Solve Enjoyable Enigmas with Mr.E's Challenge

[Ibix]

Spoiler

He climbed on a block of ice to tie the rope. It subsequently melted, leaving a water stain and a locked room mystery.

 

[collinsmark]

Spoiler

He climbed on a block of ice to tie the rope. It subsequently melted, leaving a water stain and a locked room mystery.

This is correct. (It is the solution to the hanging man enigma.)

 

[Ibix]

Regarding the dalmations, I think collinsmark's solution is correct, but can be expressed rather more simply.

Spoiler

It is indeed a proof by contradiction. Imagine a set of eleven dogs from which we cannot pick a subset divisble by eleven. Line the dogs up in a row - sit! Stay! Let us say that the ith dog has S_i spots.

In front of each dog, write the remainder when its spots plus the spots of all dogs to its left are divided by eleven. That is, in front of the ith dog, write the remainder when $\sum_{k=1}^i S_k$ is divided by 11.

You now have eleven numbers. If any two of them are the same, then the intervening dogs' spots must add to a multiple of 11 (so if the nth and mth remainders are the same then $\sum_{k=n+1}^m S_k$ is divisible by eleven.

So if no sets are divisible by eleven then you have eleven distinct non-zero remainders from division by eleven - but there are only ten such numbers. That is a contradiction - therefore you can always find a subset whose spots add to a multiple of eleven.

 

[collinsmark]

Regarding the dalmations, I think collinsmark's solution is correct, but can be expressed rather more simply.

Spoiler

It is indeed a proof by contradiction. Imagine a set of eleven dogs from which we cannot pick a subset divisble by eleven. Line the dogs up in a row - sit! Stay! Let us say that the ith dog has S_i spots.

In front of each dog, write the remainder when its spots plus the spots of all dogs to its left are divided by eleven. That is, in front of the ith dog, write the remainder when $\sum_{k=1}^i S_k$ is divided by 11.

You now have eleven numbers. If any two of them are the same, then the intervening dogs' spots must add to a multiple of 11 (so if the nth and mth remainders are the same then $\sum_{k=n+1}^m S_k$ is divisible by eleven.

So if no sets are divisible by eleven then you have eleven distinct non-zero remainders from division by eleven - but there are only ten such numbers. That is a contradiction - therefore you can always find a subset whose spots add to a multiple of eleven.

Yes. I like your solution in that it's far easier to conceptualize compared to mine, me thinks.

Spoiler

It does add some insight into what I was doing though. In your solution, the striving to keep any resulting number (in front of the dog) from being 0 or from being equal to any other dog's number, is equivalent to my solution's F_n number restrictions removing a unique, non-zero number as a valid choice (for each dog). After that, your solution and mine are pretty much identical in the fact that we run out of numbers before we run out of dogs.

 

[consciousness]

Next one-

A group of prisoners are trapped in a forcefield (like an invisible wall). These prisoners are perfectly brave, meaning that a prisoner would attempt an escape if he has any positive probability of success. The prisoners are monitored by a guard who has only one bullet in his gun, but who also has perfect marksmanship skills (he never misses).

The prisoners have overheard the guard saying "I have only one bullet left!". A maintenance technician needs to tune up the forcefield generator, and so for one second, the forcefield is released. How can the prisoners be kept detained?

 

[CompuChip]

Taking "any positive probability of success" literally, the problem is that if all n > 1 prisoners dash out, the guard can only shoot one of them so they have a probability of (n - 1)/n of escaping. So I'm thinking that this is the "problem" you need to solve / prevent.

Spoiler

Is it something as simple as saying "I will shoot the first one to come out", after which they will all wait for another one to go first?

 

[consciousness]

Spoiler

Is it something as simple as saying "I will shoot the first one to come out", after which they will all wait for another one to go first?

Correct.

Spoiler

More clear might be "the first person to lift his leg will be shot."

 

[Office_Shredder]

consciousness, if the guard does that then as a prisoner I

Spoiler

shove another prisoner, making him stumble and wasting the guard's bullet

 

[Enigman]

Or everyone agrees to go together...
1/n chance...

 

[consciousness]

consciousness, if the guard does that then as a prisoner I

Spoiler

shove another prisoner, making him stumble and wasting the guard's bullet

Spoiler

The best answer is perhaps then "the first one to make any movement in the next second will be shot!"

Or everyone agrees to go together...
1/n chance...

Spoiler

There is bound to be a small time lag so technically they can't go together. It is assumed that the guard is very alert. He can detect this small lag and know who moved first. (In hindsight a programmable turret gun would have worked better)

 

[Enigman]

It is assumed that the guard is very alert. He can detect this small lag and know who moved first. (In hindsight a programmable turret gun would have worked better)

So? The other prisoners still escape. The act of moving first is random and hence all of them as far they are concerned have 1 -1/n chance of survival.

 

[consciousness]

So? The other prisoners still escape. The act of moving first is random and hence all of them as far they are concerned have 1 -1/n chance of survival.

I always made the assumption that the prisoners would wait for others to move and ensure their survival. But now I am convinced that your post destroys this solution!

So the correct answer hasn't been posted yet. An alternative solution that doesn't warrant assumptions about the behaviour of the prisoner exists.

 

[consciousness]

(The prisoners know that the other prisoners are perfectly brave.)
Hint-

Spoiler

You can assign numbers to make the prisoners distinguishable.

 

[JorisL]

Spoiler

Shoot the technician and wait until more bullets arrive. This is assuming the force field won't break down if the tune up gets delayed a little.

 

[Office_Shredder]

Spoiler

Assuming it takes the prisoners more than one second to escape the guard's line of fire: Assign the prisoners numbers 1,...,n. Then the guard announces "When the shield goes back up I will kill the prisoner whose number is lowest that is not inside the shield.

1 can't try to escape or he will die. 2 knows this, so knows he can't try to escape either. Etc.

 

[Enigman]

Next one-

.-- --- .-- / -.-- --- ..- / -.-. .- -. / ... . . / -- .
Google search is allowed.

 

[Enigman]

Running out of ideas for Enigmas...

 

[consciousness]

Spoiler

Assuming it takes the prisoners more than one second to escape the guard's line of fire: Assign the prisoners numbers 1,...,n. Then the guard announces "When the shield goes back up I will kill the prisoner whose number is lowest that is not inside the shield.

1 can't try to escape or he will die. 2 knows this, so knows he can't try to escape either. Etc.

Correct!

Also shooting the tecnician might work but the prisoners will most probably not believe the guard.

 

[cArma]

Next one-

Spoiler

Samuel Morse. No need to search. Answer is in the URL of the picture.

 

[Enigman]

Spoiler

Samuel Morse. No need to search. Answer is in the URL of the picture.

Not the question ...
What does that post mean?
A bit 'under the belt' Enigma...
But since you quoted it, it should be easier...

 

[consciousness]

Spoiler

Wow I can see you!

 

[CompuChip]

Spoiler

Assuming it takes the prisoners more than one second to escape the guard's line of fire: Assign the prisoners numbers 1,...,n. Then the guard announces "When the shield goes back up I will kill the prisoner whose number is lowest that is not inside the shield.

1 can't try to escape or he will die. 2 knows this, so knows he can't try to escape either. Etc.

Is this fundamentally different from what I said initially?

 

[Enigman]

Spoiler

Wow I can see you!

Sold!

 

[consciousness]

Is this fundamentally different from what I said initially?

There is a subtle difference.

Spoiler

If the prisoners decide to rush out together, a prisoner can unconsciously be the first to move. His demise is certain but the probability of his dying wasn't one when he made the decision!
Now in this solution if the prisoner numbered 1 decides to go first then he makes a conscious decision to kill himself! This starts the domino effect.

 

[Enigman]

The Enigma That Killed Homer.

I generally hate it when a movie or T.V. show 'adapts' some real world event to suit itself. But just sometimes these leave me grinning ear to ear, like the last Doctor who finale- It's prophesied that at some place (don't remember the name) a question will be asked to the Doctor and it is at this place he will die- obvious conclusion don't go to that place...well, Doctor didn't oblige and so didn't Homer (Iliad guy NOT yellow toon whose guts I hate).
It was foretold by The Oracle of Delphi that Homer would die in the island of Ios and that he should beware of a riddle posed to him by some young boys. Well obviously old Homer disregarded the prophecy and while he was walking on the banks of Ios a group of fisher boys asked him a riddle:

“What we caught, we threw away; what we didn’t catch, we kept. What did we keep?”

Unable to solve the riddle, Homer eventually died on the island, refusing to leave until he discovered the answer. So can you solve it?
(Googling is not allowed)

 

[collinsmark]

Next one-

.-- --- .-- / -.-- --- ..- / -.-. .- -. / ... . . / -- .
Google search is allowed.

[Source: http://abstrusegoose.com/255]

 

[Enigman]

It itches...

 

[collinsmark]

It was foretold by The Oracle of Delphi that Homer would die in the island of Ios and that he should beware of a riddle posed to him by some young boys. Well obviously old Homer disregarded the prophecy and while he was walking on the banks of Ios a group of fisher boys asked him a riddle:

“What we caught, we threw away; what we didn’t catch, we kept. What did we keep?”

Unable to solve the riddle, Homer eventually died on the island, refusing to leave until he discovered the answer. So can you solve it?
(Googling is not allowed)

Spoiler

Is it bait?

 

[Enigman]

Spoiler

Is it bait?

Nope, what you couldn't catch you had to keep, unfortunately...

 

[collinsmark]

“What we caught, we threw away; what we didn’t catch, we kept. What did we keep?”

Hmm. How's about this guess:

Spoiler

"Mistakes." As in, if you catch your mistake in time, you can conceivably correct it (the mistake goes away), and all is good. But if you don't catch your mistake in time, you have to live with it.

 

[Enigman]

Hmm. How's about this guess:

Spoiler

"Mistakes." As in, if you catch your mistake in time, you can conceivably correct it (the mistake goes away), and all is good. But if you don't catch your mistake in time, you have to live with it.

Hmmm...*Scratches the E head* might work but not the answer the fishermen were talking about

It itches...

 

[Enigman]

Spoiler

Lice!

 

[drizzle]

Haha!

 

[collinsmark]

Okay, here's a new one. (And back to morbid.)

A man is recently dead, alone in an empty field. He died a quick and traumatic death. The man was young and healthy immediately before the time of his death. The only belongings the man had with him are the clothes on his back and an unopened package. The man would have survived had he opened the package.

So the question is, what is in the package?

 

[Ibix]

It is a little worrying how many of these puzzles involve dead guys.

Spoiler

A parachute