Math problem, last people I can think of

[dvd_master]

My psycho teacher wants us to write an ESSAY on this one problem, and it's just worth so much of our grade I have to make sure the answer is right.

Please? I know this isn't very on-topic, but I can't think of where else to ask and you can delete this thread as soon as it's done.


http://www.hamsterhouse.com/dvd_master/sa.jpg

Thanks!

[Doobie]

You don't know the Pythagorean theorem? You just find the hypotenuses of the four triangles that make up the shrouds.

What's with units in feet? I thought everyone in school was having the metric system shoved down their throats?

[Kedirekin]

You know, I started to right up the answer, but I decided that would be helping too much.

Here are some hints:
- right triangles
- pythagorean formula
- 12 and 3
- 9 minus 6 divided by two and 24

Wow! Even that feels like giving you the answer.

[dvd_master]

That's evil! I still don't get it. I only see 2 triangles, the ones at the top, or else I'd be able to get it.

As far as I know, I got the top triangle's needed length, but the bottom I'm getting messed up with.

A squared + B squared = C squared... I just don't see what goes into what for the top large part.

Just tell me if this is right:
Top part, 23.2 (rounded)
Bottom, 47.1 ? (also rounded)

[r6d2]

Whatever numbers you get, remember to add some wire for the knots.

[sysKin]

Originally posted by dvd_master
Just tell me if this is right:
Top part, 23.2 (rounded)Hm, no. 2*sqrt(12^2 + 1.5^2) =~ 24.18677
Bottom, 47.1 ? (also rounded)That is less than double 24ft alone.... if they were vertical it would be 48.000 right? Can't be less.

it's 2*sqrt(24^2 + 3^2) =~ 48.37355

[edit]Ok I just noticed Kedirekin's post and he's probably right. It's all about a question - is 3ft only one half (on the top) or two halves? 9ft is two halves at the bottom, that's obvius - but if only one on top, Kedirekin is right.

[edit2] Kedirekin lol, I had written "2.25" but then I realized that "9" for lower part might be confusing. - so I changed to 1.25^2 lol. This is exactly how I make xvid bugs, too.

[edit3]You are most likely right about upper half being 6 total. So, you're right.

Radek

[Kedirekin]

Sorry, that's not the answer I get.

Wish I could draw you a picture.

Draw a line from the bottom of the upper shroud straight down. Now how long is the little piece from that line straight across to the lower shroud? Hint: 9/2 - 3. Another hint: the line you just drew forms a new right triangle. Once you get that, you have the base and height of the lower right triangle and you can use the pythagorean formula to calculate the hypotenuse.

Here's the answers I get:
One upper shroud: ~12.369 ft (root of 153))
One lower shroud: ~24.047 ft (root of 578.25)

Total of all shrouds: 2 upper shrouds + 2 lower shrouds.


I had one too many zeros for the lower shroud. Sorry about that - always check your work.

[Kedirekin]

Oops! We're giving you conflicting answers.

Is the top bar one 3 foot section or two 3 foot sections? I thought it was two 3 foot sections.

@sysKin
One half of 3 is 1.5. I think you used 1.25.

[dvd_master]

It means the entire rod is 3 feet, not half. I'm searching through the equations and trying to see where I made my mistake.

You guys are helping a lot, thanks a billion!

[mpucoder]

If the upper rod is three feet total, then the drawing is not accurate, the shrouds would not bend over the upper rods (hint: 3:12 :: 9:36)
Are you supposed to write an essay, or solve the problem?

[bit-wise]

Having lived on a sailboat for a while, the drawing is accurate in the fact that shrouds "bend" over the stays. So, there are two stays in the drawing, (off each side of the mast), each 3 feet in length. (Instead of a 3 foot total length, each 1.5 feet in length).

dvd_master, be thankful you are not in advanced calculus, because the exact formula takes into account the number of strands in the wire, the total sail area, and some serious equations to factor in the strechability of the shroud and the amount the mast can bend (yes, it does bend quite a lot). On a 24 foot mast, you will get about 3/4 a foot of stretch in the wire when subjecting the sail to 2000 pounds of pressure (the equivalent of around 15 knot winds across the sail).

[r6d2]

Originally posted by bit-wise
the drawing is accurate in the fact that shrouds "bend" over the stays.Regarding accuracy, isn't the keel too short?

[bit-wise]

Heh. You would be surprised at some of the keel designs from the mid-seventies. The total keel area dictates a big portion of its effectiveness, so they made the keel very shallow, but had it run the length of the boat. The best way I can describe the performance of these boats is a bathtub with a sail. However, they didn't ground as much since they only took about 3 feet of water.

I was thinking more that the boat is a bit low in the water - guess there is a leak somewhere. Maybe that's the next problem, calculate the volume of water the boat has taken on based on how low it is sitting, factoring in the buoyancy (or lack thereof) of the interior items such as beds, cusions, galley, engine...

[r6d2]

Originally posted by bit-wise
guess there is a leak somewhere.Maybe the keel is not actually there and water is entering through the slot. :)

Well, this is quite off topic but isn't this whole thread too? ;)

A small divertimento:
A house has a row of square tiles in the garden put one after the other in this fashion: <> (diagonally connected). The tiles go from the house door and exactly to the border of the garden.

The lady of the house got tired of looking always at the same arrangement so she asks the gardener to put them side by side, with no space between them, and she strongly remarks she expects to have it filled using only whole tiles.

The gardener begins his work, but when he is just on the middle of the path, he notices that each time he moves a tile he has to go further for the next one. In a moment of inspiration under the burning sun, he also concludes that there will be not enough tiles to finish the path.

He goes to the lady and tells her. She asks:

-How many more tiles do you need to finish the job?So, how many more?

[mpucoder]

about 40% more. more exactly square root of 2 times as many.

[r6d2]

Close, but no cigar... :cool:

[avih]

Originally posted by r6d2
Whatever numbers you get, remember to add some wire for the knots. LOL

[Kedirekin]

You mean the tiles are layed out roughly like this (white is tile - sorry, I can't make them look square):

********************
**** ******** ****
*** ****** ***
** **** **
* ** *
** **** **
*** ****** ***
**** ******** ****
********************
**** ******** ****
*** ****** ***
** **** **
* ** *
** **** **
*** ****** ***
**** ******** ****
********************



Based on surface area alone, the path either needs to be half as wide or will require twice as many tiles.

Is that correct? Do I win a prize? ;)

[r6d2]

Originally posted by Kedirekin
You mean the tiles are layed out roughly like this (white is tile - sorry, I can't make them look square)Nope, it is only one row of tiles. You drew two. I'll post the right answer tonight if nobody gets it before then.

The prize is the position of moderator for the upcoming Math Forum. :)

[Soulhunter]

50% maybe... Damn, I hate math... :devil:

Bye

[bit-wise]

Hmmm, I would have thought .414 more tiles would be needed. I just tried it with 3 scraps of square paper 1" by 1", and I needed exactly 1.2 pieces of paper more to make the path if they are to be laid side by side.

I guess my answer is incorrect because that .2 does not satisfy the fact that she "she strongly remarks she expects to have it filled using only whole tiles".

Am I allowed to move the house?

[Kedirekin]

It requires 1.4142135623730950488016887242097... more tiles, and the path is 1/1.4142135623730950488016887242097... as wide as before.

However, in a mathematical sense, it is not possible to fill the length of the path exactly using only whole tiles. Root 2 is an irrational number. No matter how many tiles are in the original path, there is no ratio of tiles that will exactly fill it using the new layout.

[r6d2]

@Kedirekin, :goodpost:

You got it right, except that you need only sqrt(2) - 1 more tiles, just like bit-wise posted, i.e. 41.4...% more than what you have.

But the catch it precisely that there is no number of tiles that can satisfy the restriction. You are hereby named as the Official Mod of the Math Forum (coming soon to a Forum near you).

@bit-wise, :goodpost: too. However, even though you posted first, you failed to arrive to the correct conclusion. You also need to have your scissors replaced, 1.2 is only 40%. :)

Anyway, you were close to another solution (albeit impossible in real world). There was no restriction on moving the house or the street, so that would be a good theoretical solution to the problem.

But there is another practical issue to this: they usually don't sell tiles by the unit, so the lady would probably have to round up at the store. :)

Also, if you use an infinite number of tiles, and move the house to infinity, you could speculate that's a solution too.

[r6d2]

The previous one was easy, right? Let’s see this one.
Suppose the Earth is a perfect sphere with a flat surface, and you tie a rope around its longest path (let's say the Equator), so it is perfectly tense and has no slack. Now release the knot, loose the rope and enlarge it by adding 1 meter. Tie again the knot at the new position.

Then, spread the extra meter evenly along the globe by retightening the rope simultaneously from all 360 degrees.

(Use Lanczos resizer to keep the most quality, and also to not being accused of being off-topic.) :)

Now you stand by the rope at any point in the globe.

The question is: Can your hand pass under the rope, without pulling it?Think about it. The extra meter slack is no longer there at your feet. It is evenly spread around the globe.

[bit-wise]

Actually, rack the 1.2 up to pure laziness in the fact that I didn't feel like typing lots 'o decimal places...

Now this sphere thing. Question. The rope, in its first position, (before it is resized) - is it at the largest circumference of the sphere, or does it not matter? Because if you start the rope horizontally at the equator and move the entire rope north it no longer touches the globe at any point, your hand slides right under. But I don't think that is in the spirit of the question.

[r6d2]

Originally posted by bit-wise
The rope, in its first position, (before it is resized) - is it at the largest circumference of the sphere?You are right. That's the spirit of the question. Both before and after resize the rope is around the Equator. (I edited it already).

[Kedirekin]

Is this a pure math problem?

The additional meter of rope will create an infinitesimal increase in the radius of the (perfect?) circle formed by the rope. Assuming the rope isn't simply allowed to fall to the ground after retightening, there will be an infinitesimal gap between the surface of the earth and the rope.

With a sphere the size of the earth and a hand of typical thickness, one additional meter wouldn't create a gap nearly large enough for your hand to pass under the rope, assuming we're not allowed to pass our hand within the confines of the earth's surface (no digging or tunneling), and that the definition of 'under' is between the rope and the earth's surface (not simple south of the rope).

I have no idea if I've covered all the bases. Don't keep us in suspense too long.

[r6d2]

Originally posted by Kedirekin
Is this a pure math problem?Yes, it is. There is no reading between lines or infinity stuff.

Your reasoning is correct. It comes from your intuition. You did not do the actual math, but described what you think would happen.

Well, you should do the math. ;)
Don't keep us in suspense too long.I could answer the question with a simple yes or no, but let's be patient. Perhaps we can get another mod for the Math Forum showing up...

[mpucoder]

The relationship between circumference and radius is linear, circumference = 2*pi*r. Regardless of the circumference of the Earth, the radius will increase by 1/(2*pi), about 6". So yes, your hand will slide underneath easily.

[r6d2]

We have a winner! :goodpost: This time you made it, @mpucoder, and quite precisely. The result presented above contradicts our intuition. Let’s try to prove it.

Let R be the Earth radius and L the original length of the rope. Thus we have that,
(1) L = 2*Pi*R
Now we increase the length of the rope by 1 meter, getting L’. Our intuition tells us that the radius of the second circle will be somewhat bigger, right? Let R’ be the new radius. Thus we have that,
(2) L’ = L + 1m
and also,
(3) L’ = 2*Pi*R’
Let h be the radius increase. Thus we have that,
(4) R’ = R + h
Replacing (2) and (4) in (3) yields,
(5) L + 1m = 2*Pi*(R + h)
Now we replace (1) into (5) which yields,
(6) 2*Pi*R + 1m = 2*Pi*(R + h)
Distributing the right side,
(7) 2*Pi*R + 1m = 2*Pi*R + 2*Pi*h
Subtracting the term 2*Pi*R from both sides yields,
(8) 1m = 2*Pi*h
Then we get that h, the increase in radius is,
1m
(9) h = ---- ~= 16cm ~= 6”
2*Pi
As you can see, that infinitesimal increase is high enough not only a hand could pass underneath, but also a whole rabbit, a cat, even a small dog.

The most surprising part of this result IMHO, is that it does not depend on the radius of the sphere in question. You could do the same at home with a tennis ball and the result would be exactly the same. If you don’t believe it, you are welcome to try.

Now we have two mods for the Math Forum. :D

[Kedirekin]

I just did the math and it makes me question my sanity.

Why is intuition so far off here? I can see the numbers and I still don't believe it.

This is truly a wierd feeling. It doesn't matter if I wrap the rope around a bowling ball or around the circumference of the galaxy - add one meter of rope and the radius increases by the same amount.

GAH! Where's the number for Bellevue?

[jel]

GAH! Where's the number for Bellevue?
in case of further maths questions i've taken the precaution of putting the number on 'speed dial'...

[r6d2]

Originally posted by Kedirekin
Why is intuition so far off here? I can see the numbers and I still don't believe it.Ah... The Lord works in mysterious ways... :)

Intuition is an important part of our survival skills, and in spite of that it plays those tricks on us quite often. Want another ride? :)
Suppose you have a huge, and I mean huge sheet of paper, 0.1mm of thickness (10 sheets add up to 1mm).

Also suppose you can fold it in two without any problem, regardless of the number of times you do it (in practice, there is a limit of about 6 to 8 times you can fold a sheet by its half every time).

Then begin to fold it, repeatedly, in two by its half (remember the sheet is huge). With each turn you double the thickness, so the folded sheet begins to grow.

How many times do you have to fold the sheet to get to the Moon?
This time say a number, from your guts, then do the math.

[Kedirekin]

Gut, 25 times.

Now I'll do the math.

[KpeX]

I googled quickly to find the distance from the earth to the moon is 384,400 km.

d=384,400 km =384,400 * 10 ^3 m = 384,400 * 10^3 * 10^3 mm=384,400,000,000 mm


thickness =.1 mm (per sheet)*2^n (thickness doubles every fold, n=number of folds)

384,400,000,000=.1 * 2^n

n = log(384,400,000,000/.1)/log(2)=41.8, so 42 folds would be required.

My intuition didn't really have an answer, but it probably would have been more than 42.

[r6d2]

Originally posted by KpeX
My intuition didn't really have an answerFirst you cheat, then you spoil the problem. :)

Well, I wonder why NASA spends billions of dollars in R&D instead of building a sheet of paper big enough.
Derived question: How big does the sheet need to be to hold a person standing on top?Again guts, then math.

BTW, @KpeX, since you are an audio mod (and also to keep you busy so you don't spoil this one), ;) would you pay a visit to this (http://forum.doom9.org/showthread.php?s=&postid=441940#post441940) thread and clear up if sound can travel in the vaccum through a space-time shockwave?

[KpeX]

@r6d2

About the audio thing, I'm pretty sure you're correct and audio will need a medium to transmit. You can't vibrate vacuum ;)

Regarding the math - Gut instinct, assuming that we need, say, 1 square foot to stand on, i'd guess maybe a few thousand square miles. I'll do some math now.

I doubt my guess is very close however. That's why I like math, cuz my intuition isn't worth much :D


Edit:

area to stand=1 square foot

# of folds=42

area = (original area) * (1/2)^(number of folds) (i.e., each time you fold the paper the area is halved, since one side is half as long)

Original area= 1/( (1/2) ^ 42 )=2^42=4398046511104 square feet

Length of one side= 4398046511104^.5 = 2097152 feet = 397.2 miles

So using a square piece of paper one side would have to be about 400 miles long.

Fun stuff guys :D

[BoNz1]

You guys are absolutely correct in order to hear sound you need medium for it to travel in. Sound is simply the compression and rarefraction of molecules in the air. So all those big explosions that you see in movies, ie Armageddon would never happen. BTW, a couple of months ago I had a math question of the week in my sig usually they were quite hard. Possibly I should resurrect this, no? :)
EDIT: There we go, done enjoy 8D

[r6d2]

Originally posted by BoNz1
You guys are absolutely correct in order to hear sound you need medium for it to travel in.Please, BoNz1, let's keep on topic, will you? :D:D:D

Interesting question that of yours, but did they really ask you to use differential calculus to solve it? My bet is simple geometry is quite enough.

[bit-wise]

There was a recent publication that sound, does, and is, traveling through space. It appears that the Chandra X-Ray laboratory detected sound from a distant black hole as a nice B-flat 57 octaves below middle C.

From my skimming the article, it appears that bubbles in the escaping plasma jets at the edge of the black hole produce enough force to create this sound. It requires something with as much energy as a black hole to produce enough energy for the sound waves to travel to Earth. Since space is only 99.999(to some decimal place) a pure vacuum, there is enough force to produce a sound wave that can travel 250 million light years back to Earth. I leave the mathematical proof as an exercise for the reader. But no, Hollywood explosions would not be heard.

But your reasoning is correct, sound can not transfer through a pure vacuum.

Oh, on another note. r6d2: On behalf of my client, they all would like to express thanks in causing them to collectively scratch their heads and ponder the rope question. The answer caused quite a stir.

So, did we ever answer the sailboat question?

[sysKin]

Originally posted by r6d2
(...)if sound can travel in the vaccum through a space-time shockwave? Now *this* got me interested (but I haven't seen the thread yet). A sound is a wave. A wave means that the particlaes (of the medium) are accelerating. An accelerating particle (with a mass) creates gravitational wave that is 90 degrees phase shifted compared to the sound wave, and has the same direction and frequency and everything.

Such wave would have to be emmited by something thin, much thinner than the (sound) wavelength - otherwise different particles, which have different phase of the same wave, would emmit together and interfere one with another. This is because of differet speeds of sound and light.

If such wave would hit something thin again, it would make it vibrate and in the end, you would recieve the sound again (90 degrees phase shifted again, doesn't matter).

:D
Radek

[sh0dan]

You should change your description to:

"sysKin: inventing new XviD bugs and solving deep space communication problems!"

:D

[SeeMoreDigital]

Is the answer to the question 42?

In HitchHiker's Guide to the Galaxy... that's the answer for everything!

[r6d2]

Originally posted by SeeMoreDigital
Is the answer to the question 42?Yes, it is.
Originally posted by bit-wise
So, did we ever answer the sailboat question? I don’t know. I think it was left as an exercise to the reader. :)

About the spa question, the absolute error in volume is,
Pi*(r^2 + 2*r*h)*eWhere r is the radius, h the height and e the error of the instrument. In this case, the error in volume amounts to about 5 cubic feet, which is about 5% of the volume of the spa. As with any error, it is a +/- value.

You see intuition again is misleading. It’s a huge error you get, eh? Considering the instrument error is only about 1.7% of the magnitudes in question.

[r6d2]

Let’s have another stroll at a intuition divertimento:
Suppose you are among a group of people attending a class. There is a final exam approaching and there’s a motion from the students to postpone the test for another week.

(Nobody has studied enough because you all have been encoding too many movies using the university computers and the last version of AutoGK. OK, the reason is actually irrelevant, it’s just to be on topic.) :)

The teacher says,

-OK, I’m willing to postpone the test, but that makes me uncomfortable and would not look good if I just decide it on my own based on your reasons. Let’s leave it to chance. If any two of you have a birthday on the same date, I’ll do the test on schedule. If none of you has the same birthday as any other member of the class, I’ll postpone it. So take a sheet of paper and write down your birthday. I’ll collect them in one minute.

The students get very happy. Having the year as much a 365 days, and being a group of only 50 people, it is very unlikely two of them were born on the same day.

The question: What is the probability of the exam to be postponed?Now this one is hard. As usual, use your guts, then do the math.

[duartix]

Ok, 3 guys (A,B & C) check into a hotel. The owner says the room is 30$. They gather 10$ each and give them to guy A.
A goes down the stairs to pay the hotel owner.
Hotel owner tells A he got it wrong because the room is 25$.
Hotel owner gives back 5$ to A.
A thinks:
- "How am I going to distribute 5$ between 3 people?"
- "I know, I'll tell them it's 27 (9$ each), split 3$ between the 3 of us and I'll keep the other 2$."
So he does.

Now you people tell me:
They payed 27$ (9 each) right?
27$ + 2$ (kept by A) makes 29$. The room was 30$.
There is 1$ missing. Where is it?

[downimp]

[QUOTE]Originally posted by duartix
"split 3$ between the 3 of us and I'll keep the other 2$."

That means the A guy has 3$ (3/3 +2).
Then 9*3 + 3 = 30

Corect?

[duartix]

Yes, A has kept 2$ for himself + 1 (from the split).
Your alternative explanation is correct, but what I want is someone to explain why is 1$ missing from my reasoning. That is the fun part. :D

Oh, by the way, downimp, welcome to the forum !!! :D :D :D

[Wilbert]

It seems that many of you are bored :) Well, here is another nice puzzle:


Consider n number of poeple standing in a circle. All of them are numbered from 1 to n. Now, Man no. 1 is given a sword. He kills the person next to him (i.e Man No. 2) and passes the sword on to Man no.3. Man 3 in turn kills the man next to him (No. 4) and passes the sword on to the next man (i.e Man No. 5). This sequence continues until only one man is alive. Find which number is it.

For eg, take n=5

then:
|1| 2 3 4 5 (the number in | | has the sword)
1 |3| 4 5
1 3 |5|
|3| 5
|3|

Thus in the end man no. 3 remains.

Suppose we have 3141 man, which person stays alive after doing this?

[duartix]

which person stays alive after doing this? This is definitely the kind of problem where it's easier to write a computer program than to think about it. ;)

[Nic]

I managed to write a ten line piece of C++ code to solve it. Now to work out the formula. :)

[duartix]

Let's see if a factorization may show some light.
(edit) No, it doesn't. Perhaps some MOD.
(edit) Not either. There is a pattern however, numbers kept show a 2^N diference between them as we reach every Nth pass. But it always depends on the last number.
I can't get there but I bet is has involves some 2 based Power formulas.

[downimp]

Originally posted by Wilbert
It seems that many of you are bored :) Well, here is another nice puzzle:



Suppose we have 3141 man, which person stays alive after doing this?

3139?

[Nic]

I got 2187...but that may well be wrong. How did you get yours downimp?

Edit:
(The key to it is solving the sequence:
1 1 3 1 3 5 7 1 3 5 7 9 11 13 15 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 1 etc )

(Note the way the mini sequence loops at 2^n intervals)

Ahhh, for more info look up:
Josephus' Problem

Doesn't look like you can create a very simple formula for it easily...

[downimp]

Just a [stupid] guess. In the 5 man example 3 was the last odd number before 5. So...

[r6d2]

Originally posted by Wilbert
It seems that many of you are bored :)Welcome indeed. :)

On the contrary, we are having lots of fun :D
Suppose we have 3141 man, which person stays alive after doing this? I guess it is man number 2187, same as Nic found.
Originally posted by Nic
I managed to write a ten line piece of C++ code to solve it.Just a simple spreadsheet did on my case, but it is limited to groups below 65536. :)

I guess the admin position of the Math Forum would be to the one who finds the recurring relation for any n.

[r6d2]

Originally posted by duartix
Your alternative explanation is correct, but what I want is someone to explain why is 1$ missing from my reasoning.Well, nothing is missing. You are counting the 2 bucks twice.

[Kedirekin]

Someone tell me if this reasoning is correct.

Every other man is killed.

If the original number of men is even, the last man in line will be killed and the first man will get the sword. The remaining number of men will be even (n/2 were killed), so the pattern repeats. For an even starting number, the last man standing will always be the first man man in line.

If the original number of men is odd, then after the first killing the number of men remaining will be even. At that point, the third man in line will be holding the sword, and he effectively becomes the first man in a 'even number of men' scenario as above. The sword passes around, the original first man is killed and the third man gets the sword again. The pattern repeats. For an odd starting number, the last man standing will always be the third man in line.

The answer for 3141 is 3.

[Kedirekin]

I don't know how to explain were the $1 went. The original description is just whacked.

It's 25 + 2 + 3 = 30, or 27 + 3 = 30, or 27 - 25 = 2. 27 + 2 just isn't mathematically valid for this problem. There is no $1.

[Nic]

@all: The closest I can get to a good formula is:
f(n) = 2m + 1
Where,
m = n - 2^k
and
2^k <= n <= 2^(k+1)

And that's with digging around on the net... :(

@Kedirekin:
Look at the pattern for n = 7 to see why your premise does not stand true:
|1| 2 3 4 5 6 7
1 |3| 4 5 6 7
1 3 |5| 6 7
1 3 5 |7|
|3| 5 7
3 |7|
|7|

@duartix: That puzzle is well known, but is more a case of word play than a mathematical puzzle to solve. It's like a riddle. Never been keen on riddles.... :) (Kedirekin's explanation to it is good)

[r6d2]

Originally posted by Kedirekin
At that point, the third man in line will be holding the sword, and he effectively becomes the first man in a 'even number of men' scenario as above.I think your reasoning is quite correct. But it fails at this point. It is the same scenario, yes, but numbers are not the same. Men have to be renumbered... Fix that and I think you will get the admin position as well. :)

[duartix]

If the original number of men is even, the last man in line will be killed and the first man will get the sword. The remaining number of men will be even (n/2 were killed), so the pattern repeats. Not quite Kedirekin, n/2 can be odd (take n=6) after 3 men were killed there remain 3. Same for odd numbers.

What I found quite interesting about this problem (try manually solving this by displacing them in an Excel sheet) is that solutions are recursive. The structure of solving n=11 involves a sub-structure of n=7.

But Nic's sequence gives you the answer to the problem.

[duartix]

Of course the presentation is whacked :D
I cannot say:
27$ + 2$ (kept by A) makes 29$. The room was 30$.
There is 1$ missing. Where is it?

Instead I should say:
27$ - 2$ (kept by A) = 25$ (in the hotel safe).

It prooves how easy it is to mischange a sign. :p

[Kedirekin]

I think the birthday problem has me stumped. My gut reaction is there's about 99% chance that two people will share a birthday, but I keep getting an answer greater than 100%, which plainly isn't possible.

[Kedirekin]

@duartix, right you are. Rookie mistake on my part - and I thought my solution was so elegant.

[r6d2]

Originally posted by Kedirekin
I think the birthday problem has me stumped.:thanks: I was beginning to think that people in this forum found manslaughter more interesting than birthday parties. :)
My gut reaction is there's about 99% chance that two people will share a birthdayMy gut feeling when I first heard the problem was that it was extremely unlikely, provided most people in a class would be born around the same year.

But don't give up!

[Nic]

@Kedirekin: r2d6's one is a very well known one...if you get really stuck look up "birthday paradox"...but otherwise just sit and think it's quite surprising...
(i.e. as long as there is 23+ in the class the answer chances will be over 50% of having two people with the same b'day!)

-Nic

[r6d2]

Originally posted by Nic
@Kedirekin: r2d6's one is a very well known one...OK, Nick, I guess I'll have to come up with one you cannot cheat on... :)

BTW, do you think Doom9 will make an exception and let me change my user name to the way you type it? :D:D:D

/Add:
This (http://mathworld.wolfram.com/JosephusProblem.html) is the formula for the sword thing.

[Nic]

well, r2d2 (lol), you'll have to come up with an original problem I haven't heard of (like wilbert did) ;)

-Nic

ps
There's already one mod called "Nick", two called "Nick" could get confusing (reminds me of a monty python sketch ;) )

[BoNz1]

Originally posted by r6d2
Yes, it is.
About the spa question, the absolute error in volume is,
Pi*(r^2 + 2*r*h)*eWhere r is the radius, h the height and e the error of the instrument. In this case, the error in volume amounts to about 5 cubic feet, which is about 5% of the volume of the spa. As with any error, it is a +/- value.

You see intuition again is misleading. It?s a huge error you get, eh? Considering the instrument error is only about 1.7% of the magnitudes in question.

Yes, exactly the max error is 5.22ft^3 intuitively you wouldn't think it would be that large and it may not be actually it is quite likely it would not, as you illuded it is quite possible that your error may simply cancel itself a bit and it may be quite small ;). For anyone who wanted to know how you do this question basically you set up a differential equation like this:
dV = (partial derivative wrt to r)dx + (partial derivative wrt to h)dh
of course we know V = (pi)r^2h so it becomes quite easy once you know this 8D.

[r6d2]

Originally posted by BoNz1
intuitively you wouldn't think it would be that large and it may not be actually it is quite likely it would notYes, in fact there is a 68% chance of the error being below the one we calculated.
you set up a differential equationYou must really like calculus. :) I just computed this:Pi*(r + e)^2*(h + e) - Pi*r^2*hand discarded all e^2 and e^3 terms. This is in fact the same theory behind, but you can't expect a spa builder knows differential calculus. :D
Originally posted by Nic
you'll have to come up with an original problem I haven't heard ofOriginal or not, you cheated anyway by searching the net. :)

The one of the garden tiles was original. Let's see if I can bring you guys a real challenge.

BTW, one of the two nicks should get another nick, which would make this something like nick^3.

[Nic]

cheated by searching the net?! What for the birthday paradox? Everyones heard of that....Not my fault you didn't have anything original ;) lol

-Nic

[fccHandler]

Great thread! :cool:

This isn't original either, but it's my all-time favorite math problem:

A rope over the top of a fence has the same length on each side. It weighs 1/3 of a pound per foot. On one end hangs a monkey holding a banana, and on the other end a weight equal to the weight of the monkey. The banana weighs 2 ounces per inch. The rope is as long (in feet) as the age of the monkey (in years), and the weight of the monkey (in ounces) is the same as the age of the monkey's mother. The combined ages of the monkey and its mother are 30 years. One half the weight of the monkey, plus the weight of the banana, is 1/4 as much as the weight of the weight and the weight of the rope. The monkey's mother is half as old as the monkey will be when it is 3 times as old as its mother was when she was half as old as the monkey will be when it is as old as its mother will be when she is 4 times as old as the monkey was when it was twice as old as its mother was when she was 1/3 as old as the monkey was when it was as old as its mother was when she was 3 times as old as the monkey was when it was 1/4 as old as it is now.

Question: How long is the banana?

[Hiro2k]

Originally posted by fccHandler
Great thread! :cool:

This isn't original either, but it's my all-time favorite math problem:

A rope over the top of a fence has the same length on each side. It weighs 1/3 of a pound per foot. On one end hangs a monkey holding a banana, and on the other end a weight equal to the weight of the monkey. The banana weighs 2 ounces per inch. The rope is as long (in feet) as the age of the monkey (in years), and the weight of the monkey (in ounces) is the same as the age of the monkey's mother. The combined ages of the monkey and its mother are 30 years. One half the weight of the monkey, plus the weight of the banana, is 1/4 as much as the weight of the weight and the weight of the rope. The monkey's mother is half as old as the monkey will be when it is 3 times as old as its mother was when she was half as old as the monkey will be when it is as old as its mother will be when she is 4 times as old as the monkey was when it was twice as old as its mother was when she was 1/3 as old as the monkey was when it was as old as its mother was when she was 3 times as old as the monkey was when it was 1/4 as old as it is now.

Question: How long is the banana?

ROFL

Nothing, because the monkey ate it! :p

[BoNz1]

Originally posted by Nic
cheated by searching the net?! What for the birthday paradox? Everyones heard of that....Not my fault you didn't have anything original ;) lol

-Nic

I think this is one the first problems you get in introductory statistics courses, it is one of those problems that always fascinates students every year. Most students probably would guess there would have to be 365 or half of 365. But if you do the math you find that it is indeed 23 for a 50:50 chance 8D.

Originally posted by r6d2
You must really like calculus.

Well, it is one of the most beautiful subjects IMO. And as you advance it becomes more and more useful too. Calc. 1-2 are reasonably useful but calc. 3 is extremely useful for obvious reasons. Unfortunately, I always seem to end up in the 70s in math so I can't take anymore since it pulls down my GPA :(.

[fccHandler]

Originally posted by Hiro2k
ROFL

Nothing, because the monkey ate it! :p

:D Yes, I'm sure he did! But what was its length before he ate it?

FWIW, the complete solution is here:
http://math.ucsd.edu/~mathclub/games/brainteaser-archive/monkey_business.html

...but don't click on it unless you completely give up!

[r6d2]

Originally posted by fccHandler
Great thread! :cool:Yes indeed! In fact, in only 2 days it got more page views than Wilbert's "Posting in General Discussion" sticky has got in 7 months... Not to be surprised, since seems that nobody posting on this thread has read what that post says. I hope he will not wipe us out. :)
Originally posted by Nic
Ahhh, for more info look up:
Josephus' ProblemNeed I say more? :D

BTW, I doubt the net will help any of you with this one. I was about to post it with a drawing and all, but I released a new FACAR version on Sunday and my Yahoo data transfer limit was reached. :(

So, in plain and simple words, here it goes.
Suppose you have two buildings 774 ft high, separated by a distance d. From the nearest parts of the roof hangs a rope 387 ft long, which connects the buildings. You may assume that the rope is subject to a tension equivalent to the rope's weight, and that wind velocity is zero.

(For engineers out there this means the parameter of the catenary is 1, and the on-topic commentary would be that it is the same rope from the Earth/hand problem, but resized with Lanczos to 387 ft.) :)

The rope hangs in such a way that the lowest part of it is barely touching the floor, at the same point the height of the buildings is measured from.

Question: What is the value of d, the distance between the buildings?

[fccHandler]

:confused: Um, how is it possible for any part of the rope to touch the ground, since it's only half as long as the buildings are tall?

[KpeX]

@r6d2

You're going to have to post a picture on that one I think - maybe my brain is a little bit fried today but the way I'm picturing it it's not possible.

How could a 387 foot rope touch the ground hanging straight down from a 774 foot building, let alone touch the ground and another building?

[r6d2]

Men, you're quick! But you have not found the answer yet. Remember, I said "barely touching". Also ignore the wind. :)

/Add:
A tip: Remember Matrix? Well, the problem is choice.

[jel]

From the nearest parts of the roof hangs a rope the only way i can see this working is if you have a VERY large roof overhang and a VERY steep roof fall...
but it still doesnt give me an answer :confused:

[fccHandler]

I'm imagining two vertical buildings with straight sides and a rope hanging between them, with it's two ends connected at the tops of the two buildings. Is that really what you mean?

Even if I twist your first sentence to mean the combined height of the buildings is 774 ft (rather than each being 774 ft), I still can't overcome the impossibility that a 387 ft rope could touch both roofs AND the ground.

Is this really a math puzzle or just some tricky word play?

[r6d2]

Originally posted by fccHandler
Is that really what you mean?Yes, of course. No steep roofs or hangovers.
Is this really a math puzzle or just some tricky word play? It is a math puzzle indeed, with one single solution in the realm of Real numbers.

Seems that "The problem is choice" is too cryptic as a clue. But read again, which is the only part of the problem where you are given a choice?

[jel]

...well then, i 'choose' to construct my buildings such that the roof is also the walls etc (picture a pyramid). therefore the nearest part of the roof/floor would be the ground level, giving me a distance, d, of 387 feet.

[r6d2]

Wrong choice. In fact, you don't get anywhere to choose that. And I said before they were vertical buildings with straight sides... Think of... think of... a building, that's it. :) Not modern architecture or egipcian stuff.

/Add:
I was hoping Nic to come up with something but either he's still writing some C++ code or searching the net like crazy. :D:D:D

/Add:
Yahoo seems to have recovered from the hiccup. Let's see how long is lasts...
http://www.geocities.com/r6d2_stuff/Catenaria.gif

[fccHandler]

Originally posted by r6d2
which is the only part of the problem where you are given a choice?
Actually, there are two parts where we are given a choice. These are the sentences beginning with:

1. Suppose you have...
2. You may assume that...

I choose not to suppose that we have two buildings 784 ft high. :D


P.S. I'm not able to see the picture you posted.

[r6d2]

Point one is not really a choice. It's a fact you are supposed to accept. If you don't, there is no problem to solve. Point two is the only choice. Look at the picture!

/Add
Yahoo hiccup again. It releases in one hour. Sorry, the problem of free sites. Anyway, the plain and simple words are really enough, and you got the picture quite clerly already in your mind!

[fccHandler]

I'm sorry if I'm trying to turn this into a word problem. I simply can't imagine that the situation you've described is even possible. I've tried bending the earth, hanging the buildings off cliffs, and so on. I assume the 387 ft rope is not being stretched, nor are the buildings being squashed under the weight of the rope. (In those cases we would need to know the amount of squash / stretch...)

I'll wait an hour and see if your picture shows up.

[KpeX]

The diameter of the rope is not specified, so one could assume a 774 foot diameter rope, which would make d=387. Or do some weird calculation with the weight of the rope and the strech in it. That's the only choice I can think of anyways.

[dvd_master]

....

*gives up continuing reading this topic*

[fccHandler]

d = -246.372 feet

And if I'm wrong, I hope you will give us the correct answer. :angry:

[Nic]

Edit: Scratch that, bad question.... :)

[r6d2]

@Nic, since seems that everyone else has posted a solution or given up, I'll post the answer in a hour or so. Are you coming up with something on the buildings problem? (did you google it already? :D:D:D)

[r6d2]

Originally posted by fccHandler
:confused: Um, how is it possible for any part of the rope to touch the ground, since it's only half as long as the buildings are tall? Guys, my terrible mistake. :o I made a typo when transcripting the problem... I just realized, when writing down the solution.

The numbers are switched! The buildings's are 387 and the rope is 774.

Sorry guys, I didn't intend to play a trick on you. :o:o:o

So I guess by now all of you will notice the answer is trivial.

[duartix]

Sure. The buildings are close to each other (d=0).
Too easy now.

Now here is one for Nic to search. I have searched and found nothing on it.:)

A friend of mine who works in Customs found himself with this problem:
There was an unclaimed import of 19 camels from an Arab.
When he contacted the importer he found out the man had just died.
He contacted the exporter and found out that there were 3 heirs to the camels:

a) The older son who was intitled to 1/2 of the heritage.
b) The younger son who was intitled to 1/4 of the heritage.
c) The wife who was intitled to 1/5 of the heritage.

How is it possible to distribute the camels (without using croping filters, of course) by the heirs?

[mpucoder]

Pretend there are 20 camels, give 10 to the older son, 5 to the younger son, and 4 to the wife.

[duartix]

Your answer is 99.5% correct.
You just forgot to give back the camel you borrowed.;)

[duartix]

For those who still find the earth/rope problem dificult to absorb, look at this picture (http://www.xpphotoalbum.com/showphoto.php?photo=68230&password=&sort=0&cat=998&page=1).

Suppose that instead of distributing the slack around the earth, you just distribute 30 cm in 4 places N, S (Red lines) and W, E (Green lines). I guess that there is no doubt of the effect of this distribution, it creates a slack of 15cm at those cardinal points. Even though the effect is almost the same, distributing it across the entier planet is what is missleading us because we tend to believe this distribution divided across the planet is meaningless. It is meaningless in the curve shape. And also in the size increase 1m/6200000m, but it is effective.

[fccHandler]

Originally posted by r6d2
Guys, my terrible mistake. :o I made a typo when transcripting the problem... I just realized, when writing down the solution.

The numbers are switched! The buildings's are 387 and the rope is 774.
AARGH! That was plain cruel. :devil:

Nevertheless, I solved the original by assuming the earth was shaped like a torus, the buildings were at its poles, two wormholes were engulfing the buildings halfway, the tops of the buildings were poking through a black hole in the center of the torus, and the heights of the buildings were relative to the positions of two tightrope walkers standing on the rope as it crossed the event horizon.

[duartix]

Nevertheless, I solved the original by assuming the earth was shaped like a torus, the buildings were at its poles, two wormholes were engulfing the buildings halfway, the tops of the buildings were poking through a black hole in the center of the torus, and the heights of the buildings were relative to the positions of two tightrope walkers standing on the rope as it crossed the event horizon. ... and very nicely done without even resorting to VSL (http://www.amazon.com/exec/obidos/tg/detail/-/0738205257//qid=1076517059/sr=1-1/ref=sr_1_1/103-0333484-6962212?v=glance&s=books&vi=reviews) theories. Being Portuguese I'm a prime suspect for compliments, but I can't help saying besides challenging, this is also a most entertaining book. (The uncensored version, that is!)

[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...

[Kedirekin]

Here's the result I got for spheres in a pizza box.

cols: int( (2*b/d) - 1 )/2
rows: int( 2*b/(sqrt(3)*d) )

So, for example, if the (square) box has dimension 8*d:

cols: int( (2*8*d/d) -1 )/2 = 15/2 = 7.5
rows: int( 2*8*d/(sqrt(3)*d) ) = int(16/1.732) = 9

For box with dimension 8.5*d:

cols: int( (2*8.5*d/d) -1 )/2 = 16/2 = 8
rows: int( 2*8.5*d/(sqrt(3)*d) ) = int(17/1.732) = 9

And finally for box with dimension 8.7*d:

cols: int( (2*8.7*d/d) -1 )/2 = 16/2 = 8
rows: int( 2*8.7*d/(sqrt(3)*d) ) = int(17.4/1.732) = 10


PS. I had to sit down with a small box lid and a pile of nickles before I could figure this out.


For anyone who has difficulty visualizing half a column:

O O O O O O
O O O O O
O O O O O O
O O O O O
O O O O O O
O O O O O

O O O O O O
O O O O O O
O O O O O O
O O O O O O
O O O O O O
O O O O O O

[Christos]

Originally posted by 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...

Ok I made a prog that calculated all the possible sums (what "Plus" sees on his paper) that can't be expressed as a sum of 2 primes (see my post above)...
Here they are...
It should help you there are only 83:
11 93 137 177
17 95 139 179
23 97 141 181
27 101 143 182
29 103 145 183
35 105 147 184
37 107 149 185
41 109 151 187
47 111 153 188
51 113 155 189
53 115 157 190
57 117 159 191
59 119 161 192
65 121 163 193
67 123 165 195
71 125 167 196
77 127 169 197
79 129 171 198
83 131 173 199
87 133 174 200
89 135 175

[r6d2]

I think I narrowed it a bit more. These are the "plus" values whose corresponding products can be expressed in more than one way, thus suposedly mult looked at them, used his product (which he knows) and solved the problem...

11 17 23 27 29 35 37 41 47 53

[Christos]

Now lets narrow it down to pairs. All you have to do is to check for every corresponding product its corresponding sum (look at things from "Mult's" point of view). If you find one which has only one corresponding sum in the range you specified then it's possible...

For example:

11=6+5...
1st Corresponding product=6*5=30
30=15*2=10*3=6*5
Corresponding sums: 15+2=17, 10+3=13, 6+5=11
11 is in the list you specified and 17 is in so 6 and 5 are not the answer.
We want only one of the corresponding sums of the product to be in the list so "Mult" can decide which one...

11=7+4
2nd Corresponding product=7*4=28
28=14*2=7*4
Corresponding sums: 14+2=16, 7+4=11
11 is in, 16 is not...
So 7,4 is a possible answer.

11=8+3
3rd Corresponding product=3*8=24
24=2*12=3*8=6*4
Corresponding sums: 12+2=14, 8+3=11, 6+4=10
11 is in, 14 is not, 10 is not...
So 8,3 is a possible answer.

In the end the correct answer can be found if we remember Plus' last words:
"In that case I know too."

Remember that he can't see the product but just by seeing the sum and knowing that Mult reached an answer with certainty then he finds the answer also.
So the sum he sees has only one possible corresponding product that has only one corresponding sum in our list.

For example:
As we so before there are 2 possible combinations with a sum of 11 (7,4 and 8,3) so the sum can't be 11 because then Plus' last words would not be true.
So Neither 7,4 or 8,3 is the answer

[r6d2]

Either you are getting some marbles and a pizza box to do the real thing, or you are writing lots of numbers on a sheet of paper. Come on, did everybody give up? :)

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My favorites were the books of Martin Gardner. I can't remember the titles, but he wrote lots of books on "recreational mathematics" and logic puzzles. He also did a wonderful annotated Alice in Wonderland. ;)

[r6d2]

Originally posted by Kedirekin
Here's the result I got for spheres in a pizza box.The problem with this approach is that sometimes it is better to use the dice arrangement for the marbles. For instance when b = 2*d you can see that clearly.

For d << b your formula is more appropriate, but the general case is not so simple. There are some thresholds to determine.

@Manao,

Please don't post the answer to the two integers problem yet, but can you tell me if I am close with those candidates I have so far for the sums?

[Kedirekin]

I can't speak for Manao, but I did get the same set of 10 sums by (I assume) a different method.

Basically I found all the combinations of 3 prime numbers that give one and only one factoring that isn't in Christos's list of sums that can't be expressed as the sum of two primes (and is that a mouthful).

For example, the primes 2, 2 and 7 give the product 28, which has factors 2*14 and 4*7. 11 (4+7) is in Christos's list, but 16 (2+14) is not. Therefore 4 and 7 are two numbers that allow Mult to know the answer. Unfortunately they are not numbers that allow Plus to know the answer, because 2 and 9 (which add to 11) would also allow Mult to know the answer.

The possible factors I got were:
2, 9
2, 25
2, 49
4, 7
4, 13
4, 19
4, 23
4, 31
4, 37
4, 43
4, 47

There is of course the remote possibility that the two numbers factor into 4 or more primes. I didn't go that far.

And unfortunately I cannot find an easy way to factor all the products of all possible addends that add up to the 10 sums; it's just too much work.


Fixed some numbers in my list. Guess my list of sums isn't exactly the same - I don't have 29, 37 or 53, but I do have 51.

[Christos]

Originally posted by r6d2
Either you are getting some marbles and a pizza box to do the real thing, or you are writing lots of numbers on a sheet of paper. Come on, did everybody give up? :)


I have exams so my free time is close to zero... :) But I check this thread every day... :p
I Finally found some time to write a program that does exactly what I discribed in my previous post...
Here are the results...

results.txt (http://users.ntua.gr/el99605/results.txt)

So the numbers are 4 and 13 (and the sum is 17 by the way:))

It took less than a second for my P4 to find the answer :p

P.S.: As I said this was done in a hurry and I had no time to check so please verify...
I could be wrong... :)

[r6d2]

Originally posted by Christos
So the numbers are 4 and 13Congrats, Christos! You know, I still don't get it. My brain only has a 6502 CPU it seems.

[Christos]

Originally posted by r6d2
You know, I still don't get it. My brain only has a 6502 CPU it seems.

Did you read carefully my 3 previous posts? If you could tell me what isn't clear, I could explain...

[r6d2]

Originally posted by Christos
Did you read carefully my 3 previous posts? If you could tell me what isn't clear, I could explain...My 6502 is parsing one word at a time. :)

But, no, I did not understand s**t, but let me try harder. It has been a tough week.

[Christos]

Originally posted by r6d2
My 6502 is parsing one word at a time. :)

But, no, I did not understand s**t, but let me try harder. It has been a tough week.

Well here is why 4 and 13 is the right answer. Maybe it can help you understand how I got it if you reverse the process?
Plus sees 17 (4+13) on his paper and Mult sees 52 (4*13) on his.

So Mult says: "I don't know the answer" because:
52=26*2
52=13*4
(Nothing else possible)

Plus says: "I Knew it" because:
17=15+2
17=14+3
17=13+4
17=12+5
17=11+6
17=10+7
17=9+8

And so Plus knows for sure that Mult sees either:
15*2=30
14*3=42
13*4=52
12*5=60
11*6=66
10*7=70
9*8=72
For all of those possible products there are more than one factorings, so plus knows beforehand that Mult could not guess correctly. Hence he says "I knew it".

Now as we so before Mult knows that the numbers are either 4,13 or 2,28 and he says "Then I know" as soon as Plus says "I knew it". Why?
Because if it were 2 and 28 then Plus would have 30 (2+28) and he could not say "I knew it" (***meaning "I was sure that you could not find the numbers based on the sum I had", meaning "The sum I have can not be expressed as a sum of 2 primes"***) because 30 could come from 23+7 so Mult would have 161 which would cause him to say "I know the answer".

Now the hard part...
As soon as Mult says "Then I know the answer" Plus says "Then I know it too". Why?
Because for for all the possible combinations that give 17 (15+2, 14+3, 13+4, 12+5, 11+6, 10+7, 9+8) only 13,4 could lead Mult to say "Then I now the answer" as soon as Plus said "I knew it".
Why?
Well for example if we consider 15,2 Mult would have 30 (15*2) which could come from:
15*2=30
10*3=30
6*5=30

But
15+2=17
10+3=13
6+5=11
So when Plus would still say "I knew it" but Mult would not be able to decide which one was it. (Because neither can be expressed as the sum of 2 primes). Therefore he could not say "Then I know"

If the same happens with 14,3 and 12,5 and 11,6 and 10,7 and 9,8 then my answer is correct ?

That's it...
I hope I didn't make things worse... :)

P.S: Any suggestions for the hats and prisoners problem? :p

[Kedirekin]

For what it is worth, I checked 4 and 13 by hand, and it does satisfy all the criteria.


Numbers Sum Possibles Possible Factors Sum of Is Sum of Possible
From Sum Product Factors Primes? Solution

4 13 17 2 15 30 2 15 17 no no
6 5 11 no
10 3 13 yes

3 14 42 3 14 17 no no
6 7 13 yes
21 2 23 no

4 13 52 4 13 17 no yes
2 26 28 yes

5 12 60 5 12 17 no no
10 6 16 yes
15 4 19 yes
20 3 23 no

6 11 66 6 11 17 no no
3 22 25 yes
2 33 35 no

7 10 70 7 10 17 no no
14 5 19 yes
35 2 37 no

8 9 72 8 9 17 no no
4 18 22 yes
2 36 38 yes
3 24 27 no

52 is the only product that has one and only one factoring that cannot be expressed as the sum of two primes.

[Christos]

Originally posted by Kedirekin
For what it is worth, I checked 4 and 13 by hand, and it does satisfy all the criteria.

52 is the only product that has one and only one factoring that cannot be expressed as the sum of two primes.

Thanks Kedirekin...
I was right then... 4 and 13 are the right numbers :)
My P4 got it right :)
I thing your post does a much better job explaining the answer than mine...
Cheers...

@r6d2
Look at the table Kedirekin posted and look at the column titled:"Is sum of primes?"
The "hard part" I was talking about in my post is that the correct combination must have only one "no" in that column...

[r6d2]

I'll have to digest that with ease. :)

Now, ready for the marbles in a cube? Well, noboody has solved the marbles in a pizza box yet. :)

After those two, we can go for marbles in a cilinder!

[r6d2]

OK, let's leave the marbles thing aside for a while. Let's get back to this intuition thing.
Suppose you have two jars of the same volume. One contains coke and the other one contains pepsi. You also have a small glass. You fill the glass with coke and pour it into the pepsi jar. You shake the pepsi jar so the two fluids mix perfectly.

Then, you fill the glass with the contents of the pepsi jar, and pour it into the coke jar. Thus the two jars end up having the same volume again.

Which jar contains a bigger portion of the other one? The coke one has more pepsi than the pepsi one has coke or is it the other way around?

[jel]

my guess is that the pepsi jar will contain more coke than the coke jar will contain pepsi, as the liquid mixed with the coke *could* contain pepsi and coke.

[r6d2]

Originally posted by jel
as the liquid mixed with the coke *could* contain pepsi and coke. I don't quite get your reasoning, but remember you shaked the pepsi jar to get a perfect mix.

[Christos]

Since the jars had the same volume of coke (the first one)and pepsi(the second one) then both will end up having the same amound of the other in the end...

Example:
*Start
Jar1:600ml Coke
Jar2:600ml Pepsi
Glass:300ml

*Move 1 (Fill glass with coke and poor it in the pepsi jar)
Jar1:300ml Coke
Jar2:600ml Pepsi + 300ml Coke (1/3Coke-Pepsi mix)

*Move 2 (Fill glass with the mix)
Glass:300ml of mix so... 100ml Coke + 200ml Pepsi
Jar2:400ml Pepsi + 200ml Coke (1/3Coke-Pepsi mix)

*Move 3 (Pour into the Coke jar)
Jar1:400ml Coke + 200ml Pepesi (1/3Pepsi-Coke mix)

[jel]

hmmm...yup christos is right ... mathematically ... but instinctively it has the same effect as the 'rope' question a little while back ... :confused:
@r6d2 ... did you edit your previous post or have i completely lost the plot...

[r6d2]

I didn't edit the post but to fix a typo, just 2 minutes after posting it, which plot did you miss?

BTW, Christos, you're as good with beverages as you are with mysterious numbers... Solution not easily according to intuition, right?

A derivative problem...
What is the size of the glass needed to end up with 50% mix on each jar?

[jel]

which plot did you miss? all of the above :rolleyes:

...back to the marble question. are the 'boundaries' you are referring to:
if the distance from the tangental edge of the last marble in any given length, is less than the radius, then this will occur -
0 0 0 0 0 0
0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0

and if the distance is greater than the radius, this will occur -
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0

[r6d2]

Originally posted by jel
...back to the marble question. are the 'boundaries' you are referring to:It's more like this:

When an integer number of marbles fit on the first row, the second row may be better arranged diagonally. This depends on the relationship between b and d. For instance, if b = 2*d, it is better to arrange them like if they were dice, vertically. Here you have one of the thresholds.

The second one is when b = n*d + d/2, where it will always be better diagonally. The general solution takes into account any values of b and d.

[sysKin]

Originally posted by r6d2
A derivative problem...
What is the size of the glass needed to end up with 50% mix on each jar?Hah, let me give you the answer at leat once. As big as the jar :P or bigger if you like the "half-empty/half-full" problems :D

No really, you need to mix them all, not leaving anything.

Radek

[r6d2]

Originally posted by sysKin
Hah, let me give you the answer at leat once.Well, that one was easy. :)

What about this one:
How many times do you have to repeat the process with the small glass to get a 50% mix?

[sysKin]

Originally posted by r6d2
What about this one:
How many times do you have to repeat the process with the small glass to get a 50% mix?How about: infinite number of times? Any time you do it, mixture gets closer and closer to 50% but never actually gets there.

:)

[samlar]

What has all this got to do with the price of a can of beans

[r6d2]

Originally posted by sysKin
How about: infinite number of times?Well, what you're saying is that if the glass has the volume of the jar, it takes 1 time, and if not, it takes an infinite number of times? Doesn't intuition tell you that there should be a relationship between the volume of the jar and the glass, which determines the number of times?

BTW, on the can of beans... We haven't got there yet. But it would be an interesting problem to determine how many beans fit into a can. :D

[fccHandler]

All this reminds me of an article I read on CNN:

http://www.cnn.com/2004/TECH/science/02/16/science.candy.reut/index.html

Try M&M's in your pizza box, instead of marbles. :D

[r6d2]

Originally posted by fccHandler
Try M&M's in your pizza box, instead of marbles. :D LOL!

That's astonishing. For centuries mankind has thought that spheres are the optimum shape for a given volume, and thus intuitively it was considered that the arrangement of them as bulk would be optimum too!

In fact, planets are sphere shaped because it minimises the momentum of inertia and thus the energy of the physical body. So, what's true for planets isn't true for associated subatomic particles? Great discovery indeed!

[Manao]

Sorry for not answering sooner, I was in holidays.

Christos : you got it right, that's the answer.

[r6d2]

Well, seems that there is no more interest in marbles in boxes, or the solution is beyond the scope of this forum. :D:D:D

Let's go for a simpler ride?
Coke cans have a standard volume of 350 ml and are made of aluminium. Suppose the can is a perfect cylinder and ignore the stacking top and bottom irregularities.

Can you design a container which uses less aluminium and still contain the same volume?

[Manao]

A ball which have the volume of the cylinder, hence a radius of 4.37 cm ( the formula of the volume being 4*pi*r^3 / 3 )

[wmansir]

Originally posted by r6d2
Coke cans have a standard volume of 350 ml and are made of aluminium. Suppose the can is a perfect cylinder and ignore the stacking top and bottom irregularities.

Can you design a container which uses less aluminium and still contain the same volume?

Ha! I can design a container that holds almost 10 times that volume and still uses less aluminium!















http://216.58.174.219/internetclassifiedsws/ecook/images/Kyle%20and%20the%20Amazing%203%20Liter.jpg

:p

And no, that is not me.

[r6d2]

Originally posted by Manao
A ball which have the volume of the cylinder, hence a radius of 4.37 cm ( the formula of the volume being 4*pi*r^3 / 3 ) OK, but your answer is merely speculative... :) How much aluminium is saved?

Or in other words, how many more containers could be built?

And also, as a derivative problem, if you have to make it a cylinder anyway (balls wouldn't be functional really, you could not put one on the table once open) :), which would be its dimensions and how much aluminium is saved?

[rudeboymcc]

you could make it flat bottomed so it stays upright! but it will be a bitch to drink, you'll have to put it completely upside down to get to the last part!

[Manao]

I wasn't speculative, I was affirmative ;)

You should have specified we should keep a cylinder, because the sphere is known to be the optimal volume in consideration to the ratio area / volume.

Anyway, with a cylinder, let's r be its radius, h its height.

Its volume is pi*r*r*h, its area is 2*pi*r*h + 2*pi*r*r. We want 2*pi*r*h minimal, with pi*r*r*h = V constant.

So we have h = V / (pi * r * r), and we want to minimize 2*pi*r*(h + r) = 2*pi*r*(V / (pi*r*r) + r)

The derivate is 4*pi*r - 2 * V / (r*r), we want it to be zero, hence, r^3 = V / (2 * pi) and r = 3.82 cm, and h = 7.64. We remark that 2 * r = h, hence, it fits perfectly a cube ( which is not quite surprising ) and it differs from usual cans, which are thinner.

I don't know the exact height and radius of a can, but I guess it's 3.3 / 10, hence a total area of 279 cm². With 3.82 and 7.64, it's 275 cm². Not much of a gain ( 1.5 % )

[vio]

Originally posted by r6d2
Well, seems that there is no more interest in marbles in boxes, or the solution is beyond the scope of this forum. :D:D:D



I think it is beyond the scope of this forum. I think there is a prize if you can get sphere packing conjecture above a certain value.

I'm still wrapping my mind around the colored hat question.

[Manao]

To end ( I hope ) the marbles in the pizza box :

We put the marbles like that :0 0 0 0 0 3
0 0 0 0 2
0 0 0 0 0 1So we have to know :
- the number of odd and even rows ( n_o, and n_e )
- the number of marbles in an odd row ( n_mo )
- the number of marbles in an even row ( n_me )

And the total number of marbles will be n_o * n_mo + n_e * n_me

----

First, the number of rows ( n_r ) :
The distance between the middle of two consecutive rows is d * sqrt(3) / 2. But we also have to count the radius of the marbles in the first and last rows. So, if we have n_r rows, we have a total height of d * ( 1 + (n_r - 1) * sqrt(3) / 2 ). Hence, n_r = int(( b / d - 1 ) * 2 / sqrt(3) + 1). Let's check :
d = 1, b = 1 -> n_r = 1 ( height : 1 )
d = 1, b = 2 -> n_r = 2 ( height : 1.866 )
d = 1, b = 9 -> n_r = 10 ( height : 8.794 )

----

Now, from n_r, we can deduce n_o and n_e :
n_o = int( (n_r + 1) / 2 )
n_e = int( n_r / 2 )

Check :
n_r = 1 + 2 * k -> n_o = k + 1, n_e = k
n_r = 2 * k -> n_o = k, n_e = k

----

Finally, we have to compute n_me and n_mo. The length of an odd row is l_o = n_mo * d, so n_mo simply equals to int( b / d )
The length of and even row is l_e = n_me * d + d / 2, hence n_me = int(b / d - 0.5)

Check :
b = 2, d = 1 : n_mo = 2, n_me = 1
b = 2.5, d = 1 : n_mo = 2, n_me = 2

----

So the final formula ( if no mistakes were made ) :
int((int(( b / d - 1 ) * 2 / sqrt(3) + 1) + 1)/ 2) * int(b / d) + int(int(( b / d - 1 ) * 2 / sqrt(3) + 1) / 2) * int(b / d - 0.5)

But hey, I forgot another bundary effect : it may happened that for the last row, putting another odd row instead of an even row is better, like that : 0 0 0 0 0
0 0 0 0 0
0 0 0 0
0 0 0 0 0
0 0 0 0
0 0 0 0 0When does this happen ? When b is between d * (k * sqrt(3) + 2) and d * ((k+1) * sqrt(3) + 1) ( height wondition ), and when b is between k * d and (k + 0.5) * d ( in order to have even rows shorter than odd ones )

Such a number, for example, is 2 ( thanks r6d2 for pointing me to that ). That makes the formula even uglier, so I won't modify it.

[r6d2]

@manao,

Good work on the ideal can thing!
Originally posted by Manao
But hey, I forgot another bundary effect : it may happened that for the last row, putting another odd row instead of an even row is better, like that : 0 0 0 0 0
0 0 0 0 0
0 0 0 0
0 0 0 0 0
0 0 0 0
0 0 0 0 0I think this may happen with several of the top rows, not just the last one. Now that would make the formula even uglier.

[r6d2]

Consider movies like Armageddon and Sudden Impact, which are categorised by some people practically as comedies from a physics standpoint. :)

Now let's speculate about the following scenario.
The Earth (mass M) revolves around the Sun at a speed v (~30 km/s). An asteroid (mass m) is in course of collision with Earth along the tangent of its orbit, on the same plane, at the same speed and in the same direction.

Suppose the timing of the trajectories are set up in such a way that both Earth and the asteroid will collide at the precise point where the radius of the Earth orbit forms a 90º angle with the asteroid's path.

After the collision, the asteroid and Earth form one single body of mass M + m.

What happens with Earth and its movement, from a physical perspective?

(i.e., ignore the casualties, the zone of impact, and such. Just think about the orbit. Does it stay the same?)

[Manao]

We don't add any quantity of movement ( I'm not sure how it is called in english, it's the product of the mass by the motion vector ), so we won't modify the orbit, nor the period of revolution.

And for the marbles, yes you're right, you may have several such rows ( but always less than 7, because sqrt(3) * 8 / 2 < 7, so you could put 8 rows with offsetinstead of the 7 ).

[easybeat]

The proposed collision wouldn't occur. there's no way Arnie, Sly and Bruce would let that happen. (Actually I here Arnie's a bit busy at the moment banning same sex marriages so we'll have to make do with Bruce and Sly)

[duartix]

I don't understand why it shouldn't change orbit Manao.
I guess from my intuition that the new body Earth+Asteroid would suffer a route deviation of about:
(mAsteroid*vAsteroid) / (mEarth*vEarth)

And I believe quantity of movement is caled momentum. (Can someone please confirm?)

[Manao]

Maybe I misunderstood the way the asteroid is meeting the Earth, but r6d2 says " An asteroid (mass m) is in course of collision with Earth along the tangent of its orbit, on the same plane, at the same speed and in the same direction " so for me the asteroid doesn't crash on the earth, it just happens to be at the same place at the same moment, with the same speed and direction than the Earth.

[r6d2]

Well, the asteroid does crash, but it does is "softly", sort of begins to ride along with Earth. Actually the point of collision cannot be the exact point of the right angle, because both Earth and the asteroid have a radius, but the simplification does not affect the result.

Manao is right, there is no change in momentum (and yes, that's how is it called). In fact, we have that:
M*v[M] + m*v[m] = (M+m)*v[M+m]

Notation: x[y] reads property x of body y, i.e., x sub yThe left side of the equation is before the impact, and the right side is after the impact. Since the original velocity v is the same for both masses, this implies that v[M+m] = v, i.e, the orbit is maintained.

However, intuition suggests that this is only valid during the instant immediately after the impact. After that, since the Earth now weights more, the attraction force from the Sun will be higher, and since it is no moving any faster, it should begin to "fall" in a spiral into the Sun.

It this reasoning correct?

[Manao]

No. The Earth will be heavier, so indeed the attraction of the sun will be higher. But, the Earth will be heavier, so the centrifugal force will be higher, and will compensate exactly the attraction force.

The falling in spiral would occur only if the energy of the Earth wasn't constant, hence if there was something to slow it down. No planet / asteroid falls in spiral, it's always revolving in ellipse ( if it doesn't touch the sun itself )

[mpucoder]

That's what I always heard, the orbit around a body of a certain mass depends on velocity alone. This is why all geosynchronous satellites, regardless of mass, are at the same altitude.
But what about the Moon? Earth now has more mass, so the gravitational attraction between Earth and Moon increases. Causing a slight decay in the Lunar orbit, and an increase in the tidal friction, slowing Earth's rotation.

[Kedirekin]

Originally posted by r6d2
An asteroid ... is in course of collision with Earth along the tangent of its orbit, on the same plane, at the same speed and in the same direction.


You do realize that, from an orbital mechanics standpoint, you've described an impossibility. If the asteroid is in a tangent to earth's orbit and moving at the same speed and direction, it is in the same orbit and can never collide.

In effect, what you've described can only happen if the asteroid is already resting on earth's surface.

[Manao]

mpucoder : there are no loss of energy for the system Moon - Earth, so the rotation of both around the sun will not be disturbed. But the Moon will indeed be slightly closer to the Earth, and Earth's roation around itself will slightly be slower.

Kedirekin : It's possible. Orbits do not have to be circular, they can be elliptic. And such a situation could happen with the asteroid reaching it's perihelion / aphelion. At that point, it's motion vector would be perpendicular to the segment [Sun - Asteroid], and its speed would be the same as Earth's one, since distance from the Sun is the same.

[r6d2]

Originally posted by Kedirekin
If the asteroid is in a tangent to earth's orbit and moving at the same speed and direction, it is in the same orbit and can never collide. You are right! What I meant was that the asteroid was in course of collision along the tangent at the point of collision. The asteroid is going in a straight line. Sorry. Anyway, Manao got the idea. ;)

I really need a way to post pictures here. :o

Another problem based on the same idea:
What happens in the situation above if the asteroid is moving 10% faster than the Earth at the time of collision?

[Manao]

With the asteroid going faster than the Earth, the motion vector of Earth + Asteroid after collision will be (M + m*1.1) / (M + m) * v.

One of the keplerian laws says that angular speed is a constant of the movement of a planet around the Sun. Another says that the value a^3 / T^2, with a being the 'big' radius, and T the period of revolution is a constant for all planets around the Sun.

So angular speed ( radius * v when v is perpendicular to the radius ) has been raised, so the period has been reduced. So Earth will be revolving faster around the Sun. Moreover, since the period is lower, the 'big' radius of the elliptic orbit of Earth has also been reduced. Hence, if Earth's orbit was circular, it's now elliptic, with the crash point being its aphelion. So the new orbit will be inside the previous one.

[Kedirekin]

I have a feeling I'm going to get pedantic here.

It actually is not possible. If the asteroid's perihelion just touches earth's orbit, it cannot have the same velocity as the earth. It must have a higher velocity than the earth in order to move further away from the sun (barring collision of course).

Another way to put it, a body's orbit is completely described by it's velocity vector at any point in time. If it's velocity vector is the same as the earth's, then it's orbit is the same as the earth's.


PS. I don't mean to be a kill-joy. Ignoring the orbital mechanics, it's an interesting - slightly tricky - question. With the described conditions, Manao is of course correct; momentum is always preserved.

Here's another slightly tricky one.

Given two objects in the same orbit that are mutually gravitationally attracted, but separated by a distance significantly larger than their diameter, what will happen?

Note: this situation exists right here in the solar system.

[r6d2]

Originally posted by Manao
So angular speed ( radius * v when v is perpendicular to the radius ) has been raised, so the period has been reduced.Angular speed isn't v/r? It is more natural to think that if speed increases, the new orbit will have a bigger radius. Think for instance that the bigger the mass of the asteroid, the greater the momentum and thus the impact. Earth might get kicked to the outer solar system by a big asteoid, not towards the Sun.
Originally posted by Kedirekin
I have a feeling I'm going to get pedantic here.ac3 is getting used to it. :)
It actually is not possible. If the asteroid's perihelion just touches earth's orbit, it cannot have the same velocity as the earth.Consider then the asteroid is self-propelled and can compensate for the gravity of Sun. :)

Actually, if the orbit is elliptic, as Manao suggests, it is possible. I don't think the orbit is completely described by its velocity vector. This is true only for circular orbits, where speed is constant. I said the speed was constant, but to limit the scope of the problem you can consider it constant in the vecinity of Earth. We are not considering either the Earth's gravity effect on the asteroid.

[r6d2]

Originally posted by Kedirekin
what will happen?My first guess, based on Saturn's Rings, is "nothing". And the "significantly larger" part only reinforces that. But I'm sleepy. :)

[Kedirekin]

Agreed. If the asteriod is under thrust, the scenario is possible. Heck, any scenario is possible.

But I do stand by what I said. An free-falling asteriod with a perihelion the same distance from the sun as earth's orbit will have a higher velocity at perihelion than earth does. The velocity has to be higher, or the asteriod wouldn't move back outward again.

I can see all the diagrams and the machanics in my head (velocity vectors and changes over time, acceleration, translation between potential and kinetic energy), but I'm not eloquent enough to explain them with crystal clarity. I guess if you don't believe me, we'll just have to agree to disagree.

[Kedirekin]

In response to Manao's response to the 10% faster question.

The combined body will be moving faster, as Manao says, but the higher velocity immediately after impact will push the combined body outward - it will be moving at higher than orbital velocity. The point (or more accurately, the distance) of impact will become the perihelion.

As is so often true with orbital mechanics, thinking about it kind of twists your senses. In the words of one of my favorite authors -


"East takes you out, out takes you west, west takes you in, and in takes you east"
- Larry Niven

[Kedirekin]

Hint (or spoiler): Epimetheus and Janus

The universe is an amazing place.

[Manao]

r6d2 : angular speed was indeed v / r, and I said something wrong about the keplerian law, because in that case it's not the angular speed which is constant, but v * r. Which explains my second error, rightly pointed out by Kedirekin : in an elliptic orbit, speed decrease with the distance, and if angular speed was constant, it would be the contrary. So I inverted aphelion / perihelion.

Third mistake, it's indeed not possible for the Earth and the asteroid to meet with the same motion somewhere around the Sun.

Fourth, the period will be higher, not lower, because the 'big' radius will be higher, not lower.

[r6d2]

Originally posted by Kedirekin
Hint (or spoiler): Epimetheus and JanusBut these two don't fit into the "significantly larger" category, and they are not in the same orbit either. :confused: I'll give it more thought.

[duartix]

@Manao:Maybe I misunderstood the way the asteroid is meeting the Earth, No, Manao, it was me who missunderstood the colision, I thought it was a perpendicular colision...

[Kedirekin]

Okay, Epimetheus and Janus aren't in exactly the same orbit, but close enough as makes no never mind. Their orbits are so close that the body of one moon nearly (or maybe actually) overlaps the other's orbit.

And most of the time they are thousands or tens (or hundreds) of thousands of kilometers apart in their orbits, while their diameters are roughly 100 kilometers.

You can ignore Epimetheus and Janus if you want and deal with the ideal case. The mechanics is nearly the same, though in the ideal case the results are a little different. I wasn't getting at anything too intense or mathematical.

[r6d2]

Originally posted by Kedirekin
You can ignore Epimetheus and Janus if you want and deal with the ideal case.What I don't get is that if the two bodies share the orbit, they shoud have the same speed, as you pointed out above on the asteroid problem. So the "nothing" result I posted previously, meaning they stay like that forever, is the only one that makes sense... :confused:

[Kedirekin]

Sorry, I should have been clearer.

Assume two small moons 100 km in size are in orbit around a large planet. Further assume they are 10,000 km apart, traveling in the same orbit, and initially they have exactly the same velocity (tangential to the orbit at their respective positions, of course).

Taking into account that the moons are mutually gravitionally attracted to each other (in other words, don't ignore gravity), what will happen? Will the moons collide?

I fear I've already ruined the problem by giving you the spoiler.

[Kedirekin]

Here's another, perhaps more interesting way to state the same problem.

Assume you are floating in space (in a space suit) holding a rock. You are 500,000 km ahead of the earth in it's orbit, stationary with respect to the center of the earth (i.e. you are not in orbit around the earth).

You drop the rock and it very slowly begins to fall towards the earth - remember, you are stationary with respect to the earth.

Will the rock hit the ground?

[r6d2]

Just checking if I understood correctly. I'm in orbit around the Sun, at the same speed as the Earth and in the same orbit. The rock is with me but still at the same speed. I let the rock go. Is that it?

[sysKin]

Originally posted by Kedirekin
You drop the rock and it very slowly begins to fall towards the earth - remember, you are stationary with respect to the earth.

Will the rock hit the ground?As I understand, there is some engine that prevents you from falling.

OK, so why the rock wouldn't hit the Earth? It will.. if it wasn't for Earth's gravitation, the rock would be on orbit around the Sun, just like Earth currently is. Extra pull from Earth will make the rock accelerate exactly towards the centre of Earth, all the time, so eventually the rock will hit the Earth exactly in the centre (from your point of view).

Any extra forces present (Sun's pull) will make the rock and Earth accelerate, with exactly the same acceleration - so they don't matter.

:)
Radek

[RadicalEd]

Of course it won't hit the ground, it'll burn up in the atmosphere long before ;)

[Kedirekin]

@r6d2, yes, you've got it.

@sysKin, oh, that it were that simple. And yes, consider yourself magically suspended in space so you yourself aren't affected by earth's gravity.

@RadicalEd, I know your post is meant tongue-in-cheek, but actually I'm not sure the rock would get going fast enough to burn up. I know escape velocity is 11 km/sec, but I don't know if half a million km can be considered escape. I think I might be curious enough to check it out (later).