Math problem, last people I can think of

[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!)