# Forums / Discussion / General

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## KYM Mathematics General, AKA the Genius Thread

Last posted Apr 14, 2013 at 02:39AM EDT. Added Mar 31, 2013 at 10:03PM EDT
40 posts from 14 users

On quite a few occasions I’ve seen discussions on this subject appear out of nowhere- a lot of users (like myself, in case it wasn’t already obvious) seem to be really interested. Now all of that has a home. Keep it semi-serious; no branch is off-limits. Have a blast.

Mar 31, 2013 at 10:03PM EDT

You are a contestant in a quiz and reached the finale. There are 3 doors you can choose from:

[ 1 ] [ 2 ] [ 3 ]

Facts:

• One of these doors contains a car.
• The other 2 doors give you a goat.

Now you have picked one of these doors (which one isn’t relevant). After this the quizmaster opens a different door, one with a goat. Knowing one of the doors with a goat is gone, he offers you the option to change your choice to the remaining door you didn’t pick.

Question: If you want to win the car, is it better to switch doors or not?

I know the answer and the explanation for this, I will request both from you if you are going to answer.

Last edited Mar 31, 2013 at 10:09PM EDT
Mar 31, 2013 at 10:07PM EDT

I like math, but I’m not great at it on the whole.

I have some experience with some statistics from school, but the most complex testing I can do involves linear regression modeling. I can’t say I’m super interested in talking about it in my free-time, but it’s interesting to do within the time that I am working.

Mar 31, 2013 at 10:18PM EDT

RandomMan wrote:

You are a contestant in a quiz and reached the finale. There are 3 doors you can choose from:

[ 1 ] [ 2 ] [ 3 ]

Facts:

• One of these doors contains a car.
• The other 2 doors give you a goat.

Now you have picked one of these doors (which one isn’t relevant). After this the quizmaster opens a different door, one with a goat. Knowing one of the doors with a goat is gone, he offers you the option to change your choice to the remaining door you didn’t pick.

Question: If you want to win the car, is it better to switch doors or not?

I know the answer and the explanation for this, I will request both from you if you are going to answer.

It’s best to switch your answer because at first your odds were 1 out of 3, and after he opens a door it’s 1 out 2. Might as well switch.

Mar 31, 2013 at 10:34PM EDT

I love meth math!

@RandomMan

I’m not sure about the explanation, cause it’s a one in three chance you get it right. After the first door opens with a goat, it becomes fifty fifty. What if the two doors I didn’t choose had goats in them all along? I’m guessing I’d better not switch doors, because I might have been right the first time. Did I win my goat?

Mar 31, 2013 at 10:45PM EDT

RandomMan wrote:

You are a contestant in a quiz and reached the finale. There are 3 doors you can choose from:

[ 1 ] [ 2 ] [ 3 ]

Facts:

• One of these doors contains a car.
• The other 2 doors give you a goat.

Now you have picked one of these doors (which one isn’t relevant). After this the quizmaster opens a different door, one with a goat. Knowing one of the doors with a goat is gone, he offers you the option to change your choice to the remaining door you didn’t pick.

Question: If you want to win the car, is it better to switch doors or not?

I know the answer and the explanation for this, I will request both from you if you are going to answer.

Switch.
If you don’t, then the fact that one door was opened didn’t change anything, so you retain the 1/3 probability you had at the start. However, when you switch you’re working with an entirely new scenario. Now, out of the three doors, only one will make you lose- it’s of course impossible to pick the one the quizmaster opened, but nonetheless you have to factor it into the new probability. This time, it’s presence is in your favor because you wouldn’t pick it anyway- you would know that it would cause you to lose. So now there’s only one you can choose that will not give you the car. Thus, your odds become 2/3 instead.

Mar 31, 2013 at 10:49PM EDT

Okay, let me try to explain again because I don’t think I made myself clear. When you choose a door for the first time, it’s one in three. However, when the first door opens with a goat, it’s becomes fifty fifty. But here’s the thing, if the first door opened always has a goat in it, then it’s always fifty fifty. No matter what happens, you always end up having to pick between two doors.

Mar 31, 2013 at 11:17PM EDT

RandomMan wrote:

You are a contestant in a quiz and reached the finale. There are 3 doors you can choose from:

[ 1 ] [ 2 ] [ 3 ]

Facts:

• One of these doors contains a car.
• The other 2 doors give you a goat.

Now you have picked one of these doors (which one isn’t relevant). After this the quizmaster opens a different door, one with a goat. Knowing one of the doors with a goat is gone, he offers you the option to change your choice to the remaining door you didn’t pick.

Question: If you want to win the car, is it better to switch doors or not?

I know the answer and the explanation for this, I will request both from you if you are going to answer.

I pick Door Number 3!

Mar 31, 2013 at 11:18PM EDT

Here’s something that’s always great for discussion- Goldbach’s conjecture. How long to you think it will take for it to be proven (or disproven)? 271 years is a pretty intimidating length of time, not to mention all the hours of work that are certainly put into it today, considering that every mathematician knows they’ll become practically immortal if they finally solve it. I looked into it and found some patterns, such as how, when you graph with even numbers equal to or greater than four as x and the number of possible combinations of primes that add up to those numbers as y, it definitely trends upwards. This suggests that the conjecture is true, but by no means proves it up to infinity.

Apr 01, 2013 at 01:27AM EDT

Normally I like to keep pony stuff restricted to pony general, but I’m going to make an exception for this video given how aptly it expresses what is going on in my mind whenever I listen to people talk about the maths.

Last edited Apr 01, 2013 at 02:14PM EDT
Apr 01, 2013 at 02:11PM EDT

RandomMan wrote:

You are a contestant in a quiz and reached the finale. There are 3 doors you can choose from:

[ 1 ] [ 2 ] [ 3 ]

Facts:

• One of these doors contains a car.
• The other 2 doors give you a goat.

Now you have picked one of these doors (which one isn’t relevant). After this the quizmaster opens a different door, one with a goat. Knowing one of the doors with a goat is gone, he offers you the option to change your choice to the remaining door you didn’t pick.

Question: If you want to win the car, is it better to switch doors or not?

I know the answer and the explanation for this, I will request both from you if you are going to answer.

Hey, the Monty Hall problem!
Classic!

As it has been said, it is always better to switch as you will improve your chances.
Probably the easiest way I’ve seen it explained is this:
On your first pick, you have a 1/3 chance of picking correctly and a 2/3 chance that your choice is wrong.
Since the host will always reveal one of the incorrect doors, you are left with a 2/3 chance that the door you did not pick has the car.
Switching will literally double your odds, not just make it 50-50.

Apr 01, 2013 at 03:05PM EDT

That moment when you miss 2 points on a dumb mistake and get a 98 on a Differential Equations test and feel retarded.

Edit: Actually I missed 8 points, but I got all 6 of the bonus points. This was just one of the dumber mistakes I made on it and it was right in-front of me when I got it back. lol.

Last edited Apr 03, 2013 at 09:03PM EDT
Apr 03, 2013 at 08:57PM EDT

So uh, are we just circlejerking over our interest in math or what? :o

Maybe we could direct it to more recreational math instead, or maybe general non-googleable questions we don’t feel like asking on stackexchange because they’re scary, or project euler (anyone want to be friends on project euler btw?), or whatever, or else i feel this thread will die quickly ¯\(-_o)/¯

I’m not sure what everybody’s level of math knowledge is, but here’s a question my friend and I were discussing. Suppose you have n points, and you don’t know their particular coordinates. However, you do have an n x n matrix A that contains all the pair-wise distances between the n points – e.g. A_ij = 0 if i == j, and A_ij = distance between ith point and jth point if i != j.

Two questions we couldn’t figure out:
1) Given such a matrix A, are all sets of points that can result in that particular matrix isomorphic to each other (like, basically the same besides rotation, reflection and translation of points)?
2) (actually the question the original question we were originally considering) Given such a matrix A, is there a good algorithm for figuring out which pairwise distances correspond to points “close” to each other (“close” using some logical mathematical criteria). But you’re not given the n points, just A.

If Question 1) is affirmative: all sets of points corresponding to such a matrix are isomorphic, then answering question 2) is not too hard, but i couldn’t figure it out :s

Last edited Apr 04, 2013 at 03:38AM EDT
Apr 04, 2013 at 02:15AM EDT

0.9999...=1 wrote:

Here’s something that’s always great for discussion- Goldbach’s conjecture. How long to you think it will take for it to be proven (or disproven)? 271 years is a pretty intimidating length of time, not to mention all the hours of work that are certainly put into it today, considering that every mathematician knows they’ll become practically immortal if they finally solve it. I looked into it and found some patterns, such as how, when you graph with even numbers equal to or greater than four as x and the number of possible combinations of primes that add up to those numbers as y, it definitely trends upwards. This suggests that the conjecture is true, but by no means proves it up to infinity.

I know it’s been beaten to death into number theory students, but conjectures like Merten’s conjecture, Polya’s conjecture, skewes number, euler’s sum of powers conjecture, show how conjectures can hold up for really large numbers and still be wrong.

Ex: it would seem that li(x) < pi(x) for all x, but the first counterexample is at most e^(727.95133)…. which is like… 300 digits o_e

That is a pretty cool trend though.
I know someone mentioned that the collatz conjecture, on a deeper level, suggests that there is a nontrivial relationship between the factorizations of n and n + 1, since dividing by 2 and multiplying by 3 has small effect on the factorization, but adding 1 doesn’t effect the fact that we can always bring back the number to 1 using the collatz process… which is interesting. It’s like a conserved quantity.

I suspect that if the mathematics between the collatz conjecture could be more fully fleshed out or proven, then that would be the way to attack the goldbach conjecture, although the undecidability of the general collatz problem scares me.

Also, that graph of the ways to write an even number as the sum of two primes is pretty sexy:

Last edited Apr 04, 2013 at 02:49AM EDT
Apr 04, 2013 at 02:31AM EDT

Apr 04, 2013 at 03:29AM EDT

Thanks, OGW for letting me know this was here; did you remember me mentioning that my degree is in math, or that I was once a high school math teacher, or did you just vaguely recall I was a math guy who always jumps in to discussions involving statistics?

@RandomMan
RE: The Monty Hall Problem

This is indeed a classic math problem, but it actually has several (well, at least two) answers that are conditional upon real-world conditions.

0.9999…=1 got it right. The classic answer is that you should switch, because the probability that the remaining door contains a car is 2/3. Why? When you first pick, you have a 1/3 chance of finding the car, 2/3 of the time, it’s behind one of the doors you did not choose. When Monty shows you the goat, all of that 2/3 probability gets collapsed into the last door. This is counter-intuitive to most people, so like to illustrate by adding more doors: imagine there are 20 doors with one car and 19 goats. Choose your door, let’s say #1. Now Monty shows you that there is a goat behind doors 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, and 20 and asks you if you want to take door #7 instead. Now do you feel it?

Now, the practical answer: You can’t know, and you may want to assume that Monty is trying to deceive you. There’s an important question that needs to be answered: does Monty always show you a goat after you pick? If Monty only does this some of the time, he can twist the odds. In a scenario where Monty always shows you a goat if you pick right, and shows you a goat 50% of the time when you’re wrong, the odds go back to 1 in 2. An extreme example to illustrate this is to assume Monty only shows you a goat if you pick the car, which would mean switching is a guaranteed loss. Of course, that would be a poor strategy for him, since contestants would quickly figure it out.

@The Stare Master:
Differential equations was the only math class I failed in college, so you impress me, anyway.

@0.9999…=1 and ogreenworld:
I’m always fascinated by things like Goldbach’s conjecture; I recently became obsessed over Fermat’s theorem on sums of two squares because it was brought up in a book I was reading, and I’d somehow never heard it before. I ended up examining all integers from 2 to 1000 that can be expressed as the sum of two squares to find some interesting properties. (Including the fact that the statement about the Brahmagupta–Fibonacci identity at the top of that article is incorrect; I should fix that sometime.) The Collatz conjecture article was not only interesting, but amusing in noting that for n=27, the sequence climbs to over 9000 before descending to one.

Anyway, thanks to 0.9999…=1 for starting this thread. I saw your comments on the Pascal’s Triangle entry on April 1st, and was going to comment on how the article barely scratched the surface of the properties of the Triangle; maybe later in this thread when I get back to it.

Apr 04, 2013 at 05:02AM EDT

Re: Brucker/In Soviet Russia, meme force you!

I think I just remembered you as being mathy :<

Re: pascal’s triangle
I recently found a sequence of sequences while working on a combinatorics problem on project euler (having to do with counting numbers whose digits left-to-right strictly increase or strictly decrease), and again in calculating geometric-like infinite series that sum up to 1/(1-x)^n for some integer n. What I somehow got was a lesser known way to create pascal’s triangle o_o;

Let the sequence s_1 = 1, 1, 1, 1, 1, 1, 1…
Let the ith number (starting with i = 1) in the sequence s_k be the sum of the first i terms in s_k-1
so:
s_2 = 1, 2, 3, 4, 5, 6, 7…
s_3 = 1, 3, 6, 10, 15, 21, 28…
s_4 = 1, 4, 10, 20, 35, 56, 84…
s_5 = 1, 5, 15, 35, 70, 126, 210…
and so on…

These sequences form the diagonals of pascals triangle.

Last edited Apr 04, 2013 at 05:58AM EDT
Apr 04, 2013 at 05:44AM EDT

The Cute Master :3 wrote:

That moment when you miss 2 points on a dumb mistake and get a 98 on a Differential Equations test and feel retarded.

Edit: Actually I missed 8 points, but I got all 6 of the bonus points. This was just one of the dumber mistakes I made on it and it was right in-front of me when I got it back. lol.

>Multiple-choice Math test. Those tricky ones where they make certain that simple mistakes are always amongst the options.
>Forget a negative in the second to last question.
>Dumb mistake costs me 0,6 point.
>Mfw:

There were no survivors.

OGW wrote:

Let the sequence s_1 = 1, 1, 1, 1, 1, 1, 1…
Let the ith number (starting with i = 1) in the sequence s_k be the sum of the first i terms in s_k-1
so:
s_2 = 1, 2, 3, 4, 5, 6, 7…
s_3 = 1, 3, 6, 10, 15, 21, 28…
s_4 = 1, 4, 10, 20, 35, 56, 84…
s_5 = 1, 5, 15, 35, 70, 126, 210…
and so on…

These sequences form the diagonals of pascals triangle.

Also known as the triangular numbers, it is simply one of many ways to create a Pascal’s Triangle, and one of multiple that makes use of polygonal numbers.

Another nice way to create a Pacal’s triangle with polygonal numbers is through square numbers, which can be found in the same diagonal as the triangular numbers. Their name explains them perfectly, with the first square number being 1 (1 2, assuming there’s a 0 outside of the triangel), the second square number being 4 (2 2), the third being 9 (3 2), and so on.

Last edited Apr 04, 2013 at 09:27AM EDT
Apr 04, 2013 at 09:23AM EDT

@ogreenworld:
Yeah, that’s essentially a way to construct the Triangle in more of a sideways manner. In the past, I’ve used a similar method to create Pascal’s Triangle in MS Excel. However, I don’t think of those as the “diagonals”, but merely the “rows”, which can be also defined by successively higher-order polynomials:
s1(n) = 1
s2(n) = n
s3(n) = (n 2 + n)/2
s4(n) = (n 3 + 3n 2 + 2n)/6
s5(n) = (n 4 + 6n 3 + 11n 2 + 6n)/24
…etc.

What I consider the “diagonals” have an interesting property I discovered by accident in high school. Fairly trivially, the sum of each successive layer of the Triangle is 2 n-1, but I wanted to see if there were other interesting series hiding in there, and I found them, basically like this:

The sums of these “diagonals” form the Fibonacci series.

@RandomMan:
A lot of people point out the “1’s, counting numbers, triangular numbers” thing and stop there, but as someone who’s also been interested in higher-dimensional math, I realized that the next row {1, 4, 10, 20, 35,…} is the tetrahedral series (When I discovered this on my own, I called them “pyramid numbers”), and the next must be a fifth-dimensional series for which I had no name (but Wikipedia calls them “pentatope numbers”).

Oh, and the reason the square numbers show up like that is because a square is, of course, two triangles stuck together.

I’ve always felt that Pascal’s Triangle has just about every meaningful mathematical series in it so long as you know where to look. I might try and find some expression of Pi in there as a challenge.

Apr 04, 2013 at 12:35PM EDT

You also get the Sierpinski’s Triangle fractal if you shade the odd/even numbers differently on Pascal’s Triangle.

Sadly to say I have no idea why. I wonder what other patterns one can create with Pascal’s triangle.

Last edited Apr 04, 2013 at 05:43PM EDT
Apr 04, 2013 at 05:42PM EDT

KI....D wrote:

You also get the Sierpinski’s Triangle fractal if you shade the odd/even numbers differently on Pascal’s Triangle.

Sadly to say I have no idea why. I wonder what other patterns one can create with Pascal’s triangle.

It has to do with number theory.
ODD + ODD = EVEN
EVEN + EVEN = EVEN
ODD + EVEN = ODD

The outsize frame will always be odd because it is the addition of 1 + NULL, which produces another 1. Interior structures are the only ones that can be EVEN, but are always reductive in nature because they either touch a frame 1, or are bordered by a number that was produced with an ODD number. Because the addition structure favors EVEN with a 2-to-1 ideal, it is impossible for ODD numbers to form any structure other than skeletal. Add in the reductive nature of EVEN, and the way that they finish their lower points within a single line, the only shape they can make is triangular (a reversed image of the primary structure as a whole).

Did that make sense?

Last edited Apr 04, 2013 at 06:18PM EDT
Apr 04, 2013 at 06:16PM EDT

Sonata Dusk wrote:

It has to do with number theory.
ODD + ODD = EVEN
EVEN + EVEN = EVEN
ODD + EVEN = ODD

The outsize frame will always be odd because it is the addition of 1 + NULL, which produces another 1. Interior structures are the only ones that can be EVEN, but are always reductive in nature because they either touch a frame 1, or are bordered by a number that was produced with an ODD number. Because the addition structure favors EVEN with a 2-to-1 ideal, it is impossible for ODD numbers to form any structure other than skeletal. Add in the reductive nature of EVEN, and the way that they finish their lower points within a single line, the only shape they can make is triangular (a reversed image of the primary structure as a whole).

Did that make sense?

I think so. Thank you.

Apr 04, 2013 at 07:18PM EDT

Here’s a fun one. Euler’s formula:

Which then leads to Euler’s identity:

Such an interesting equation. Irrational number to an imaginary irrational product equals -1.

Apr 05, 2013 at 10:23AM EDT

KI....D wrote:

You also get the Sierpinski’s Triangle fractal if you shade the odd/even numbers differently on Pascal’s Triangle.

Sadly to say I have no idea why. I wonder what other patterns one can create with Pascal’s triangle.

Color-coding the triangle based on different moduli will yield a variety of patterns:

Mod 3

Mod 4

Mod 5

The Stare Master wrote:

Here’s a fun one. Euler’s formula:

Which then leads to Euler’s identity:

Such an interesting equation. Irrational number to an imaginary irrational product equals -1.

Ah! That reminds me:

Thoughts, anyone? (I did find a series in Pascal’s triangle that converged on Pi/8 as I said I would try above, but it wasn’t as interesting as I hoped for, IMO.)

Last edited Apr 05, 2013 at 02:15PM EDT
Apr 05, 2013 at 02:15PM EDT

Here’s a question for ya’ll it’s applying maths to physics at the same time, if the universe is everything, must it be infinite? If it is not, can you ‘fall off’ or ‘bump into’ the edges and if it is, how can it be expanding because surely if all the space in the universe is there, there is no space for it to expand into.
This is all purely theoretical (no-one really knows the answers) and I have my ideas on it myself (shown below) but it would be interesting to see some opinions and thoughts from others.
MY IDEAS
First of all, what we call ‘the universe’ isn’t quite that at all, it’s the problem with physicists romanticizing their study to make it seem more exciting. What you think is the universe isn’t, it’s just the observable universe; the one that we know is there because we can see it (this is literally how they do it, they have a look). Why can we only see some of it? Because light © travels at 50,00 mp/s and it takes time to reach us, not all of it has reached us. So that deals with the expanding bit, it has room to expand because it’s only our universe that is doing so (this is a problem for big bang hardcore supporters because if only one expanded in such a way, what were the others doing? They could be doing the same but that messes up the space to expand thing). So is the universe infinite, not necessarily because to prevent it having edges it must have infinite edges and what shape has that? A sphere, the universe could well be spherical and we can work it out (I can’t personally, I don’t have the right facilities) but what makes it spherical, it owns gravity? Would that mean all of space-times is bent? For now we must be content to love the questions themselves.

Apr 05, 2013 at 05:52PM EDT

eloquent atheist wrote:

Here’s a question for ya’ll it’s applying maths to physics at the same time, if the universe is everything, must it be infinite? If it is not, can you ‘fall off’ or ‘bump into’ the edges and if it is, how can it be expanding because surely if all the space in the universe is there, there is no space for it to expand into.
This is all purely theoretical (no-one really knows the answers) and I have my ideas on it myself (shown below) but it would be interesting to see some opinions and thoughts from others.
MY IDEAS
First of all, what we call ‘the universe’ isn’t quite that at all, it’s the problem with physicists romanticizing their study to make it seem more exciting. What you think is the universe isn’t, it’s just the observable universe; the one that we know is there because we can see it (this is literally how they do it, they have a look). Why can we only see some of it? Because light © travels at 50,00 mp/s and it takes time to reach us, not all of it has reached us. So that deals with the expanding bit, it has room to expand because it’s only our universe that is doing so (this is a problem for big bang hardcore supporters because if only one expanded in such a way, what were the others doing? They could be doing the same but that messes up the space to expand thing). So is the universe infinite, not necessarily because to prevent it having edges it must have infinite edges and what shape has that? A sphere, the universe could well be spherical and we can work it out (I can’t personally, I don’t have the right facilities) but what makes it spherical, it owns gravity? Would that mean all of space-times is bent? For now we must be content to love the questions themselves.

Actually, we pretty much know the universe is expanding because we can directly observe all the large systems of matter drifting apart from each other. But of course there’s certainly much more to it than that. I recommend you watch this. The show’s fucking amazing, I tell ya.

Oh, by the way, that’s not quite math. It’s a fascinating topic, but it probably doesn’t belong here. One of us could create a “Theoretical Physics General”, though.

Apr 05, 2013 at 06:30PM EDT

Actually, while these sorts of issues are questions for physics, they’re the sort which understanding them depends very highly on mathematical principles. If I recall correctly, Einstein’s theory of relativity was based on recent discoveries in physics that showed the speed of light to be a constant. Starting with that assumption, Einstein simply expanded on current knowledge of physics using mathematical calculations. Special Relativity, for instance, actually can be understood with high school algebra; it’s Special Relativity that requires higher-level maths.

However, eloquent atheist, let me address your question using some simple mathematical principles. You’re probably familiar with the number line? This is the line that goes from zero through one and on to infinity (which just a short way of saying it doesn’t come to an end), and through negative one on to negative infinity (see previous parenthetical). Now, ignoring complex numbers for simplicity’s sake, one can say that the number line contains every individual number that ever was used or ever will be used, and thus is a sort of one-dimensional “universe” of numbers. Yet, I can assure you that “every individual number that ever was used or ever will be used” is, although unimaginably large, a finite set. Thus the number line is an infinite one-dimensional space that contains a very large, but finite set.

Similarly, I believe that the universe has (or may have; special relativity throws us a curve here, literally) infinite space, but that infinite space is filled by a finite amount of matter. So, when one talks about the “edge of the universe”, there needs to be a clarification of what is meant. If you’re talking about the edge of the “known universe” or the edge of this supposed finite, ever-expanding cloud of matter that we sometimes call the universe, what is beyond that is simply more empty space. If you are talking about the edge of space itself, it doesn’t have an “edge”, at least not in any meaningful way.

Apr 05, 2013 at 08:12PM EDT

Hmmm… so Ogreenworld suggested “recreational math”, suggesting math riddles and/or interesting math-based paradoxes and the like; maybe I could try to get this thread restarted with one of each.

1) Three men are sharing a fancy hotel room, and the manager charges them \$300. The bellhop sees the men up to their room and comes back to the front desk, where he sees the manager looking upset. The manager tells the bellhop that he feels he really overcharged the men for their room, and wants to refund them \$50. He gives the money to the bellhop and tells him to run it up to the room. On the way up, the bellhop starts thinking about the situation. Any refund will please the guests, and in the end, wouldn’t a \$30 refund be easier for the men to divide among themselves? When he gets to the room, he gives the men \$30 and walks away having pocketed \$20. So here’s the question: The men have now paid \$90 each for the room. This adds up to \$270, plus the \$20 the bellhop takes: \$290. Where did \$10 disappear to?

2) Not really a riddle, but there is an arguably fancier hotel which, as it happens, is fancy because of a special feature: it has an infinite number of rooms. One night, the hotel has managed to be completely booked, and every room has an occupant, but a man shows up that evening looking for a place to stay. What does the manager do? He goes to room #1 and tells the guest there that he needs to move to room #2, and tell the guest in room #2 to move to room #3 and carry on the message, etc. The new guest is now moved into room #1, and once again, everyone has a room.

If you do want a puzzle, there are more complicated situations for the manager to deal with:

A) Suppose a bus with infinite seats, all occupied, pulls up to the hotel and the driver asks for rooms for each of his passengers. What can the manager do?

B) The next night, the manager is greeted by the sight of an infinite number of similar buses pulling up to the hotel. What now?

Anyone interested should try and solve these without Googling the answers, which I will save you the trouble of doing.

Anybody else have any good math puzzles, or thoughts on the “Pi is (still) wrong” video?

Apr 07, 2013 at 03:37PM EDT

1)

Well, the three men are paying for both the hotel’s expenses and the bellhop’s greed.
Together, the hotel has \$250 and the bellhop has \$20, so in total they have \$270.
The men have paid \$90 each, so in total, they paid \$270. So clearly no money has actually disappeared.

I’m so confused as to why you’d add the money that you’ve spent to the money that was given to the bellhop; it would seem that the proper way to account for totals is to consider the money given to the bellhop as part of your expenses. I feel like there is no missing \$10, but that instead of adding \$20 dollars, you should be accounting for the money in the way i did above.

2A)

Move the person in room 1 to room 2, person in room 2 to room 4, person in room 3 to room 6, person in room 4 to room 8, and so on…
{1,2,3,4,5,…} -> {2,4,6,8,10…}
All the odd rooms have now opened up.

2B)

Well shit.

Here’s one way i think.
Move the person in room 1 to room 2, person in room 2 to room 3, person in room 3 to room 5, person in room 4 to room 7, and so on…
{1,2,3,4,5,6,7…} -> {2,3,5,7,11,13,17…} (prime numbers).
Now all rooms of the form 2^k for k > 1 have opened up for the first bus, all rooms of the form 3^k (k>1)have opened up for the second bus, all rooms of the form 5^k (k>1) opened up for the third bus and so on…

I don’t see why pi and tau can’t both be used (which in engineering it is) :<
There are a lot of cases where tau is more revealing about the properties of equations (like the even answers of the zeta function or the derivation of the formulae for areas/surfaces/volumes of sphere), but there are a couple where pi is slightly more convenient though, like fourier series.

I mean, we use both the golden ratio and it’s conjugate using different symbols, i don’t really see why pi and tau should be any different ¯\(-_o)/¯
I also wish we were raised using base 12, had a standard for what the natural numbers are, had a satisfactory reason as to why we shift the gamma function down, and had a better symbol for the legendre symbol… more than using tau >_>

…quadratic reciprocity, my butt, that’s no reciprocal >:T

Here’s a rather easy puzzle:

Slice the T below with 3 straight line cuts to give three triangles and two squares with no left over pieces.

And here’s a harder and more interesting one:
Determine, without numerical calculations, which of these is larger:

or

Last edited Apr 07, 2013 at 06:27PM EDT
Apr 07, 2013 at 05:51PM EDT

@Ogreenworld

Your solutions to the problems were fine, so I really have nothing to add there.

@Eloquent Atheist
You wrote:

First of all, what we call ‘the universe’ isn’t quite that at all, it’s the problem with physicists romanticizing their study to make it seem more exciting.

I… I just..
You haven’t studied physics much beyond high school or basic college, have you?

You have.. several.. things wrong.
I have no idea what unit you’re trying to use (milliparsec? what?).
The reason we know our universe is expanding is because we’ve observed the most distant galaxies become increasingly redshifted: closer galaxies are redshifting slowly, and ones further away are redshifting more quickly.
The Universe is not actually expanding into anything. What is happening is the space between matter is expanding. I’d do the whole raisin loaf thing, but I hate that explanation. Because the universe is its own limit, when the fabric of it expands, so does the universe itself.

I can’t understand where you’re going with everything else. You start with a half comment on the big bang, then suddenly jump to the multi-universe idea, then to.. universal shape? I really don’t know what you’re trying to say.

0.9999…=1 is right, however. This is a place more for mathematics and not physics. That post just grated at my nerves, though.

Last edited Apr 07, 2013 at 07:10PM EDT
Apr 07, 2013 at 06:58PM EDT

@Ogreenworld:
Your answer to #1 is correct, of course. There’s really no mathematical conundrum, but rather the question is phrased in what is really a nonsensical manner. Your solution to 2A is the usual classic solution, although certainly others exist; your solution to 2B is not only clever and straightforward, but interestingly leaves more empty rooms than occupied ones!

I’m curious as to why you think base 12 would be a preferable system. Base 60 seems very nice, assuming one could find an easy notation system more elegant than the Sumerians’. Many people seem to have ideas as to which numbers would serve as good bases, and they usually choose a composite number, but I wonder if there might be some utility in a prime number base, such as 5 or 13. (I’m fond of 13 because my senior “thesis” was on gambling theory with an emphasis on card games.)

Regarding the Legendre symbol, my memory of quadratic reciprocity is unfortunately very fuzzy after all these years, but I do recall thinking that there just came a time in math when we seemed to run out of clear and meaningful methods of notation. For instance, there seemed to be a half-dozen things that “(x,y)” could mean, so you always had to pay close attention to context.

I actually found your second puzzle easier, as it happens. The first quantity is larger because for sufficiently large numbers, if a > b then a b < b a . Which reminds me of a funny story. I was a grading assistant in a math class which was on the subject of using certain programming languages for mathematics, and on one assignment, there was the above statement phrased as a question: “For sufficiently large numbers a and b where a > b, which will be larger…” One student plugged in the values a = 3 and b = 2, calculated the results, and gave the incorrect answer. I remarked on his homework, “So you consider 2 and 3 to be ‘large’ numbers? Exactly how long have you been doing mathematics?”

@Crazy☾:
Sure, eloquent atheist was fairly off-base, but I thought it made for interesting discussion nonetheless. Actually, since I brought up mathematics and relativity theory, there are some interesting questions that relativity equations bring up if you look at them purely mathematically. Perhaps in a purely physical sense, one cannot travel faster than light speed, but velocities greater than c can be plugged into various equations and solved nonetheless.

Apr 08, 2013 at 04:49AM EDT

Ehh? Really?
I was looking at 1), seriously wondering what was wrong with my logic, and you’re telling me it’s just a terrible question? `@_@` araegaerg

I figure that base 12 would be best since it’s only two off from our current base, and i’m… pretty sure it’s better than 10 in most respects because of it’s divisors :o
There’s always the fact that our weights and measurements were in base 12 (12 inches per foot, 12 months), and for good reason. 60 is also nice, but 60 symbols is not so fun to deal with…

I don’t think most people are too fond of prime number bases (the consideration of base 11 as a compromise between base 10 and 12 was apparently a popular joke. ohoho). The two exceptions that come up are usually base 2 (for binary and irrational bases) and base 3 (balanced ternary). Base 3 is also particularly interesting for having the lowest radix economy (well for integers, anyway).

Also, that’s really unfortunate. Especially with computers and latex i suspect it’ll only be harder to implement new notations. I’ve always wished mathematicians would be more diverse in their letters too; maybe use some letters from foreign languages instead of greek letters? :o

Hm, the a^b vs b^a problem was something i found in training tests for the putnam exam. I suppose if you knew that, it’s pretty easy, but proving it is unintuitive since at surface level, it seems to be a question regarding the particular choice of a and b, but really it just matters that a < b. :S
(also, now i’m really curious what the heck has to do with the choice of programming languages for mathematicians).

Meh, I suppose this thread is not very full of people, so if this thread must die, so be it u_u;

Nevertheless, I’ll just post one more puzzle, in case anyone wants to figure it out.

Prepare to witness the most amazing card trick in the world! Here I have a standard deck of cards, completely shuffled. Go ahead, pick any 5 cards you want. Any of them! But don’t show them to me, give them instead to my lovely assistant, who will show me 4 of the 5 cards, one at a time: the 7♠, then the Q♥, the 8♣, the 3♦!

There’s only one card left, known only to you and my assistant! Or is it? The hidden card, my friend, is the K♠!

How can I know the last card? Keep in mind there is no illicit communication between the assistant and I, other than the four cards shown. All I have is a sequence of four of the five cards you chose, and I can name the fifth one.

Last edited Apr 09, 2013 at 01:39PM EDT
Apr 09, 2013 at 02:01AM EDT

Oh boy, we’re taking about bases now? I typed up a big-ass post a couple days ago on the subject, but right towards the end my iPad decided that it just wants to watch the world burn. So I’ll sum it up quick: Basically, a graph with positive integers in base 10 (or decimal) as x and their representations in base m ( where m equals 2/3/4/…/9) as y has a fractal pattern. This is because, for every set of integers from 0 to ((10^i(nteger))-1) in base 10, there are (10-m)*(10^(i-1)) unique ones that are removed with the conversion into base m. This is self-similarity, which practically defines fractals. If you don’t believe me, do this exercise yourself.
On a side note, base e has the true lowest radix economy. Wouldn’t it be fascinating if that became the standard notation of the future? I’ll do some conversions myself later to see how it looks.

Apr 09, 2013 at 02:33AM EDT

1 base ∞

Wait…for…it…

There we go.

Edit: Oh gawd! What have I done! I swear I just hit post once and it double posted! I broke KYM with MATH!!!

Last edited Apr 09, 2013 at 11:14AM EDT
Apr 09, 2013 at 11:12AM EDT

The Cute Master :3 wrote:

1 base ∞

Wait…for…it…

There we go.

Edit: Oh gawd! What have I done! I swear I just hit post once and it double posted! I broke KYM with MATH!!!

Actually, one in base ∞ would just be one, because 1*x^o=1 for ALL values of x by definition. So no, math and KYM are just fine.

Apr 09, 2013 at 01:40PM EDT

@Ogreenworld:
A lot of math riddles are actually based on misdirection, the more subtle the misdirection, the tougher the riddle. I was looking at this one the other day, and I had to read it something like four times before I figured out the problem:

-1 = -1
Then, if I divide both sides by 1, I get
-1/1 = -1/1
Now, we know that -x/y = -(x/y) = (-x)/(y) = (x)/(-y). It doesn’t matter where you put the minus sign. So, from that we get
-1/1 = 1/-1
And, if we take the square root of both sides, we get
root(-1/1) = root(1/-1)
But we can split the square roots out, so
root(-1) / root(1) = root(1) / root(-1)
Now, we can cross multiply (to get rid of the fractions), and get
root(-1) * root(-1) = root(1) * root(1)
But surely root(x) * root(x) = x. That’s the definition of root(x), so
-1 = root(-1) * root(-1) = root(1) * root(1) = 1
Which leaves us with
-1 = 1
Right?

Ogreenworld wrote:

I figure that base 12 would be best since it’s only two off from our current base, and i’m… pretty sure it’s better than 10 in most respects because of it’s divisors :o

Arrrrgh… Just because this is a math thread doesn’t mean you can throw grammar out the window. This sentence was painful to read.

Anyway… Your example of 12 inches to a foot is arbitrary, isn’t it? There are a few of things that are naturally 12s, such as months (although it’s really closer to my favored 13) and the number of edges on either a cube or octahedron. Still, I’m sure some people would like a base 12 system.

(I can’t believe I forgot that television was the first to teach me duodecimal!) Actually, I suppose 12 does have a lot of things going for it both mathematically (being a highly composite number and superfactorial, mostly) and culturally (12 months, 12 zodiac signs in both the western and eastern traditions, and 12 hours on a standard clock face). The latter may actually be of particular significance, since apparently there was once an attempt to apply the metric system to time, and make the day 20 hours with 100 minutes each, but nobody could accept it. It might actually be easier to get people to accept a change in mathematics than in fundamental culture.

The relationship between a^b and b^a might be another interesting concept to look into, but it has nothing in particular to do with programming languages, that just happened to be the subject of the class. Perhaps the idea behind posing the question was that it would be easy to check a variety of numbers quickly and/or make a graph of some sort.

Your card trick puzzle is a complete stumper. I have no idea what relationship exists between those cards whatsoever.

@0.9999…=1:
I couldn’t figure out what the heck you meant by your graph description at first, then after fiddling around, I came up with this:

Is that the sort of thing you meant?

Oh, and I’m sure The Stare Master either meant base zero…or that’s actually a sideways eight, which would imply who knows what.

Apr 10, 2013 at 01:02PM EDT
-1 = 1

Hm, I think the error’s in √(-1/1) = √(1/-1) --> √(-1)/√(1) = √(1)/√(-1). Everything before that point is consistent, but √(-1)/√(1) = √(1)/√(-1) is already inconsistent:
√(-1)/√(1) = √(1)/√(-1)
i / 1 = 1 / i
i = -i

However, normally you can split radicals √(xy) into √(x)√(y) unless both x and y are negative; not sure why it’d be different for division, but I can’t see anything else wrong with the “proof.”

Arrrrgh… Just because this is a math thread doesn’t mean you can throw grammar out the window. This sentence was painful to read.

come at me bro

Your card trick puzzle is a complete stumper. I have no idea what relationship exists between those cards whatsoever.

Hm. I probably phrased the question poorly. The point is that the assistant somehow needs to use the 4 cards to “pass a secret message” to the magician, so he can figure out the 5th one unambiguously.

The naive way is to take 4 of the cards, order them biggest to smallest (or if they’re the same suit, decide on which suits will be bigger and which will be smaller). You can present the smallest, the second smallest, the second biggest, and the biggest in 4! = 24 ways.
This alone is, of course, not enough information to determine the 5th card.

It’s sufficient to find some sort of scheme in which you can use 4 of the 5 cards in a way such that the magician can guess the 5th.

Re: bases
My friend and I were arguing if unary was a base or not. I argued that it has none of the property of the other (integer) bases:
-You can’t calculate the number of digits using floor(log_b(n))+1, where b is the base (else you get a singularity).
-All other bases can only use digits 1 through (b-1), or least nonnegative residues mod(b).

Although it seems that ∞ is not a base, though, it acts more like a base than unary.
-If you think about it intuitively, base ∞ would just have a different symbol for every integer. So base ∞ can only use digits from 1 through ∞, which makes sense.
-n mod(∞) would be the remainder r after dividing n by ∞, which is r = n. So the “last digit” of n base ∞ would be a symbol denoting n.
-floor(log_∞(n))+1 = ??
Well a^2 = a*a, a^1 = a, a^0 = a/a.
This is, of course, not defined for ∞/∞, because ∞/∞ can equal anything. ∞^0 = n, for any n it wants to. But ∞ raised to any other power creates a singularity or zero. So unless n = 0, log_∞(n) is probably unambiguously 0.
So floor(log_∞(n)) + 1 = 0 + 1 = 1.

That’s how i see it anyway…
…although dealing with infinities means I definitely made some recklessly irresponsible manipulations with equations.

edit

Actually, one in base ∞ would just be one, because 1*x^o=1 for ALL values of x by definition. So no, math and KYM are just fine.

Not sure if this gives my conclusions any validity, but Dr. Math seems to agree that ∞^0 is undefined…

So you’re probably right, but for a dubious reason.

Last edited Apr 10, 2013 at 11:42PM EDT
Apr 10, 2013 at 11:20PM EDT

Unary is the same as tallying, AKA |||| for the number four. It’s a non-positional notation system, so in a way you’re right, but one is still acceptable as a base.
On another topic, the function f(x)=x^x has fascinated me for quite a while- it’s so simple and elegant, and yet I can hardly find any information on it and others like it. Does anyone have something? Specifically, a way to calculate it’s inverse would be great. (And yes, I know it can’t have a true inverse function because it’s not one-to-one.)

Apr 11, 2013 at 12:32AM EDT

Gosh darn it, Ogreenworld, I was writing a perfectly good reply yesterday, and then somehow you made me start thinking about your card trick puzzle, and instead of being “completely stumped”, I would say I’m now “annoyingly perplexed”, as I’ve analyzed the crap out of this puzzle for hours without finding the answer, but finding various coding methods that are almost enough to give the answer! Without detailing everything I’ve found in my analysis, here are a few observations:

1) A deck of cards has 52 different cards, so as you said, 4! = 24 ways of arranging your four displayed cards is not enough to cover the deck by nearly half.

2) If you assume the first card is in some way a key card, that can narrow the field of cards needed to consider, but then you lose the numerical expressiveness of the remaining three cards to 3! = 6, which will not be enough because you can’t assume any two of your five cards will be within a numerical distance of six cards in any method of reckoning. There is a sense in which the choice of first card means you have five six-card ranges to choose from, but you can’t guarantee that your fifth card will be among those 30 cards

3) I could be wrong, but I don’t think you have the ability to send any secondary message based on incidental properties of the cards, such as the sum of the cards being odd/even, the number of different suits displayed, etc., because as they say, “you play the cards you are dealt,” and even with choice of four out of five, there is limited control of these sorts of things. The only meaningful surface information you can invariably get from the four cards shown is that the fifth card is not one of those four.

4) Finally writing that out of course revealed the answer to me as I somehow suspected would be the case. Since I wasn’t 100% sure about #3, I checked it, and found I was indeed wrong. Since I was wrong about that, I reexamined #2 thinking about the assumptions I had made about #3, and realized I was wrong there as well. Since I was wrong about #2, that meant I had the solution, which also meant I was in a sense wrong about #1, I suppose.

That probably didn’t make any sense, but your puzzle was driving me nuts. I’ll put my solution in another post under a spoiler tag.

Anyway, you were right about the erroneous step in the proof. Apparently √(-x) is meaningless, but mathematics allows one to change √(-x) into i√(x) and proceed. This essentially removes the ambiguity of √(-1), which has two equivalent imaginary solutions. Of course, a step or two down, you could just as easily highlight that ambiguity by pointing out √(-1) * √(-1) = i * -i = 1 is just as valid.

The phrase “just have a different symbol for every integer” seems to me to be meaningless. Either you assign arbitrary meaning to a finite cardinality of symbols, or you hold an infinite amount of information in your head in order to decode an infinite cardinality of symbols. If you find some recursive method of creating symbols that have non-arbitrary meaning, then you’re either working with a base, or using some form of unary in essence, which you previously denied as being a base. If I’m wrong about this, then please explain why. Most of the rest of what you’re saying about “base ∞” is making the mistake of treating the symbol ∞ as if it stood for an actual number. Despite L’Hospital, ∞/∞ can’t “equal anything”; it is not numerically meaningful in itself. Indeed, you “definitely made some recklessly irresponsible manipulations with equations.”

@0.9999…=1

I did some Google and Wikipedia searches on x x but didn’t find anything useful. You’re right that it’s oddly mysterious for such a simple function. You’d think there would be an inverse for x > e -1 but I couldn’t find anything.

EDIT: And once again, the act of posting seems to have led to me finding something immediately afterwards. You could check out Wikipedia’s info on the square super-root or the Lambert W function.

Last edited Apr 12, 2013 at 06:21PM EDT
Apr 12, 2013 at 06:10PM EDT

Okay, seems to be dying for real, but as promised:

Since you are taking five cards out of the deck, at least two of them have to be the same suit. So your assistant chooses two cards out of the five with the same suit, making one of them the card to be guessed, and the other the first card revealed.

The two cards have to be chosen so that the card to be guessed is somewhere within the next six cards of the disclosed card by rank, with the Ace being the one and the thirteen cards of that suit acting like mod 13 arithmetic.

In your example, the 7♠ was the first card shown, indicating that the last hidden card will be in the set {8♠, 9♠, 10♠, J♠, Q♠, K♠}. (If your assistant had shown the 10♥, the hidden card would be in the set {J♥, Q♥, K♥, A♥, 2♥, 3♥}.)

The suits of the remaining cards are essentially irrelevant (beyond the fact that you’d need to rank the suits in case your remaining three cards contained a pair) and what matters is the order of the cards by rank. If the cards in order by rank were (A B C) it would be simplest to assign values to their permutations as follows:

(A B C) = 1st card
(A C B) = 2nd card
(B A C) = 3rd card
(B C A) = 4th card
(C A B) = 5th card
(C B A) = 6th card

When your assistant shows Q, 8, 3, then that corresponds to the sixth card, in this case, the K♠! It’s actually easier to do than to explain, I’d say.

Something I really like about this is that it displays an interesting property of the structure of a deck of cards. A four-suit, 13-rank deck of 52 cards is perfectly suited for this trick. Any change in those numbers would imply a need for a completely different structure of the trick. Five cards gives you a guaranteed card to identify one out of four suits, and the remaining three give you 3! = 6 possibilities which is perfect for 13 ranks. If there were five suits, you’d need six cards, which would leave you with 4! = 24 which is the optimal number for picking from 49 ranks, and therefore overkill. If there were 14 ranks, then 3! wouldn’t suffice.

All of which reminds me of some old classic math puzzles that over the years, I’ve revised into what I call “Brucker sock math”.

Puzzle 1: Brucker has a drawer full of socks in which there are five pairs of black socks, three pairs of brown socks, and five pairs of blue socks, all piled loosely. He gets up for work before sunrise, and since he doesn’t want to turn on the light and wake his wife, he decides to pick out in the dark just enough socks to ensure he has a matched pair of some color. What is the minimum number of socks he will need to grab to guarantee this?

Four. Once he has picked out three socks, he might have one of each color, but the fourth sock will have to match one of the first three.

Five. If Brucker picks out only four socks, once he examines them in the light, he will discover one of them is a white sock the existence of which he had been unaware.

Puzzle 2: Brucker, who has managed to purge his sock collection of all but black socks, has two piles of socks: a clean pile and a dirty pile. Upon examination, the clean pile has an odd number of socks. The dirty pile is also found to have an odd number of socks. Brucker takes all his dirty socks, (along with other dirty clothing items) runs them through the washer and dryer, and puts all of his now-clean socks together in his sock drawer. Will he have an even number of socks, or an odd number?

Even. An odd number plus an odd number is an even number.

Odd. Brucker has discovered through years of observation that any number of socks greater than two is always odd.

Apr 14, 2013 at 02:39AM EDT

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