Math problem, last people I can think of

[r6d2]

Originally posted by duartix
The buildings are close to each other (d=0).Well, if that was the case, then you could not place the rope between them, would you? The correct answer to the “fixed” problem is that d equals the thickness of the rope.
Originally posted by fccHandler
AARGH! That was plain cruel.Man, I recognized my mistake, then assumed it, and then apologized for it. :)

Now I hope to vindicate myself, so here you are: :D

The problem can be solved even with the numbers switched, believe it or not! The key is to give the word "barely" a loose meaning, which in fact it has if you look it up at Webster's (http://www.webster.com/cgi-bin/dictionary?book=Dictionary&va=barely). Barely meaning is wide open from "almost there" to "no way". So let's choose "no way".

Remember it was about choice? Well, you were given the freedom to assume the parameter of the catenary is 1. If you choose that, then,
L = 2*sinh(d/2)
where L is the length of the rope and d the distance between the buildings. In this case, d is 11.92 ft.

So, that’s two problems solved by the price of one. :)

[fccHandler]

Originally posted by r6d2
Well, if that was the case, then you could not place the rope between them, would you? The correct answer to the “fixed” problem is that d equals the thickness of the rope.
Shouldn't d be equal to twice the thickness, since the rope is folded? You said the rope hangs "from the nearest parts of the roof," which would be an exactly perpendicular line between the parallel walls of the buildings (thus, no room for the rope to squeeze in sideways). Unless the walls face each other at an angle, in which case d varies depending on where you measure it from, and the puzzle has no clear solution.

[r6d2]

Originally posted by fccHandler
Shouldn't d be equal to twice the thickness, since the rope is folded?You have a point. That seems to be a solution too, though not minimal.

But I guess that the nearest thing on the sentence was just to clarify it was not the opposite one.

Are you still mad? ;)

[fccHandler]

Nah, I forgive you. :)

Here's an easy one. What is the pattern of this sequence:

8, 11, 5, 4, 9, 1, 7, 6, 10, 3, 2...

[Kedirekin]

Alphabetical?

[Kedirekin]

Is it okay to post one that isn't math?

What distinguishing characteristic separates these two groupings of letters?

BCD G J OPQRS U

A EF HI KLMN T VWXYZ

[jel]

the use of curves in the character?

[Kedirekin]

Wow! That was fast. I had to be told the first time I saw it - never did figure it out on my own.

[jel]

thanks! by trade, i am an architect, so i tend to get visual questions quicker than complex maths - which i havent used since high school, and that was a loooong time ago....
and btw r6d2, at no point in the original definition of the last question did you state that the buildings must have straight, vertical sides, so i still stand by my answer of using a pyramid (or geodesic dome):D

[fccHandler]

How many F's are in the sentence:

FINISHED FILES ARE THE
RESULT OF YEARS OF SCIENTIFIC
STUDY COMBINED WITH THE
EXPERIENCE OF YEARS.

P.S. I swear I actually wrote a program once to count them, because I couldn't trust my eyes. :p

[jel]

first answer 3
reread it and found 6 ... i think it has to do with the size and common use of the word 'of' :confused:

edit/:actually i think its the phonetics of reading the scentence to yourself rather than looking at individual letters.

[downimp]

Originally posted by jel
first answer 3


I've got the same result :).
And English is not my native language (obviously).
I think you'll have the same (first) result no matter your English skills.
Nice one.

[r6d2]

Originally posted by jel
at no point in the original definition of the last question did you state that the buildings must have straight, vertical sidesYou're right, but I clarified that aftewards (and the picture shows it). BTW, I knew you were an architect! :D

[r6d2]

OK, let's get to the basics to see how far we can go.
Problem 1: What is the maximum number of dice you can fit into a cubic box?

Notes: Let d be the size of the dice edge, and b the size of the box edge. Consider d < b.

[Kedirekin]

Are those 6-sided dice?

Just kidding.

[mpucoder]

(int (b/d))^3

[r6d2]

That was easy, right? Only 7 minutes! Now let's got to the next one:
Problem 2: What is the maximum number of marbles you can fit into a pizza box of the same height as the marbles?

Notes: Let the box be dimensions b*b*d, and d the diameter of the marbles. Consider d < b./Add:
20 views and no contenders so far? If you're stuck, we can simplify a little by letting b be an integer multiple of d. That would be like Problem 1.5...

[Christos]

@r6d2
(int (b/d))^2

@all
Here is a "simple" one but try to find the logic behind your answer...
You will need it for the really hard one at the end...

Let n be any positive integer. Imagine n prisoners standing in the yard single file, one behind the other:
1 2 3... l
Each of the prisoners can see the prisoners standing in front of him but not the ones standing behind him. That is, for any integers j and k:
1 <= j < k <= n
prisoner k can see prisoner j but prisoner j cannot see prisoner k. The dreaded warden strides into the yard. From a bag containing n red hats and n blue hats, he selects n hats at random. His assistant places the first hat on the head of prisoner 1, the second hat on the head of prisoner 2, and so forth. Now all the prisoners are wearing hats (red or blue). Each of the prisoners can see the hats on the heads of the prisoners standing in front of him but not the hat on his own head or the hats on the heads of those behind him. The warden tells the last prisoner in line to speak one word, either "red" or "blue". If he speaks the word which matches the color of his hat then the assistant immediately sets him free. If not then the assistant immediately shoots him. Next, the warden tells the prisoner in front of him to speak one word, either "red" or "blue". If he speaks the word which matches the color of his hat then the assistant immediately sets him free. If not then the assistant immediately shoots him. And so forth. However, unknown to the warden and his assistant, the prisoners were aware, earlier, of the warden's depraved game. Of course, they had no way to know what would be the order in which they would be forced to stand or what would be the colors of their hats. What should they do, when they all meet the night before, to ensure that most of them will be set free? (Strategy)

Got it??? Good...
Now what if there are also n green hats in the bag???

Got it??? Good...
Now what if there are m different colors of hats (And the prisorers don't know how much m is beforehand, but they are all damn good at math)

[r6d2]

Originally posted by Christos
(int (b/d))^2Close, but no cigar. You are leaving lots of marbles out.
What should they do, when they all meet the night before, to ensure that most of them will be set free?Tell the prisoner ahead the color of his hat. The last prisoner with a hat flips a coin or calculates the most likely color remaining.

BTW, You only provided hats for k prisoners, and there are n of them...

[Christos]

Originally posted by r6d2
BTW, You only provided hats for k prisoners, and there are n of them... [/B]
OK I edited it...
Originally posted by r6d2
Tell the prisoner ahead the color of his hat. The last prisoner with a hat flips a coin or calculates the most likely color remaining.

Read again...
Of course, they had no way to know what would be the order in which they would be forced to stand or what would be the colors of their hats.
They found out what was going to happen the night before and talked about their strategy...
The next morning they were taken to the yard...
Starting from the last in line each one is only allowed to speak one word when his turn comes...
One color...
If it matches the color of the hat he is wearing he is set free.
If not he is shot...
Then the man in front of him is asked...

[r6d2]

Originally posted by Christos
Read again...But they're not deaf, and they can speak while the hats are being placed on their heads... There is no restriction on that, is there? If there is, they can develop some sort of key, like coughing, kicking the guy left/right foot or something.

[Manao]

You have to put the marbles like this :
O O O O ...
O O O ...
O O O O ...
With assuming b = k * b, you have k marbles on the lowest row, and you have int(b / ((sqrt(3)/2)*d)) rows, half of them having (k-1) marbles, the other half havinf k marbles.

That makes something like k * int((int( b / ((sqrt(3)/2)*d)) + 1)/2) ( that's for the totally filled rows ), +
(k-1) * (int(int(b / ((sqrt(3)/2) * d))) / 2) ( for the other half.

[r6d2]

Originally posted by Manao
That makes something like [...]You're on the right path. You must consider also that the height allows for more marble rows that the width, since it is a square box. There ar certain factors that create border conditions.

You might enter your formula into a spreadsheet to see if it gives the right values for known small numbers.

I hope you finish that so you can elaborate for the general case too!

/Add:
This forum should have an equation editor! :D

[Christos]

Originally posted by r6d2
But they're not deaf, and they can speak while the hats are being placed on their heads... There is no restriction on that, is there? If there is, they can develop some sort of key, like coughing, kicking the guy left/right foot or something.

No no cheating is allowed everybody plays by the rules or gets shot...
No triks just logic and math.

[Christos]

It's amazing but there is a solution even for the last problem where everybody can guess with absolute certainty the color of the hat he is wearing based only on the answers the men behind him gave. Except for the last one who has 1/m chance of been saved...

[Manao]

r6d2 : that's why I use sqrt(3)/2 : because there are more rows than ''columns''. To correct the border phenomenon, let's be more precise :

The first row's marble's center is at d/2 ordinates. The second row's center is at d/2 + d*sqrt(3)/2, and so on. You also have to let the last row of marbles fit completly in haight, so you have to take into account another d / 2

So let's call a = 1 + max_k( d + 2 * k * d * sqrt(3)/2 < b ), and b = max_k ( d + (2 * k - 1) * d * sqrt(3)/2 < b ). You have a = int((b-d)/sqrt(3)), and b = int((b-d*(1-sqrt(3/2)))/sqrt(3))

Let also call n1 = int ( b / d ), and n2 = int ( (b + 0.5) / d ).

You have a the number of even rows, b of odd rows, n1 the number of marbles on even rows and n2 = the number of marbles on odd rows.

And finally, you can put a*n1 + b*n2 marbles.

---------------

Now, it's my turn :

Two mathematicians have been given a paper on with is
written, for one of them, the sum of two integers ranging
from 2 and 100 ( we'll call him 'plus' ), and for the other
one the product of the same two integers ( he will be called
'mult' ). They both know that the integers are between 2 and
100.

They, of course, can't see what the other's got on is paper.


Then, 'mult' says : I can't know what are the two numbers.
'plus' answers : I knowed it.
'mult' then says : In that case, I know what are the numbers.
And finally, 'plus' also says : In that case, so do I.

What are the two numbers ?

[Christos]

Originally posted by r6d2
Problem 2: What is the maximum number of marbles you can fit into a pizza box of the same height as the marbles?

Notes: Let the box be dimensions b*b*d, and d the diameter of the marbles. Consider d < b.
/Add:
20 views and no contenders so far? If you're stuck, we can simplify a little by letting b be an integer multiple of d. That would be like Problem 1.5...

If b is an integer multiple of d and the box is the same height as the marbles (d) then its the same as asking how many circles can you put in a square with side b so the answer is (b/d)^2...

Please explain if not...

what Manao suggested applies to when marbles can be on top of each other...

[r6d2]

Originally posted by Christos
what Manao suggested applies to when marbles can be on top of each other...Manao is looking the pizza box from above, and he is right.

[Christos]

Originally posted by r6d2
Manao is looking the pizza box from above, and he is right.

exactly so he sees a square with circles inside...
Think about it

@Manao

The numbers are 2 and 11 i think...
I 'll double check it...

[Kedirekin]

It isn't abudantly clear which end of the line the warden is starting at.

I'm going to take a leap of faith and assume the warden starts with the man who can see every one in front of him (is this man # 1 in our 1-to-n arrangement?).

The best strategy is probably for each odd man to state the color of the hat on the man in front of him, and every even man to state the color he just heard from the man behind him. At least half the men will be set free, and probably more than half (odds are there will be two men in a row with same colored hats). I don't see a way to increase the odds beyond that.

[Kedirekin]

The marbles (circles) aren't packed in a square arrangement though - they're arranged hexagonally. That's the tightest packing for circles on a plane. I believe the unit cell is an isosolese traingle.

Now to get out my graphing paper to help me visualize...

[Christos]

Originally posted by Kedirekin
It isn't abudantly clear which end of the line the warden is starting at.

I'm going to take a leap of faith and assume the warden starts with the man who can see every one in front of him (is this man # 1 in our 1-to-n arrangement?).

The best strategy is probably for each odd man to state the color of the hat on the man in front of him, and every even man to state the color he just heard from the man behind him. At least half the men will be set free, and probably more than half (odds are there will be two men in a row with same colored hats). I don't see a way to increase the odds beyond that.

Yes the warden starts asking the last in line and everybody sees what everyone in front of him is wearing.
But no there is a better solution where everyone is saved exept the last one who has 1/2 chance of been saved...
For the first case that is...
For the general case he has 1/m chances and all the others are saved also.
Good try, you are on the right track...

[r6d2]

Originally posted by Christos
Yes the warden starts asking the last in line and everybody sees what everyone in front of him is wearing.Oh, now I get it. You mean each guy knows the hats of the guys ahead and also knows the hats behind, since he could count the shots/free guys. But that's too easy... No, actually I don't get it yet.

[Manao]

For the prisonner problem, with two colors :

Let's call 'nrk' and 'nbk' respectively the number of red hats and blue hats seen by the kth prisonner to speak. 'dk' will be nrk - nbk.

They have all agreed to say, if they are in first position to speak :
- if they are an even numper : d1 will be odd, but it may be 1[4] or 3[4]. If it's 1[4], they'll say blue, else red.
- if they are an odd number : d1 will be even, it may be 0[4] or 2[4]. If it's 0[4], blue, else, red.

We'll assume they are an even number. The first says blue ( hence, d1 = 1[4] ).

d2 is even. If the 2nd is blue, d2 = d1 + 1[4] = 2[4], else d2 = 0[4]. So the second knows his color, let's say blue ( hence 2[4] ). The third can count d3, which is either 1 or 3[4]. He knows d2 was 2[4], so he knows his color, and so on.

With more than two colors, I'll think of that while sleeping ( 3 A.M already, and have to get up at 7. Damn you, christos ;) )

[r6d2]

Hats may be all the same color, right? They are chosen at random, and originally there were twice as many hats as prisoners in the bags...

P.S.: it's only 11 pm here, but I cannot even understand Manao's notation! Better go to sleep. :)

[Kedirekin]

I don't understand the notation either.

I think he's basically saying the first man codifies how many red hats he sees, even or odd. The next man in line merely needs to count how many red hats *he* sees (even or odd) to know what color hat he is wearing.

Each man after that needs to count the red hats in front of him, and keep track of how many times "red" was shouted out behind him, to know what color hat he is wearing.

Seems to work for two colors - I doubt I would have thought of it. I don't see how to extend the strategy to 3 or more colors though. You could use color to codify the modulus of the count of any one color, but the other two colors would be unknown.

[Christos]

@Manao
Is 11 and 2 the correct answer to your problem?

@all
Here is the solution for the two colors in simple english...
When told to speak the last prisoner, he would say "red" if the number of blue hats in front of him was even; but would say "blue" if that number was odd.
Unfortunately for prisoner n, he would have only a fifty/fifty chance to go free. However, all the others could call out the correct colors of their hats, one by one in succession. They would all go free.
Why? Well, remeber the next prosoner to speak is the one before last so he sees evereone in front of him so he counts the number of blue hats of the prisoners in front of him.

*If the number is even and the last prisoner said "red" (so he also saw an even number of blue hats in fron of him) then he concludes he is wearing a red one...
*If the number is even and the last prisoner said "blue" (so he saw an odd number of blue hats in fron of him) then he concludes he is wearing a blue one (so that the last prisoner would count an odd number)...
*if the number is odd and the last prisoner said "red"...
*if the number is odd and the last prisoner said "blue"...

You catch the drift???


Now the hard part...
What if there were three diferent colors of hats the guard could select from with equal chances???
Got it??? Its not easy but some can find a solution...

Now the really hard part...
What if there could be m different colors of hats???
Can you save them all exept the last one???

Hint:
Think of the answer to first one and ask yourself how did you reach to the solution???
Understand the process in mathematical terms and you 'll find the general case...

[Christos]

Originally posted by Kedirekin
I don't understand the notation either.

I think he's basically saying the first man codifies how many red hats he sees, even or odd. The next man in line merely needs to count how many red hats *he* sees (even or odd) to know what color hat he is wearing.

Each man after that needs to count the red hats in front of him, and keep track of how many times "red" was shouted out behind him, to know what color hat he is wearing.

Seems to work for two colors - I doubt I would have thought of it. I don't see how to extend the strategy to 3 or more colors though. You could use color to codify the modulus of the count of any one color, but the other two colors would be unknown.

We posted the same time...
I must say you are on the money...
Understand the strategy and you 'll see how to extend it...

[Kedirekin]

I've been trying to find the two numbers too. No luck so far.

I don't think 2 and 11 work though. Initially Mult says he can't know the numbers, which (I think) means his product has multiple factors (example: 24, which could be 6*4 or 8*3 - no way to know which).

2 and 11 are prime numbers, so if Mult has 22 as his product, he'll immediately know that the two numbers are 2 and 11.

[Christos]

Yes you are right, I dont know what I was thinking...
Need to sleep now...

EDIT: Can't sleep...
OK so x,y are not primes or the product of primes and x+y can't be expressed as the sum of 2 primes (prime1+prime2).

Trying my luck...
16,45???

[Manao]

Damn, I overslept...

So, first, 2 and 11 are not the answers. Neither are 16 and 45. Sorry for not answering it yesterday, I was to absorb by your problem. But here is a little hint : 16 + 45 = 61, which is also 2 + 59, which are both primes. Hence if the sum has been 61, the numbers could have been 2 and 59 and so deduced from their product. Which negates what said 'plus' in his first sentence.

---

For my notations, 1[4] was meaning one modulo 4. ( ie , we can write the number 1 + 4 * k, with k integer ).

The k-th prisonner sees in front of him nr_k red hats and nb_k hats, so he can compute the difference d_k = (nb_k - nr_k).

If (nb_k + nr_k) is even, we have either nb_k and nr_k even, or nb_k and nr_k odd. So d_k is even. So it can be written (4 * k), or (2 + 4 * k), which I note d_k = 0[4] or d_k = 2[4].

If (nb_k + nr_k ) is odd, one of the two number is odd, the other is even, and so d_k is odd : it can be written either (1 + 4 * k) or (3 + 4 * k) : 1[4] - 3[4].

If d_(k-1) was 1[4] ( for example ), and the k-th hat is blue, we have d_k = nb_k - nr_k = nb_(k-1) + 1 - nr_(k-1) ) 1 + d_(k-1) = 2[4]. If it was red, we have d_k = 0[4]. So if we know d_(k-1) and d_k, we can know whether we are wearing a red or a blue hat.

Let's assume they are an even number of prisonners. So the first one sees un odd number ( nb_1 + nr_1 is odd ). The others know he sees an odd number, hence, they know that d_1 is either 1[4] or 3[4]. They have decided a code, for exemple blue means 1[4], and red means 3[4]. So they know precisely whether d_1 is 1[4] or 3[4] as soon as the first prisonner spoke.

So the second prisonner knows d_1 and d_2, so he can know his color, and when he says it, all the others know d_2. So the third can deduce his color, and so on.

I hope I was clearer. My answer was close to yours, but uselessly complicated.

(Edited some typos, try be explain in a better way )

[duartix]

Is 'plus' bluffing when he first says: 'plus' answers : I knowed it. ? Because he confirms he knows it later.

[Manao]

That wasn't clear : when 'plus' says 'I knowed it', he means 'I know you didn't know'.

[duartix]

Oh yes, thank you very much :devil:, you've just made it clear that the problem is harder than I thought it was.

[Christos]

Originally posted by Manao
Let's assume they are an even number of prisonners. So the first one sees un odd number ( nb_1 + nr_1 is odd ). The others know he sees an odd number, hence, they know that d_1 is either 1[4] or 3[4]. They have decided a code, for exemple blue means 1[4], and red means 3[4]. So they know precisely whether d_1 is 1[4] or 3[4] as soon as the first prisonner spoke.


Ok am up...
Damn 59 is prime... I was really sleeppy...

Anyways about your answer. Din't you say that d_k=(nb_k - nr_k).
just a small typo but what you say about nb_k + nr_k applies for the difference as well... So your answer is correct...
What they have to do is say for example that if d_k is 0[4] or 1[4] the last one says "red" and if d_k is 2[4] or 3[4] he says "blue" right?
A bit more complicated but good thinking... you are on a good track for the general solution... simplify your answer to find it...

[r6d2]

Originally posted by Manao
That wasn't clear : when 'plus' says 'I knowed it', he means 'I know you didn't know'. Oh...! Whole different story then... Thanks for the tip.

[r6d2]

Originally posted by Manao
r6d2 : that's why I use sqrt(3)/2 : because there are more rows than ''columns''. [...]And finally, you can put a*n1 + b*n2 marbles.@manao, I don't know if this is your final word, but does your formula work for known results? Did you check it for b = 2*d?

It would be great to have the solution for the general case, so we can move on to the marbles in a cubic box problem. And then, marbles in a cilinder! :eek:

[Kedirekin]

Let a and b be two numbers. Let their product be P and their sum be S.

Mult says he cannot know the two numbers. This means the product P has multiple possible factorings, which in turn means that at least one of the two numbers, a or b, is not prime. One can be prime, but not both.

Plus says he knows that Mult cannot know. That means that all possible sums that give the same result as a+b (ex. (a-1)+(b+1)) contain at least one non-prime number. If any of the possible sums was two prime numbers, Plus could not *know* the Mult couldn't know.

Mult says he knows the answer once Plus makes the above statement. By inference, this means that out of all the possible factorings of product P, only one of them is not two primes. In other words, all possible factorings are two prime numbers except one. If more than one factoring had a non-prime, Mult couldn't *know* which one Plus was working from.

Once Plus knows that only one of Mult's factorings contains a non-prime, he also knows the answer. This means that only one of Plus's possible sums results in a product that has one and only one factoring that contains a prime. I believe this means that only one of Plus's addend pairs is one prime and one non-prime. [comment: this is hard to explain].

So, what are two number a and b such that:
- a < b
- a is not prime or b is not prime or both are not prime
- for all n between 1 and int((b-a)/2), a+n is not prime or b-n is not prime
- for all factorings f1*f2=a*b where f1!=a, f1 and f2 are prime
- one of a or b is prime
- for all n between 1 and int(b-a/2), a+n is not prime and b-n is not prime


revised rules:
1: a < b
2: a is not prime or b is not prime, but not both
3: for all n between 1 and int((b-a)/2), a+n is not prime and b-n is not prime
4: for all factorings f1*f2=a*b where f1!=a, f1 and f2 are prime

Rule 2 greatly reduces the combination of numbers we have to check, since there are a limited number of primes between 2 and 100.

Using rule 3, we can also eliminate any combination of numbers where b is greater than the next prime above a (3 and 6 for example, because 6-1 is prime, or 3 and 10 as another example, because 3+2 is prime).

This leaves a limited number of possibilities to check:
3 and 4
5 and 6
7 and 8,9,10
11 and 12
13 and 14,15,16
17 and 18
19 and 20,21,22
23 and 24,25,26,27,28
29 and 30
31 and 32,33,34,35,36
37 and 38,39,40
41 and 42
43 and 44,45,46
47 and 48,49,50,51,52
53 and 54,55,56,57,58
59 and 60
61 and 62,63,64,65,66
67 and 68,69,70
71 and 72
73 and 74,75,76,77,78
79 and 80,81,82
83 and 84,85,86,87,88
89 and 90,91,92,93,94,95,96
97 and 98,99,100

Still, that's a fair number of possibilities to check. I've about used up my patience on this one. Can anyone see a shortcut to determine the answer.

[Christos]

Since Plus knows that Mult could not know the answer it means that the sum he has can't be expressed as a sum of 2 primes because if it could then there would be a possibility that mult could guess the numbers.
So from the primes between 2 and 10o witch are...
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
Someone could write a program combining them by two and adding them to see what the possible sum could be...
For example a+b!=61 because 59+2=61 so if Plus had 61 written on his paper he could not say "I knew it" because Mult could have 59*2=118 on his paper and then he could say that the numbers were a=59 and b=2.

[r6d2]

I created a spreadsheet to tally all possible combinations of numbers from 2 to 100 (only one instance each pair), a total of 4950 rows.

Then added the product and the sum as columns. Then I counted how many times all possible products appear. I got 173 different products. I filtered out all pairs for which the products appear only once in the table.

I'm stuck there...