Solve Enjoyable Enigmas with Mr.E's Challenge

[collinsmark]

Someone sent me this a while back, I'm sure it's searchable, but don't cheat! It's a tough-ish one but not so bad.

You are given eight cards with numbers written on them: 4,4,3,3,2,2,1,1. Your task is to arrange the cards in such a way that the ones are separated by one digit, the twos are separated by two digits, the threes are separated by three digits, and the fours are separated by four digits. You must use all the cards. What is the number you create?

Spoiler

23421314...

Spoiler

I think Gad's way is valid (as is the symmetrical counterpart 41312432).

But I think the problem, as stated, is ambiguous in what is meant by being separated by a certain amount of digits.

One way to interpret being "separated by one digit" is the way Gad did, in that the as in xxxxaxax are separated by by one 'x' (where x is some other number besides a.

But another way to interpret "separated by one digit" is being adjacent to each other. In other words they are one digit away from being right on top of each other. For example in xaxaxxxx the as are separated by 2 (as in the second a is positioned two units away from the first). If that's the case, a valid solution could be 42324311, or its symmetrical counterpart 11342324.

 

[lisab]

Spoiler

I think Gad's way is valid (as is the symmetrical counterpart 41312432).

But I think the problem, as stated, is ambiguous in what is meant by being separated by a certain amount of digits.

One way to interpret being "separated by one digit" is the way Gad did, in that the as in xxxxaxax are separated by by one 'x' (where x is some other number besides a.

But another way to interpret "separated by one digit" is being adjacent to each other. In other words they are one digit away from being right on top of each other. For example in xaxaxxxx the as are separated by 2 (as in the second a is positioned two units away from the first). If that's the case, a valid solution could be 42324311, or its symmetrical counterpart 11342324.

I interpreted it the way Gad did, I think she got it . Travis, is that right?

 

[consciousness]

I had to write out and solve 3 equations. What's "hit and trail"?

Spoiler

The trick is to do everything backwards. Start with 3, add 2 and double to get 10. Then add three and double...oh and I meant "trial and error"

 

[Enigman]

Another FRUITY puzzle...
And no, you can't eat them Gad, not yet.
You have been given two oranges one unpeeled the other peeled and an aquarium which is full of water- nothing else.
Your job- make them both float. You can't use anything other than the things given.
---------------------------------------------------------------------------------
mmm...the thread got quite a few enigmas while I was away...delicious...

 

[Travis_King]

I took it the way Gad did as well, so we'll call that the correct answer. Though I admit when I first read the problem I too wondered if they meant that "separated by one digit" meant that the two 1's were adjacent. However, after reading it a couple times, I think it was clearly intended to mean that the two 1's must be separated by another digit between them. Congrats

 

[drizzle]

*cuts both pealed and un-pealed oranges into two half, and eats half of each leaving the other halfs for PFers to solve*

 

[Office_Shredder]

OK, let's roll with Gad's solution. I eat both oranges entirely, then do a backfloat in the aquarium (which I assume was designed to hold dolphins).

A more real attempt at a solution which is probably wrong

Spoiler

OK, I didn't even know which one was supposed to float normally so I looked it up on youtube. I found a couple videos showing the peeled orange is the one that sinks. They both claimed that the reason why is that the rind is less dense than the orange - for the purposes of this solution I disbelieve their claims because neither showed the rind itself floating, so really they shouldn't be calling themselves science videos if they don't even test their hypothesis.

Instead I'm going to assume the rind is less dense than water, but the orange itself has air pockets in it.. When peeled, the orange's air holes are filled by water and it sinks. So if you peel half of the unpeeled orange, and place that under the peeled orange, that orange should be able to float as long as it only sinks half of its volume into the water. If the peel is heavier than water, instead place the peel on top of the orange to prevent air from escaping and then place two oranges with rind-hats in the water.

If I had an orange I would test to see if either of these work, but alas I do not.

 

[Enigman]

OK, let's roll with Gad's solution. I eat both oranges entirely, then do a backfloat in the aquarium (which I assume was designed to hold dolphins).

You violated the first rule of fruit puzzles:
Gad always gets the fruit!

Spoiler

the rind is less dense than water, but the orange itself has air pockets in it.. When peeled, the orange's air holes are filled by water and it sinks. So if you peel half of the unpeeled orange, and place that under the peeled orange, that orange should be able to float as long as it only sinks half of its volume into the water. If the peel is heavier than water, instead place the peel on top of the orange to prevent air from escaping and then place two oranges with rind-hats in the water.

Correct.

Spoiler

The first assumption is correct. And if they are the same size (a real snug fit) it doesn't matter where you place the rind it will always return to rind on top position.

Spoiler

I probably should have mentioned the size constraints- unpeeled orange cannot be too much smaller than peeled one.

 

[collinsmark]

So we don't have a complete answer yet to the oranges; is that right?

Spoiler

The question that remains in my mind is how does one attach the previously peeled orange to the half rind, without the peeled orange falling out, and down to the bottom of the aquarium. (The orange/rind combo will naturally tend to "capsize.")

 

[Enigman]

So we don't have a complete answer yet to the oranges; is that right?

Spoiler

The question that remains in my mind is how does one attach the previously peeled orange to the half rind, without the peeled orange falling out, and down to the bottom of the aquarium. (The orange/rind combo will naturally tend to "capsize.")

Spoiler

As I said size constraints- the peeled orange has to be slightly larger than unpeeled orange...(should have mentioned it) then you can just stretch it to fit the orange and conversely if its larger you can manage it with some clever balancing- you will need to use slightly more than half a rind...isn't very stable though.

 

[lendav_rott]

Spoiler

Throw the peeled and the unpeeled orange into the water tank, they will float.

 

[inotyce]

Does adding salt into water help float the fruits ?

 

[Enigman]

Spoiler

Throw the peeled and the unpeeled orange into the water tank, they will float.

peeled one sinks- already have done this at home. Office shredder gave the correct anser.

Does adding salt into water help float the fruits ?

Yes, it would but you can't use salt, sorry.

 

[Enigman]

Four people need to cross a rickety rope bridge to get back to their camp at night. Unfortunately, they only have one flashlight and it only has enough light left for seventeen minutes. The bridge is too dangerous to cross without a flashlight, and it’s only strong enough to support two people at any given time. Each of the campers walks at a different speed. One can cross the bridge in 1 minute, another in 2 minutes, the third in 5 minutes, and the slow poke takes 10 minutes to cross. How do the campers make it across in 17 minutes?

 

[Ibix]

Deleted because spoiler tags didn't work right - I'll post again later.

 

[Ibix]

Try again.

Edit: fluff! Because the first few characters of the post shown in the post list does not respect the spoiler tag - in the app at least.

Spoiler

2 & 1 cross, 2 returns (4 minutes)
5 & 10 cross, 1 returns (11 minutes)
2 & 1 cross (2 minutes)
Total - 17 minutes

That's about the third time I've seen the problem, but the first time I've solved it. :D

 

[zoobyshoe]

Spoiler

2 & 1 cross, 2 returns (4 minutes)
5 & 10 cross, 1 returns (11 minutes)
2 & 1 cross (2 minutes)
Total - 17 minutes

Spoiler

I think this must be right! Excellent job! It had me going in circles all morning.

 

[Ibix]

I puzzled over it for hours - fifteen years ago. For some reason it clicked in about two minutes today - maybe because I'd had time to think about it.

Here's one that took me ages (then I asked a mathematician friend who stared at me for about five seconds before reeling off the answer, which made me feel a bit dumb).

I have eleven dalmations. Prove that it is always possible to select some or all of them such that the sum of the spots of the selected dogs is an integer multiple of eleven.

You may consider eleven bunches of grapes, if that makes the problem more familiar.

 

[zoobyshoe]

What's an "integer multiple," as opposed to just a multiple?

 

[Enigman]

Try again.

Edit: fluff! Because the first few characters of the post shown in the post list does not respect the spoiler tag - in the app at least.

Spoiler

2 & 1 cross, 2 returns (4 minutes)
5 & 10 cross, 1 returns (11 minutes)
2 & 1 cross (2 minutes)
Total - 17 minutes

Correct.
And the app's buggy- it doesn't show spoilers properly...

 

[Ibix]

What's an "integer multiple," as opposed to just a multiple?

The total number of spots is 11n, where n is an integer.

"Integer" is a redundant modifier in this context because the problem is trivial if you interpret "multiple of eleven" as 11x where x is real. Sorry.

 

[Ibix]

Correct.

Woohoo!

And the app's buggy- it doesn't show spoilers properly...

So I realized. You need to reload the page after posting to not see them (as it were), but then you can't always see the full text when you want to.

 

[collinsmark]

I have eleven dalmations. Prove that it is always possible to select some or all of them such that the sum of the spots of the selected dogs is an integer multiple of eleven.

You may consider eleven bunches of grapes, if that makes the problem more familiar.

Stayed up till 1:00 AM last night trying to figure this one out. Much of that time was brushing up on Galois fields, since I suspect the subject might be related, together with the "characteristic" of a finite field. I haven't been able to prove anything yet though.

[Edit: I also started coding up a program to prove that the assertion is true (I don't doubt that it is true) or show that it is false. But then I realized that the inner loop would need to go through about 2¹¹11¹¹ iterations (584,318,301,411,328 iterations). Even with my fast computer, it would take a very long time. I could knock that number down significantly by taking advantage of symmetry, but even then it would still take a long time (just not quite as long). So I stopped that effort.]

 

[consciousness]

I also stayed up for this problem, managed to simplify it somewhat but the "basic" logic behind the question eludes me. I have abandoned it for now, because the solution might require some number theory theorem which I don't know and have little chance of stumbling upon.

 

[Enigman]

...I have postponed solving it for a week...got tests coming up...Last idea was it holds even if the 11s are exchanged by 1, 2 or 3...perhaps its a general trend?...Need to focus on physics now.
Sayonara

 

[Enigman]

mathematical induction may help if the previous post is correct...

 

[collinsmark]

I can prove it conclusively for 3 dalmatians and total spots of selected dogs being an integer multiple of 3 (proof involves proof by contradiction). But I haven't yet been able to carry that over to 11.

 

[CompuChip]

Actually I'll put my reply in spoiler tags, it's not a complete answer but it's my thoughts about the problem so far.

Spoiler

I'm thinking more along the lines of a group theoretic proof, where there is probably some cool result about $\mathbb Z_{11}$ that will show it.
If all dalmatians had equally many spots $a$, it would be fairly straightforward, because since $p = 11$ is prime and any element of $\mathbb Z_p$ generates the whole group, you need to multiply your generator $[a]$ by at most $p$ (and if I recall correctly, exactly $p$, since the order of any element is the order of the group) before at some point you hit the identity $[0] \in \mathbb Z_p$.

The problem is that all the dalmatians have a different number of spots, so you need to take $p$ arbitrary elements and show that a subset of them gives the identity. Proof by contradiction may work. I was also thinking about "reverse" induction: you cannot have all $p$ elements be the same (otherwise you are back to the easy case) so there must be at least one different one - if we can show that a subset of the remaining $p - 1$ identical elements produces its inverse you might be able to continue this line of reasoning ($p - 2$ identical elements, $p - 3$, etc.).

 

[collinsmark]

I have eleven dalmations. Prove that it is always possible to select some or all of them such that the sum of the spots of the selected dogs is an integer multiple of eleven.

You may consider eleven bunches of grapes, if that makes the problem more familiar.

Solved it! Solution below.

Btw, before I give the solution, it might be necessary to rephrase to the puzzle such that each dalmatian is required to have at least one spot. That makes sense with the bunches of grapes (it's difficult to call a bunch of grapes a bunch of grapes if the bunch has no grapes) but it probably should be explicitly stated for the dalmatians version.

Spoiler

This solution relies on an overall scheme of "proof by contradiction" combined with some smaller steps involving "mathematical induction."

And before I start I should apologize for my poor rigor. I'm not a mathematician and have never taken a math class beyond what is required for engineering.

Before getting to the proof, I need to introduce the concept of Galois Fields. They're not difficult to comprehend: they're just a finite number system that "loops" around as one continues to count up or down. For example, Galois Field 11, denoted as GF(11) contains the natural numbers 0 through 10, and that's it. 11, 22, 33 are all the same thing as 0. If you try to count beyond 10 in GF(11) you wrap back around to 0.
0,1,2,3,4,5,6,7,8,9,10,0,1,2,3,4,5...

The addition operator is pretty straightforward. We define "+" such that in terms of standard integers, it is the sum of a and b MOD 11. Some examples for GF(11) are:
1 + 1 = 2,
2 + 5 = 7,
8 + 2 = 10,
8 + 3 = 0,
8 + 4 = 1,
8 + 5 = 2.

The negation operator maps the number to a new number between 0 and 10 [for GF(11)]. Examples are:
-1 → 10
-2 → 9
-3 → 8

So the subtraction operator examples [for GF(11)] are,
5 - 4 = 1
5 - 5 = 0
5 - 6 = 10
5 - 7 = 9
5 - 8 = 8

We could also define multiplication and addition operators, but its not necessary for this puzzle.

Why Galois fields? Because we're trying to ensure that the total number of spots on the selected dogs, after dividing by 11, has a remainder of 0. We've succeeded in solving the problem if we can prove that there is always a selection of dogs such that (Total number of spots of selected dogs) MOD 11 = 0. And modulo arithmetic is what Galois fields are good at.

So now on to the meat of the solution. We're going to use a "proof by contradiction." What we're going to do is assume that the puzzle statement is false. All we need to do is find a configuration of dog spots -- any configuration -- such that it is impossible to choose a set of dogs such that the spots can be divided by 11 [Same as saying the spots sum to 0 in GF(11) math]. We will later show this to be impossible.

Let s₁ be the number of spots on dog 1, s₂ be the number of spots on dog 2, and s_n be the number of spots on dog n, where n goes from 1 to 11 (there are 11 dogs).

The approach we will take is we get to pick the number of spots on the dogs. The goal is to restrict our picks such that is impossible to divide the total spots by 11 with just one dog. Then moving on, restrict our selection such that is impossible to divide by 11 with either one or two dogs with dog 2; one, two or three dogs with dog 3, and so on.

It should be noted here that in terms of our math, 12 spots is the same thing as 1 spot, which is equal to 23 spots, 34 spots, etc. Since we are only interested in the remainder after dividing by 11, the number of spots on each dog ranges only from 0 to 10 (where '0' doesn't really mean '0' in normal integers; it can be any integer multiple of 11. But for the math here, we call it 0).

Now for the first dog. We restrict the number of spots on this dog, s₁, such that it's not equal to 0. (remember 0 is the same thing as 11, 22, 33, 44, etc). That's because if any given dog has a number of spots that are an integer multiple of 11, one could just pick that dog alone, and call it quits. Therefore no individual dog can have a number of spots that are an integer multiple of 11 (in our Galois field math we say the dog's number of spots cannot be 0). For reasons which should become apparent later, allow me to define the function F₀(n), such that
F₀(n) = 0,
for all n.
Now we formalize our restriction on s₁,
s₁ ≠ F₀(1).
Again, I hope that the function notation will be more clear later. An interesting thing to point out is that we've reduced the number of possibilities for the first dog's spots by 1.

Now let's move on to our second dog. It's number of spots needs to have the same restrictions as the first dog, plus a new one. The new one is that the spots on the second dog, plus the spots on the first dog, cannot add to 0 (same thing as "cannot add to an integer multiple of 11"). In other words, s₂ ≠ -s₁. Let me formalize that with a new function, F₁(n),
F₁(n) = F₀(n-1) - s_n-1
and our restrictions on the second dog are:
s₂ ≠ F₀(2)
s₂ ≠ F₁(2)
Don't forget that F_n(m) is ultimately just a number that can range from 0 to 10. It's just some number. And it can be easily shown that F₀(2) and F₁(2) are unique. If they weren't unique but instead equal, it would mean that
F₁(2) = F₀(2), and expanding both sides gives,
-s₁ = 0, or simply s₁ = 0.
But that's not possible since s₁ cannot be 0 (restriction on the first dog) so F₀(2) and F₁(2) must represent unique numbers.

You might notice something neat happening here. We've reduced our possible choices by 1 again. There are 10 numbers to choose from for the first dog, but only nine for the second. Now let's move on to the third dog.

The third dog has all the same restrictions as the first dog and second dog plus a new restriction. Well, a couple new restrictions. Firstly, we need to apply the F₁ restriction twice, one for each of the previous dogs. The new restriction is that s₃ ≠ - (s₁ + s₂). Formalizing that with a new function,
F₂(n) = F₁(n-1) - s_n-1
our restrictions become,
s₃ ≠ F₀(3)
s₃ ≠ F₁(2)
s₃ ≠ F₁(3)
s₃ ≠ F₂(3)
Here I'm going to just remove the s₃ ≠ F₁(2) restriction. It turns out to be redundant for this particular puzzle. Remember, we are going to use a "proof by contradiction" approach. And later, if we can show that it is impossible to choose spots that falsify the puzzle statement with fewer restrictions, it is still impossible with greater restrictions. So the only restrictions we need to worry about for the third dog are:
s₃ ≠ F₀(3)
s₃ ≠ F₁(3)
s₃ ≠ F₂(3)
Now we know that F₁(3) and F₀(3) are not equal. But we can explicitly show that here. If they were equal,
F₁(3) = F₀(3), and expanding both sides gives,
-s₂ = 0, or more simply s₂ = 0, which is impossible due to earlier restrictions.
Similarly we can show that F₂(3) and F₁(3) are not equal. If they were,
F₂(3) = F₁(3), and expanding
F₁(2) - s₂ = - s₂, and expanding more and simplifying,
s₁ = 0, which is impossible.
But what about F₂(3) and F₀(3)? They are also not equal. If they were equal,
F₂(3) = 0,
F₁(2) - s₂ = 0,
-s₁ - s₂ = 0, which is impossible due to previous restrictions on s₂.

So finally, when choosing a number for s₃, there are only 8 choices: down one from number of choices from s₂.

So now is where our mathematical induction comes in. With each new dog, we place the same restrictions as with previous dogs plus an additional restriction which keeps the sum of all dogs, up to that dog from summing to 0. The new restriction comes in the form, for dog m
s_m ≠ F_m-1(m)
where
F_n(k) = F_n-1(k-1) - s_k-1.
[Edited above equation.]

So when we get to the eleventh dog, the restrictions on its spots are,
s₁₁ ≠ F₀(11)
s₁₁ ≠ F₁(11)
s₁₁ ≠ F₂(11)
s₁₁ ≠ F₃(11)
s₁₁ ≠ F₄(11)
s₁₁ ≠ F₅(11)
s₁₁ ≠ F₆(11)
s₁₁ ≠ F₇(11)
s₁₁ ≠ F₈(11)
s₁₁ ≠ F₉(11)
s₁₁ ≠ F₁₀(11)

Here's the kicker: There are 11 restrictions (starting at 0 and ending at 10, making 11 total). Each restriction removes a possible number from the choosing. But there are only 11 numbers total that we are working with. It is impossible to choose a number of spots for the eleventh dog, such that one can't sum the spots to 0 (where 0 is the same thing as an integer multiple of 11). Our "goal" of trying to disprove the puzzle statement is impossible.

[Edit: There are actually many more restrictions than just the 11 above. For example, there is only a single restriction involving the spot sum of exactly two dogs. And those are dogs 10 and 9, involving F₂(11). But there could be restrictions involving the spot sum of any two dogs. But including those restrictions make the contradiction of the puzzle statement even more impossible. The eleven restrictions given above are alone sufficient to prove this puzzle by "proof by contradiction."]

Therefore, there will always be a way to select some or all of the dogs such that the total number of spots on selected dogs adds to an integer multiple of 11.

 

[CompuChip]

Btw, before I give the solution, it might be necessary to rephrase to the puzzle such that each dalmatian is required to have at least one spot.

If there is a dog with no spots, you pick that one, and you have 11n spots in total, with n integer

 

[Enigman]

This belongs more to O_S' Math challenge thread rather than my thread...Congrats collinsmark!

 

[lisab]

If there is a dog with no spots, you pick that one, and you have 11n spots in total, with n integer

I think a Dalmatian with no spots is called a "White Dog".

 

[collinsmark]

Okay, here's a rather morbid one. (Then again, many enigmas/riddles are rather morbid. So this one is in good company.)

You are investigating what might be a murder, or maybe it's a suicide. You haven't figured it out yet.

A man was found dead by hanging.

The body was found hanging by a rather short rope attached to a chandelier (the rope was just long enough to be tied to the chandelier on one end and to make a noose on the other). The room has particularly high ceilings and the body is high enough off the ground such that one couldn't jump up to grab the rope used as the noose. (Even if the rope hadn't yet been tied into a noose, it's still too high off the ground to jump and grab.)

The room he was found in was securely locked from the inside (you and the police had to break the door down to gain entrance). The room has no other doors, windows or any other entrances or exits (no tunnels either, or any holes in the ceiling or anything like that). (Presumably the room isn't air-tight though; you don't need to be concerned with suffocation or carbon dioxide poisoning).

The room is also devoid of furniture. As a matter of fact, it's nearly an empty room. The only things in the room are the man/body, the rope, the chandelier the rope is attached to, and a water-stain on the floor.

Was this man murdered? Did he commit suicide? How did he die?

 

[CompuChip]

I know this one. It's pretty cool!

 

[JorisL]

I see what you did there CompuChip. Know it as well