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Puzzle in real analysis

Started by Luara. Last reply by Luara Apr 19. 4 Replies

1+2+3+4+5+... = -1/12

Started by Luara. Last reply by Luara Mar 4. 16 Replies

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Comment by Yannick on November 1, 2009 at 10:20pm
Bruce: fixed :P
Comment by Yannick on November 1, 2009 at 10:19pm
Ajarn: Well I don't know if you put the problem so that some one would post the solution, but I found the problem interesting so there it is (a partial solution).
It seems to me that the only solution is every table with 2^n, n positive integer seats (in the sense that in this case all people will have a different seat); but I haven't find a way to show that there aren't other solutions (I have to sleep now :P, and of course I don't know if I'm able to show this. I'm saying "only" because I simulated the problem, from 1 to 2^12 seats, with a computer)

So this is what I'm suggesting
In a table with N places numbered by 0, 1, ..., N-1. Let a_k, k = 1, ..., N , denote the place where the kth guy/girl will seat, with
a_1 = 0
a_k = (a_k-1 + k - 1 ) mod N, then

if N = 2^p , p non-negative integer, then a_k != a_j for every j,k = 1,...,N ( j!= k)

The general term of the recurrence solution is a_k = (k/2)(k-1) mod N
This means that now we have to show that the following equality doesn't hold
k²/2 - k/2 - aN = j²/2 - j/2 - bN , where a and b are positive integers
(here I'm assuming, without loss of generality, k > j)
(k - j)(k + j) - (k - 1) = (a-b)2N
(k - j)(k + j - 1) = (a-b)2N
since N = 2^p
(k - j)(k + j - 1) = (a - b)*2^(p + 1)

1. k < N and j < N-1
if k - j is even, then k-j < 2N = 2^(p + 1) and since k+j-1 is odd then the equality doesn't hold
if k + j - 1 is even, then k+j-1< 2(N - 1) = 2^p and since k-j is odd then the equality doesn't hold

2. k = N and j = N-1
(N - N + 1)(N + N - 1 - 1) = 2(N-1) , and again the equality doesn't hold

So we have that a_k cannot be equal to a_j, if j is different than k and N is a power of 2.

Well I'm really tired now, I hope that I haven't written any nonsense (excluding my lousy english)
Comment by Ajarn Robert on October 5, 2009 at 7:54am
Hello all. Here's a lovely problem I worked on for a while:

You are seating people n people at a circular table with n chairs in the following manner: You choose a seat for the first person, the next person is in the next seat clockwise, the third person you again go clockwise, but skip one seat. The fourth person you skip two seats (always clockwise), the fifth person you skip three seats, the sixth four seats, and so on.

In some cases, each person will be in a unique seat (n=1 or 2 are examples) and all seats will be filled, in other cases people will end up in each other's laps, and some seats will be empty (n=3 where person 1 and 3 share the first seat chosen, and the third clockwise seat is empty).

The question is, for which n are all the seats filled and for which n are there some people sitting in each other's laps with some seats empty?
Comment by Kevin on January 16, 2009 at 8:09pm
Evilgenius20, I love that cartoon. It certainly reminds me of my high school maths teacher but not me... umm no, definitely not me!
Comment by Cory on January 12, 2009 at 5:21pm
I do love the image. Like so many of us so many times, this poor chap has written emself into a corner.
As i view it, ey refuses to turn around until ey's sure the direction ey's going is fruitless. Is it this skeptic ability to eventually turn around that allows us to abandon religions? (I wouldn't know.)
Comment by Evilgenius20 on September 6, 2008 at 8:25am

Comment by Parallax on August 3, 2008 at 5:05pm
I feel wlecome!
 

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