have been included. By subtracting the actual weight from the expected weight (all in grams), you will get a difference that is some whole number. That is the arbitrary number that has been assigned to the dwarf who has been making the underweight bars. E.g., the sum of the sequence 1-1000 is 500,500. At 1000 grams each, this many bars should weigh 500,500,000 grams. If the actual weight is 500,499,258, then it is 742 grams underweight. Dwarf #742 must have put 742 bars weighing only 999grams each on the scale.
 
  • #255
Yes indeed that is correct.
 
  • #256
Enigman said:
Yes indeed that is correct.
Happy to hear that! That one was hard! Took me hours to sort it out.
 
  • #257
The real question is how did the dwarf know he was only smelting 999 grams each time if the king couldn't tell with his own scale technology?

OK here's a riddle. One hundred people are to be lined up tomorrow, all of them facing the front of the line (so they are looking directly at someone's back, except for the person in front). On each of them will be placed a black hat or a white hat. In any order they each can call out either "black" or "white" once, and only once, and make no other noise. If at least 99 of them call out the correct color of the hat on their head, everyone wins one hundred dollars.

Describe how they can guarantee a victory.

EDIT: Corrected for missing details
 
Last edited:
  • #258
Office_Shredder said:
The real question is how did the dwarf know he was only smelting 999 grams each time if the king couldn't tell with his own scale technology?

OK here's a riddle. One hundred people are to be lined up tomorrow, all of them facing the front of the line (so they are looking directly at someone's back, except for the person in front). On each of them will be placed a black hat or a white hat. In any order they each can call out either "black" or "white" once, and only once. If at least 99 of them call out the correct color, everyone wins one hundred dollars.

Describe how they can guarantee a victory.

The person standing last will call out the colour of the hat of the person who is standing in front of him. The 99th person will thus know his colour. Now 99 people will follow the following rules-
1)If their colour is same as that of the person standing in front of them they will call out their colour immediately.
2)If their colour is different they will wait.
By doing this the nth person will know their colour provided that the (n+1)th person knows their coulour. By a "reverse mathematical induction" All the 99 people left will know their colours. This will reqiure a lot of coordination on the part of the 99 people though.
 
  • #259
Office_Shredder said:
The real question is how did the dwarf know he was only smelting 999 grams each time if the king couldn't tell with his own scale technology?

OK here's a riddle. One hundred people are to be lined up tomorrow, all of them facing the front of the line (so they are looking directly at someone's back, except for the person in front). On each of them will be placed a black hat or a white hat. In any order they each can call out either "black" or "white" once, and only once. If at least 99 of them call out the correct color, everyone wins one hundred dollars.

Describe how they can guarantee a victory.
A "correct" color means?
 
  • #260
zoobyshoe said:
A "correct" color means?

The colour of the hat on their head I assume.
 
  • #261
Yes, correct means they call out the color of the hat on their head, sorry.

consciousness, that looks correct but as an added challenge
it's possible for them to solve the problem without needing to time the statement of their answers (i.e. there is a deterministic algorithm for who goes next that everybody can run, and not wonder if the person behind them has finished waiting to state their color or not - a problem which could occur if five or six people in a row are supposed to wait). Alternatively you could just have them state their color in weird accents to transmit the additional information :p
 
  • #262
Office_Shredder said:
The real question is how did the dwarf know he was only smelting 999 grams each time if the king couldn't tell with his own scale technology?

OK here's a riddle. One hundred people are to be lined up tomorrow, all of them facing the front of the line (so they are looking directly at someone's back, except for the person in front). On each of them will be placed a black hat or a white hat. In any order they each can call out either "black" or "white" once, and only once. If at least 99 of them call out the correct color, everyone wins one hundred dollars.

Describe how they can guarantee a victory.
They work out a code whereby each tells the one in front of him what color hat he's wearing. The code might be anything. Let's say it's clearing the throat before calling out, vs no throat clearing. A throat clearing means, "Your hat is the opposite color of the one I'll call out." No throat clearing means, "Your hat is the same color as the one I'll call out." They could use clearing the throat, or not clearing it, coughing or not coughing, whatever audible vocal tick they all agree to.. They start at the back of the line and work forward in order. The guy at the very back simply calls out the color of the guy in front of him. Even if that's not his own hat color, 99% will still be right. Each will correctly call out the color of his own hat and simultaneously tip the person in front of him off to their correct hat color.
 
  • #263
I edited the original question to prevent wiseassery :-p
 
  • #264
Office_Shredder said:
Yes, correct means they call out the color of the hat on their head, sorry.

consciousness, that looks correct but as an added challenge
it's possible for them to solve the problem without needing to time the statement of their answers (i.e. there is a deterministic algorithm for who goes next that everybody can run, and not wonder if the person behind them has finished waiting to state their color or not - a problem which could occur if five or six people in a row are supposed to wait). Alternatively you could just have them state their color in weird accents to transmit the additional information :p

Okay I have another solution that solves the 5-6 persons waiting problem.
The 100th person calls out the colour of the 99th person. Now the 99th person compares the colour of the 98th person with the 1st person (who presumably everyone can see).
1)If the colours are same he calls out his own colour. The 98th person now knows his colour and plays the role of the 99th person from next loop.
2)If the colours are not same he doesn't say anything. Then the 98th person immediately calls out his own colour.
Cycle is repeated till there are only two persons left, the first who doesn't know his colour and another who does. By an agreement if the other person doesn't call then the colour of the first is opposite to that of the other person. If he calls then the colours are same.
Probably my last solution because my ideas keep on getting more and more convoluted. :biggrin: Pretty soon my solution will be longer than the proof of Fermats last theorem. :smile:
 
  • #265
Office_Shredder said:
I edited the original question to prevent wiseassery :-p
What wiseassery?
 
  • #266
Office_Shredder said:
I edited the original question to prevent wiseassery :-p

You are forgetting that 'wiseassery' is often the point of this thread. Anything goes- within the limits of the question. Most of the answers depend on how the question was worded; what was said and what was omitted- which makes framing the question correctly quite a pain in the back.

Next one:[/size]
Y[/size]ou are at your home sitting on the sofa reading a book. Suddenly a snow ball crashes through the window, you look out of the broken* window to see the three brothers from next door -John , Mark and Fred run around the corner and me standing there enjoying the show :devil:
You ask me who did it. I write the following thing to you on a scrap of paper "?, he broke your window."
So who would you question/scold/have your yard work done by for free?
(Except me, I am off limits. Oh, and I didn't break your window and am telling the truth- enigmatically though it may be...)
EDIT:*broken.
 
Last edited:
  • #267
Office_Shredder said:
OK here's a riddle. One hundred people are to be lined up tomorrow, all of them facing the front of the line (so they are looking directly at someone's back, except for the person in front). On each of them will be placed a black hat or a white hat. In any order they each can call out either "black" or "white" once, and only once, and make no other noise. If at least 99 of them call out the correct color of the hat on their head, everyone wins one hundred dollars.

Describe how they can guarantee a victory.

EDIT: Corrected for missing details
If the person in front of you has a white hat, place your hand on their left shoulder. Otherwise place your hand on their right. Then it doesn't matter what order they get called to reveal their hat color.
 
  • #268
Enigman said:
You are forgetting that 'wiseassery' is often the point of this thread. Anything goes- within the limits of the question. Most of the answers depend on how the question was worded; what was said and what was omitted- which makes framing the question correctly quite a pain in the back.

Next one:[/size]
Y[/size]ou are at your home sitting on the sofa reading a book. Suddenly a snow ball crashes through the window, you look out of the window to see the three brothers from next door -John , Mark and Fred run around the corner and me standing there enjoying the show :devil:
You ask me who did it. I write the following thing to you on a scrap of paper "?, he broke your window."
So who would you question/scold/have your yard work done by for free?
(Except me, I am off limits. Oh, and I didn't break your window and am telling the truth- enigmatically though it may be...)

A snow ball?.. I'd say the window was open.

Unless the snow ball hides a small rock in it, that'll be a different scenario. :devil: Oh I miss those days. :biggrin:
 
  • #269
Gad said:
A snow ball?.. I'd say the window was open.

Nope, your window definitely broke.
Unless the snow ball hides a small rock in it, that'll be a different scenario. :devil: Oh I miss those days. :biggrin:
Me too...:cry:
 
  • #270
Enigman said:
Nope, your window definitely broke.

It was broken before? :biggrin:
 
  • #271
Snowball broke the window. You are :mad:.
Who broke it? -I already answered that o:).
 
  • #272
Well, move on to the next Enigmas, I'll do the yard work. :biggrin:
 
  • #273
Solve this one first...
Saying what I wrote aloud may help
AV2.gif
 
  • #274
Enigman said:
Next one:[/size]
Y[/size]ou are at your home sitting on the sofa reading a book. Suddenly a snow ball crashes through the window, you look out of the broken* window to see the three brothers from next door -John , Mark and Fred run around the corner and me standing there enjoying the show :devil:
You ask me who did it. I write the following thing to you on a scrap of paper "?, he broke your window."
So who would you question/scold/have your yard work done by for free?
(Except me, I am off limits. Oh, and I didn't break your window and am telling the truth- enigmatically though it may be...)
EDIT:*broken.
"?, he broke your window" = "Question Mark, he broke your window."
 
  • #275
Yep, ZBS that is correct.
 
  • #276
Borg said:
If the person in front of you has a white hat, place your hand on their left shoulder. Otherwise place your hand on their right. Then it doesn't matter what order they get called to reveal their hat color.
Comment and variation of your solution:
They're not called on to give their hat color. They decide when to speak up. Regardless, I think your solution should be counted as correct. I'm afraid Office Shredder will condemn it as "wiseassery," though.

Anyway, your thinking outside the box gave me an idea for another, similar, solution. Hitherto, I've been stuck thinking the last person has to be the sacrificial one who does not call out his hat color with 100% certainty. We could make the first person the sacrificial one, though, by exploiting even more non-prohibited options.

After the line is formed and all the hats are in place, everyone could take a second to check out the hat color of the person in front of them, then they could reach forward, grab the hat off that person's head, and put it on their own head. The first person would be left hatless, and there'd be an extraneous hat at the end of the line, but 99 of them would all now know the color of the hat on their head with 100% certainty. They could call out from front to back or visa versa, depending on which they decided.

I get the sense Office Shredder has some much more mathematical solution in mind. I've been thinking all day and nothing like that has occurred to me.
 
  • #278
Another one that I love-

You die and the devil says he'll let you go to heaven if you beat him in a game. The devil shows you a round table (perfectly circular). He gives himself and you a huge pile of quarters. He says "ok, we'll take turns putting quarters down, no overlapping allowed, and the quarters must rest on the table surface. We can put the quarters anywhere on the table.The first guy who can't put a quarter down loses." You guys are about to start playing, and the devil says that he'll go first. However, at this point you immediately interject, and ask if you can go first instead. You make this interjection because you are very smart, and you know that if you go first, you can guarantee victory. Explain how you can guarantee victory.

Take your time on this one, easy to understand but challenging to solve :devil:.
 
  • #279
I think I got it but I don't have the mathamatical proof for it and am just going on intuition...
The first goes into the center (as that is the only unique position) then the rest go diametrically opposite to the 2nd player's ie. devil this will ensure(I think) that all the possible places are already taken up when the final turn of devil comes
I will need some time to come up with the proof (that is of course if I can and if the answer is correct...)
BTW- I am prepared to meet the devil, but whether he is prepared to meet me is an entirely different matter.*
}:-D
*Paraphrasing Churchill
 
  • #280
Only an odd number of quarters complete the circle, the odd one being at the center. But I'm too lazy to do the math. :p

On a side note, I don't know why stoichiometry jumped to mind? *shakes head*
 
  • #281
$$ [ \frac {\pi R^2} {\pi r^2}]=2k+1$$
$$ [ \frac { R^2} {r^2}]=2k+1$$
for any k in N
mmm..why?

EDIT: Actually the above whole thing is wrong.:redface:
I implicitly assumed that the gaps between coins won't sum up to area of more than one coin...
 
Last edited:
  • #282
conciousness had essentially the solution I was thinking of, but that solution of yours zooby is pretty awesome. I just figured it was silly if you could literally tell the person in front of you what color they should say.

For the devil one
Put the first quarter in the middle, then just mirror his moves across the table

I actually took a geometry elective in high school where this was the very first question the teacher asked. There are also two out of the box solutions I see from the wording of this particular puzzle:
Steal the devil's quarters, then make your move (I know this one is lame)
The devil said you take turns putting quarters down, so if the table is small enough, put down all your quarters in a manner such that no additional quarter may be laid. The devil then takes his turn and loses

As for the original solution I was thinking of for the black/white hat one:
The person in the back calls out black if he sees an odd number of black hats in front of him, white if he sees an even number. The person in front of him can deduce from that which color he is wearing, and calls it out. The person in front of him can then deduce which color he is wearing, and calls it out. Etc.
 
  • #283
Here is one just for a bit of fun- Get Off The Earth(1898) by Sam Loyd
It was printed on two pieces of card and sold more than 10 million copies. When the discs are set one way there are 13 characters, but when they move, one of the people disappears!
getofftheearth.gif

-http://www.murderousmaths.co.uk/
 
  • #284
for the original solution I was thinking of for the black/white hat one:
The person in the back calls out black if he sees an odd number of black hats in front of him, white if he sees an even number. The person in front of him can deduce from that which color he is wearing, and calls it out. The person in front of him can then deduce which color he is wearing, and calls it out. Etc.

ohhh...:approve: awesome.

Enigman, your intuition is essentially correct. A mathematical proof is not required, the solution can be given by making statements.

Gad,
I don't think that it is compulsory that the number of quarters will be odd if one is kept at the center. If we do exactly what Enigman has written then certainly the number will be odd though. :wink:
 
  • #285
Devil's game:

Each row of coins around the center coin will be a hexagon. A hexagon will always be comprised of an even number of coins. By insisting on setting the first, center coin, you force the devil to start each hexagon leaving yourself as the person who will complete it. When the table is too full for another whole hexagon, you are still forcing him to start the filling in, and you will still be the last person who can lay a coin, since the number of fill coins will be some multiple of 6; an even number.
 
  • #286
Actually ZBS, the devil could manipulate the distance between two coins making it impossible to make a regular hexagon.
 
  • #287
Enigman said:
Actually ZBS, the devil could manipulate the distance between two coins making it impossible to make a regular hexagon.
Alas, that is true, but it doesn't matter: the hexagon scenario (most possible space covered) proves that, regardless of what pattern he starts, you will be the one to finish it.
 
  • #289
Next one-
If this is a question, what is the answer?
:biggrin:__probably the silliest puzzle I know
 
  • #290
Enigman said:
Next one-
If this is a question, what is the answer?
:biggrin:__probably the silliest puzzle I know

If that is a question then this is an answer!

A man was sitting in a room, reading a book, when his wife entered the room and switched off the light. Although this occurred at night, and the room was now pitch dark, the man continued reading as though nothing had happened. How could he do this ?
 
  • #291
Blind.Braille.
 
  • #292
Right you are! BTW is my answer what you were expecting?

A classic one-

How many places are there on the Earth that one could walk one mile south, then one mile west, then one mile north and end up in the same spot? to be precise, let's assume the Earth is a solid smooth sphere, so oceans and mountains and other such things do not exist. you can start at any point on the sphere.

Read this after you think you have got it-
There is more than one
 
  • #293
consciousness said:
Right you are! BTW is my answer what you were expecting?
If that was a question you gave the correct answer.

consciousness said:
A classic one-

How many places are there on the Earth that one could walk one mile south, then one mile west, then one mile north and end up in the same spot? to be precise, let's assume the Earth is a solid smooth sphere, so oceans and mountains and other such things do not exist. you can start at any point on the sphere.

Read this after you think you have got it-
There is more than one

Crooked and Evil...I like that.
I got it(I think) but I will let the others ponder it a bit...
 
Last edited:
  • #294
consciousness said:
If that is a question then this is an answer!

A man was sitting in a room, reading a book, when his wife entered the room and switched off the light. Although this occurred at night, and the room was now pitch dark, the man continued reading as though nothing had happened. How could he do this ?
He must have been blind and was reading braille. Issues like this come up in the movie Ray[/], about Ray Charles.
 
  • #295
You seem to be missing an i...*
EDIT:* in the italics tag in your post I mean.
There are infinite solutions to the enigma(#292) (aside from the obvious one point) the solution becomes easier when you think in terms of spherical coordinates- but that's just my opinion.
 
Last edited:
  • #296
He didn't have to be blind, just reading Braille.
 
  • #297
consciousness said:
A classic one-

How many places are there on the Earth that one could walk one mile south, then one mile west, then one mile north and end up in the same spot? to be precise, let's assume the Earth is a solid smooth sphere, so oceans and mountains and other such things do not exist. you can start at any point on the sphere.

Read this after you think you have got it-
There is more than one

Quoting it just for clarity...
(Originally by conciousness.)
(making sure people have seen this one.)
 
  • #298
consciousness said:
A classic one-

How many places are there on the Earth that one could walk one mile south, then one mile west, then one mile north and end up in the same spot? to be precise, let's assume the Earth is a solid smooth sphere, so oceans and mountains and other such things do not exist. you can start at any point on the sphere.

Read this after you think you have got it-
There is more than one
There is some circle north of the south pole whose circumference is one mile. With the south pole as the center of that circle, the distance from the south pole to that circle is exactly 1/pi/2 miles, and approximately .159 miles from the south pole.

Starting at any point one mile north of that circle, you can walk one mile south, one mile west around the circle, and then one mile north, and arrive back at the same point you started from. All the points that are one mile north of that circle constitute another circle which has the south pole as its center and an exact radius of 1 + 1/pi/2 miles (an approximate radius of 1.159 miles). Any point on this second circle fulfills the conditions of the riddle, and, since there are infinitely many points in the circumference of a circle, there are infinitely many "places" on Earth from which you can walk south one mile, west one mile, north one mile, and end up at the same point from which you started. The other case is the north pole, of course.
 
  • #299
Enigman said:
You seem to be missing an i...*
EDIT:* in the italics tag in your post I mean.
There are infinite solutions to the enigma(#292) (aside from the obvious one point) the solution becomes easier when you think in terms of spherical coordinates- but that's just my opinion.

I went out to have coffee, solved it while I was having coffee, then returned to find out you'd blurted out the answer without spoilers.

So, I want to state for the record, I found my answer legitimately before seeing your indiscretion.

I think additional clues should be the sole prerogative of the poster of the enigma, and should only be in response to explicit expressions of frustration by at least two respondents.
 
  • #300
lisab said:
He didn't have to be blind, just reading Braille.

Good point! And one I should have considered, because I know a woman who can see but who works with blind people and can read braille perfectly well.
 

Similar threads

Challenge Intermediate Math Challenge - August 2018
Replies
20
Views
6K
Challenge Are You Ready for the Basic Math Challenge This August?
2
Replies
57
Views
11K
Challenge Basic Math Challenge - September 2018
Replies
38
Views
8K
Challenge Intermediate Math Challenge - September 2018
Replies
28
Views
6K
Challenge Math Challenge - December 2018
2
Replies
67
Views
14K
Challenge Can You Solve These Math Challenges?
2
Replies
82
Views
15K
Codebreaker Wanted~Find the diamonds
Replies
15
Views
5K
Admissions Comments on PhD Statement of Purpose
Replies
6
Views
2K
Challenge Intermediate Math Challenge - May 2018
2
Replies
61
Views
12K
Challenge Intermediate Math Challenge - July 2018
Replies
16
Views
6K

Hot Threads

Recent Insights

Back
Top