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Logical solving problems! (brainy and tricky)

  1. Dec 16, 2011 #1
    1. The problem statement, all variables and given/known data

    1.During a certain movie, each person of a set of five audience members fell asleep exactly twice. For each pair of these people, there was some moment when both were sleeping simultaneously. Is it true that, at some moment, three of them must have been sleeping simultaneously?

    2. Given a set of two million points randomly placed on a piece of paper. Can you always draw a straight line on the paper so that on each side of the line there are exactly one million points?

    3. A group of friends play a round-robin chess tournament, which means that everyone plays a game with everyone else exactly once. The tournament rules do not allow draws. Is it always possible to line up the players in such a way that the first player beats the second, who beats the third, etc.? Would this way of lining up the players be unique?

    4. At the beginning of the morning the conference room in a large hall stands empty. Each minute, either one person enters or two people leave. The guestbook for the conference room shows that after exactly 31999 minutes, the room contained exactly 31000 + 2 people. Is the guestbook record accurate?

    5. An alien lock has 16 keys arranged in a 4 × 4 grid, each key is either pointing horizontally or vertically. In order to open the lock, you must make sure that all the keys must be vertically oriented, by switching the orientation of one key at a time. When a key is switched to another position, all the other keys in the same row and column automatically switch their orientations too (i.e. vertical to horizontal, horizontal to vertical). Is it true that no matter what the initial positions of the 16 keys are, it is always possible to open this lock?

    3. The attempt at a solution
    1. I don't get number 1 but what I think is this problem is relevant to Eulerian path/circuit! I think that we can call 5 people A, B, C, D, E and then connect them like a star shape. After that, we can test the "Seven Bridges of Königsberg" theory.
    2. I think no because there's too much point and it's nearly impossible to estimate each side must be 1 million points
    3. I don't get how to set up the round-robin tournament and solve this problem :O
    4. I predict this question is relevant to Congruence relation (mod) but I'm not sure
    5. Can someone get this question?

    These are very brainy and tricky questions, and they aren't advance math or something. The way to solve them is your thinking skill? (my teacher said that). I hope someone can help me. Thank you very much
     
    Last edited: Dec 16, 2011
  2. jcsd
  3. Dec 16, 2011 #2
    Problem 2 is yes, there exists a line that can put the points in half either side...

    Because: We can do that for each of 2 million's factors:

    2 Million prime-factored = 2000000 = 2 * 1000000 = 2*2*500000 = 2*2*2*250000 = 2*2*2*2 etc...

    = 26 * 54

    And we know that 2 can be evenly divided by a line

    And we know that if we have 5 randomly placed points and the line goes through the middle point then (technically) we still have either 3 or 2 points on either side, the point is we have a line that evenly divides the points.

    Because 2,000,000 is even we don't have worry about 5's problem and since both are dividable down the middle 2,000,000 has to be dividable down the middle
     
  4. Dec 16, 2011 #3
    For Problem 4 (lets work it through)

    If 31999 minutes have passed then we could have had 31002 people come into the room and now work from there to figure out how many people have left.

    For 997 minutes (which is 31999-31002) we have to see if it is possible to add 1 or substract 2 per minute such that once the 997 minutes are done there should be NO net change in people (cuz that will go back to the our original number)

    So lets see In order for a round to break even we need to add 2 people and substract 2 so every 3 minutes 2 people must come in with the third move being that they move out.

    997/3 = 332 + 1/3 so we know 997/3 is not divisible by 3 and therefore it is not possible to break even in 997 minutes meaning that the guest book is clearly lying...
     
  5. Dec 16, 2011 #4
    For Problem 1: There MUST NOT be a time when 3 of them are sleeping simaltenously because they could all sleep at the same time twice in a row meaning more than 3 of them are sleeping simaltenously... Its a no-brainer

    That doesn't mean that there CAN be a situation when 3 of them are sleeping simaltaneously, Therefore we have not answered the question:

    Must there be ATLEAST 3 of them sleeping simaltaneously...

    Lets think if each person gets 2 rounds to sleep then in that 2 rounds they have to sleep at the same time as (4-k) people and on the next round sleep with k different people. In order to reduce K as small as possible would mean that they sleep with 3 other people the first/second round and 1 person the second/first round, if k = 2 then they sleep with 2 other people the first round and 2 others the second round meaning that at least 3 are still sleeping simaltaneously. if k > 2 then the first example goes back into effect so to answer your question:

    YES, their must be a moment when ATLEAST 3 people are sleeping simaltaneously although there can be more than three in other scenarios.
     
  6. Dec 16, 2011 #5

    SammyS

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    If 4 are sleeping simultaneously, then (at that time) it is also true that 3 are sleeping simultaneously.
     
  7. Dec 16, 2011 #6
    So for problem 5 it seems possible because it appears that orienting all the keys horizontally is possible from a particular situation, but vertical still seems unknown... It may be that different arrangements have a chirality (like even and odd) such that an arrangement can allow for an absolute orientation either horizontal or vertical but not the other, but then that seems impossible because changing from vertical to horizontal should be possible (it works in 1 x 1 arrangement, and a 2x2 arrangement) so (4 x 4) should also follow the same mechanics.
     
  8. Dec 16, 2011 #7
    Thank you very much guys! I really appreciate your help :)
     
  9. Dec 21, 2011 #8
    can anyone solve #5? Thank you
     
  10. Dec 21, 2011 #9
    (2) Given ANY even number of points scattered randomly, a line can be drawn that divides the group into two equal numbered portions.

    Proof: each pair of points defines a line. Drawing all such lines will not fill the plane - there will be places remote from the points that are not on any such line. Pick such a point, then project a line from that point off to one side of the group. Rotate the line across the group - it will pass through one point at a time and when it has passed through half the points, it is a line as required to solve the puzzle.
     
  11. Dec 22, 2011 #10

    LCKurtz

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    That is an example of a "Merlin Machine". It isn't trivial and you can read about it at:
    http://maths.dur.ac.uk/~dma0jf/JoynerRubikbook.pdf [Broken]
     
    Last edited by a moderator: May 5, 2017
  12. Dec 22, 2011 #11
    @LCKurtz, I'm guiding niwaone through #5 on another thread. This particular machine is solvable.

    An interesting if slightly intimidating link, thanks!
     
  13. Dec 22, 2011 #12

    Dick

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    These are indeed brainy thinking problems. They also seem to be a 2012 entry exam for St Mary's College http://www.google.com/url?sa=t&rct=...TuRtskKCcbXHdaouQ&sig2=Q-RsWl6-0l9fyHcBH5-DAw, where they are looking for brainy thinking attempts. I'm not sure niwaone is making a lot of progress given all of the hints, as there have been almost no contributions back, just requests for more details. I'd refrain from giving any complete solutions here. As most people seem to be doing. I've got a proof of the first one, but it seems kind of awkward and I'd like to know if there is something more elegant. But I'm not sure I should discuss it here.
     
    Last edited: Dec 22, 2011
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