Math Fun

[Lev_Muishkin]

@phoneutria

Sauwelios:
[tab]It also can’t be any product greater than 198.[/tab]

Eh?

Is that in the problem?

[Lev_Muishkin]

@Carleas

That’s the one, Phoneutria.

Here’s another I just came across, I haven’t had time to try to figure it out but it looks like fun, and mathier than most problems of this kind.

@"[url=http://www.washingtonpost.com/news/volokh-conspiracy/wp/2015/04/15/beyond-cheryls-birthday/

Sasha Volokh[/url]"]Two numbers a and b are between 2 and 99. [Note: They’re not constrained to be different from each other. [S.V.]]

Peter is given the product of the numbers, ab (and knows he is given the product).

Sarah is given the sum a+b (and knows she is given the sum).

They also know the numbers are between 2 and 99.

They are UVa math majors, so they are great at math and completely honorable!

Peter says, “I don’t know the numbers.”

Sarah says, “I knew you didn’t know the numbers.”

Peter then says, “I know the numbers now.”

Sarah then says, “Ah ha! I know the numbers now.”

What are the numbers?

This one could take a while.

No matter what the solution to this one, the distintion between a and b is mute. Once again - badly worded, as not matter how cleverly you come up with some two numbers that work, without a subtraction in the mix, you can’t determine which of the numbers is a and which is b.

[James_S_Saint]

@Lev Muishkin

@phoneutria

No no…

[tab]A says B can’t know the answer, so it can’t be any number that occurs only once. 19 and 18 are out along with may and june.

May 15
May 16
May 19
June 17
June 18
July 14
July 16
August 14
August 15
August 17

I don’t get this bit?

That clue solves the problem of the month for B, so it had to be a number that repeats for either may or june. july 14 and august 14 are out.

May 15
May 16
May 19
June 17
June 18
July 14
July 16
August 14
August 15
August 17

Knowing that B knows solved the problem for A. Therefore it has to be a date with only one month option.

May 15
May 16
May 19
June 17
June 18
July 14
July 16
August 14
August 15
August 17[/tab]

Lev, was the red print yours? …that you don’t get that part?

If so;
[tab]Einstein knew the month.
Bernard knew the day.

If Bernard could not deduce possibly the month from knowing the day, the month could not be May or June.
If it was even possible that perhaps Bernard could know the month after only hearing the day, that month would have to have a unique day mentioned. Because he could not possibly know the month regardless of whatever day he heard, the month could not have been May or June.[/tab]

[James_S_Saint]

@Lev Muishkin

@Sauwelios

[tab]I thought I was well on my way to solving it–I had just reasoned out what I considered the final key to the solution–, when I suddenly realised a very basic fact had escaped my attention. The fact that the numbers are between 2 and 99 means their product cannot be a prime number. But then it’s completely superfluous for Peter to say he doesn’t know the numbers (which he could, from the information given him, only know if the product was a prime number). But then it’s just as superfluous for Sarah to say she knew he didn’t know the numbers. For then Peter could have known Sarah could have known he didn’t know the numbers! So their saying those things need not give either of them any new information. So if Peter knows the numbers after those things have been said, he could have known them as soon as he was informed of the product. In fact, he would have known them, considering that he’s great at math. So the problem has now become completely obscure to me, and I suspect that it doesn’t add up.[/tab]

Prime numbers between 3 and 99:

5      7     11     13     17     19     23     29 
 31     37     41     43     47     53     59     61     67     71 
 73     79     83     89     97

Why can’t they be prime?

They can’t be prime because prime numbers can only be a product of 1, which is disallowed.

[Sauwelios]

Carleas, I think what’s commendable about your solution is not the Excell sheets but your grasping things like [tab]Sarah’s statement, “I knew you didn’t know” tells us that whatever sum she has, every product of every pair of integers that sums to that is non-unique.[/tab]

@Carleas

@James S Saint

The very first thing that threw me off was that I was concerned about the order of the numbers proposed, later deciding that they weren’t bothering with which order, but merely which two numbers.

[tab]I had that thought too, and I considered (and fortunately rejected) focusing on the square numbers (since if we actually had to say which number is A and which is B, we could only do that if they were the same number), especially since the problem statement made explicit that they could be the same number.

I haven’t spent too much time on it, but I don’t think the problem is solvable with that assumption.[/tab]

[tab]If the question was “What is a and what is b?”, then Peter would have known the numbers immediately.[/tab]

[Sauwelios]

@Lev Muishkin

@Sauwelios

[tab]I thought I was well on my way to solving it–I had just reasoned out what I considered the final key to the solution–, when I suddenly realised a very basic fact had escaped my attention. The fact that the numbers are between 2 and 99 means their product cannot be a prime number. But then it’s completely superfluous for Peter to say he doesn’t know the numbers (which he could, from the information given him, only know if the product was a prime number). But then it’s just as superfluous for Sarah to say she knew he didn’t know the numbers. For then Peter could have known Sarah could have known he didn’t know the numbers! So their saying those things need not give either of them any new information. So if Peter knows the numbers after those things have been said, he could have known them as soon as he was informed of the product. In fact, he would have known them, considering that he’s great at math. So the problem has now become completely obscure to me, and I suspect that it doesn’t add up.[/tab]

Prime numbers between 3 and 99:

5      7     11     13     17     19     23     29 
 31     37     41     43     47     53     59     61     67     71 
 73     79     83     89     97

Why can’t they be prime?

[tab]A prime number is a number that can only be divided into integers by 1 and by itself. But that would mean at least one of the two numbers was 1, which is impossible, as both numbers lie between 2 and 99.[/tab]

[phoneutria]

I think that there might be a way to narrow the pool of numbers to a much smaller amount right from the start. I’m brushing up on my number theory, haven’t played with any of this in over 10 years

[tab]What I am trying to remember is the particularities of adding and multiplying prime numbers. Correct me if I am wrong, but I think that all non-unique products need to have A as a prime number, and B as a product of two prime numbers. Then, you can break down the multiplication to (A1*A2)*B, where all 3 are prime. Your spreadsheet would be tridimensional, but it would look only at prime numbers, so the possible results would be a much smaller amount of numbers.

Hm… Actually, I think that non-unique products would be the product of at least three prime numbers, but not necessarily 3. Still, might be a good place to start.

Unless I’m completely wrong. Maybe one of you math dorks can confirm this and save me the trouble of actually studying the problem [/tab]

[James_S_Saint]

@phoneutria

I think that there might be a way to narrow the pool of numbers to a much smaller amount right from the start. I’m brushing up on my number theory, haven’t played with any of this in over 10 years

[tab]What I am trying to remember is the particularities of adding and multiplying prime numbers. Correct me if I am wrong, but I think that all non-unique products need to have A as a prime number, and B as a product of two prime numbers. Then, you can break down the multiplication to (A1*A2)*B, where all 3 are prime. Your spreadsheet would be tridimensional, but it would look only at prime numbers, so the possible results would be a much smaller amount of numbers.

Hm… Actually, I think that non-unique products would be the product of at least three prime numbers, but not necessarily 3. Still, might be a good place to start.

Unless I’m completely wrong. Maybe one of you math dorks can confirm this and save me the trouble of actually studying the problem [/tab]

Numbers Theory was the only math course I didn’t take many years ago (I didn’t know that I would end up in philosophy) and just a practical note;
You don’t want to be preferring more than 2 dimensional arrays in Excel, unless they are very small or you are merely using the VBA.

[tab]I thought this one was interesting because it involved 10 prime numbers except for only one of them.
Why only prime except for one?!?![/tab]

[Sauwelios]

@Sauwelios

@Carleas

@James S Saint

The very first thing that threw me off was that I was concerned about the order of the numbers proposed, later deciding that they weren’t bothering with which order, but merely which two numbers.

[tab]I had that thought too, and I considered (and fortunately rejected) focusing on the square numbers (since if we actually had to say which number is A and which is B, we could only do that if they were the same number), especially since the problem statement made explicit that they could be the same number.

I haven’t spent too much time on it, but I don’t think the problem is solvable with that assumption.[/tab]

[tab]If the question was “What is a and what is b?”, then Peter would have known the numbers immediately.[/tab]

[tab]No wait, that’s only if he knew that it was possible to know what was a and what was b. But there’s no reason to think that Peter and Sarah knew the question posed to us, let alone before they started talking. Anyway, the problem cannot be solvable with that question, as the numbers can only be 4 and 13; nor is it possible, from the information given, to tell which is which.[/tab]

[James_S_Saint]

More than the Sum of Its Parts

Just think about it.

Count them.

[Carleas]

Both very clever illusions, especially the second.

First:
[tab]The hypotenuse is not quite strait. The visual effect is subtle, but you can calculate the slopes easily: for the red triangle it’s 3/8, and for the green triangle it’s 2/5. The result is the combined hypotenuse bends slightly in for the first triangle, and slightly out for the second, and that bend accounts for a 1x1 unit difference.

I’m sure there’s a more mathematically precise way to put this, but this satisfies me that the laws of mathematics aren’t being locally suspended [/tab]

Second:
[tab]This is a very impressive illusion. Even once I saw it, it’s still hard to believe that someone could come up with it and execute it.

The trick is that, rather than one person being completely removed between one version and the next, different parts of different people are removed by combining and splitting. So e.g., one person’s bottom half of a foot becomes another’s whole foot. It’s hard to identify who’s losing or creating what because it seems like everyone gives a little bit and the result is another full person.

Totally fascinating, thanks for this James.[/tab]

[James_S_Saint]

Right on both counts.

It is interesting to me how inventive some humans can be while still so devastatingly blinded.

[Arminius]

@James S Saint

More than the Sum of Its Parts

Just think about it.

Count them.

Well done, James. =D>

This thread is obviously interesting.

[Carleas]

More on #2, which I’ve been thinking about too much:
[tab]A few additional small insights into how it works:

• Each person is affected by the swap.
• Each person is shorter in the 13 person case than in the 12 person case.
• If you trace who ‘gives’ pieces of themselves to who, you will find that there is a chain of A->B, B->C, C->D etc. through every person (start in the lower left, you will end with the 2nd from the left along the top row in the 13 case).

I think a few modifications to this gif might make it clearer what’s happening:

• if the transition were done by moving the whole top half of the image to the right, as though it were wrapped around a cylinder.
• if the people were replaced with solid blocks or lines.
• if different people were different colors, so that you could see the break points more easily.

Super fascinating.[/tab]

[Carleas]

Learned of this one recently, and it’s related to a problem already discussed ad nauseam in this thread:

Alice at the Convention of Logicians:

At the Secret Convention of Logicians, the Master Logician placed a band on each attendee’s head, such that everyone else could see it but the person themselves could not. There were many different colors of band.

The Logicians all sat in a circle, and the Master instructed them that a bell was to be rung in the forest at regular intervals: at the moment when a Logician knew the color on his own forehead, he was to leave at the next bell. They were instructed not to speak, nor to use a mirror or camera or otherwise avoid using logic to determine their band color.

In case any impostors had infiltrated the convention, anyone failing to leave on time would be gruffly removed at the correct time. Similarly, anyone trying to leave early would be gruffly held in place and removed at the correct time.

The Master reassures the group by stating that the puzzle would not be impossible for any True Logician present. How did they do it?

[James_S_Saint]

[tab]

@Carleas

Learned of this one recently, and it’s related to a problem already discussed ad nauseam in this thread:

Alice at the Convention of Logicians:

At the Secret Convention of Logicians, the Master Logician placed a band on each attendee’s head, such that everyone else could see it but the person themselves could not. There were many different colors of band.

The Logicians all sat in a circle, and the Master instructed them that a bell was to be rung in the forest at regular intervals: at the moment when a Logician knew the color on his own forehead, he was to leave at the next bell. They were instructed not to speak, nor to use a mirror or camera or otherwise avoid using logic to determine their band color.

In case any impostors had infiltrated the convention, anyone failing to leave on time would be gruffly removed at the correct time. Similarly, anyone trying to leave early would be gruffly held in place and removed at the correct time.

The Master reassures the group by stating that the puzzle would not be impossible for any True Logician present. How did they do it?

I’m pretty certain there is something missing in this one and substantially different than the previous one like it.

There is no limit to the number of possible colors. Everyone could have a different color than everyone else.[/tab]

[Carleas]

A hint, responsive to James
[tab]The Master’s reassurance that the problem is not impossible for any true logician constrains the possibilities. If the problem would be impossible to solve with one possible configuration of colors, that configuration is excluded[/tab]

A hint-like clarification, not necessary but constrains the answer space:
[tab]The problem doesn’t have a specific answer, only a general one. We can’t say how many logicians there are, but we can say a lot about minimums and how the numbers affect when the logicians leave.[/tab]

[phoneutria]

[tab]The only way that this problem can be solved is by assuming that the visual field provides the constraint for the group of colors involved. I can imagine this as if I am one of the logicians and I see that around me all of the colors appear more than once. Therefore, the color of my headband has to be the same as one of the colors I can see. Otherwise the problem is unsolvable since you wouldn’t have a closed group of colors, and the options to chose from are infinite.[/tab]

Alright, someone take it from here. I’m a little busy

[James_S_Saint]

@Carleas

A hint, responsive to James
[tab]The Master’s reassurance that the problem is not impossible for any true logician constrains the possibilities. If the problem would be impossible to solve with one possible configuration of colors, that configuration is excluded[/tab]

A hint-like clarification, not necessary but constrains the answer space:
[tab]The problem doesn’t have a specific answer, only a general one. We can’t say how many logicians there are, but we can say a lot about minimums and how the numbers affect when the logicians leave.[/tab]

Nope:[tab]We know that there must be at least three band colors merely because of the word “many”. The problem is that there are any number of variety of potential colors. If there was only two other logicians, your band could be one of 100 other colors. You might not even know the name of the color. And it doesn’t matter what you see other than to let you know that if there were only two others, you could know that your color is different than theirs. Any more than 3 members and you cannot know that your color is any different.

A logician cannot deduce a color that he perhaps has never even seen before.[/tab]

@phoneutria

[tab]The only way that this problem can be solved is by assuming that the visual field provides the constraint for the group of colors involved. I can imagine this as if I am one of the logicians and I see that around me all of the colors appear more than once. Therefore, the color of my headband has to be the same as one of the colors I can see. Otherwise the problem is unsolvable since you wouldn’t have a closed group of colors, and the options to chose from are infinite.[/tab]

Alright, someone take it from here. I’m a little busy

[tab]You are showing your feminine thinking. In logic puzzles you have to think in terms of what is certainly true, not what might be true, even if almost certain. The fact that all of the others were duplicated doesn’t allow for you to conclude that your color is duplicated.[/tab]

[Carleas]

Phoneutria is on the right track.

[tab]In the terms of the problem, we are told that it is not impossible. So any assumption that would lead to an impossibility is false.

One can progress from there: If X leads to an impossibility, we know ~X. What can we conclude from ~X?

If assuming that my headband can be any imaginable color makes the problem impossible, then my headband cannot be any imaginable color. The set of possible answers is constrained.[/tab]
I don’t know what “feminine thinking” means (thinking-while-a-woman?), but whatever it is, it’s working.