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Posts posted by rocdocmac

  1. 3 hours ago, flamebirde said:
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    There are four remains: four corpses. Well worded!

    Four bodies is correct! "Remains" is used as a noun, so the wording in the OP was not a grammatical error or typo. If the question read "remain" (as a verb), then two left. Or possibly one as CaptainEd wrote. There was no shoot out, TimeSpaceLightForce!


  2. CaptainEd ...


    Spot-on with your most recent octahedral attempt and congrats for solving this one too!

    Are you going to try deciphering the dodecahedral case as well? It does take some time if done "by hand", no doubt, but certainly not a too difficult exercise!


  3. Spoiler

    A difference of two days (1 in between) is easy to explain ...

    At the the time the mother of the twins went into labor, she was traveling by boat, plane or ambulance and crossed (any) time zone in the process . The older twin was was born just before midnight on March 1st of a non-leap year, followed by the second just after midnight (5 minutes later) on 28 February after crossing the time zone. During a leap year the younger twin celebrates her birthday two days before her older twin.

    But, could there ever be a time-zone difference of more than 24 hours to make one twin celebrate three days after the first (i.e. two days in between according to the OP)? Unless traveling was in space during labor. Otherwise, the twins decided not to celebrate their birthdays on the same day!



    I am changing " (any) time zone" to the International Date Line since "(any) time zone" won't work!



    Crossing will be from West to East


  4. 35 minutes ago, CaptainEd said:

    Rocdocmac, we are clearly well matched: I had a misenumeration  and a miscount, and I’ll raise you my typographical error.


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    yes, the 8-cycles were duplicated, now I count 12 of them, for a total of 24 out of 3^8.

    and, yes, 21658347 is a typo for 21658743, which is a duplicate.



    CaptainEd got all three right! The probability of no collision for 8 ants starting at cube corners is 24/6561 = 8/2187.

    Does anyone wish to try a similar OP with ants on the vertices of a regular octahedron and (pentagonal) dodecahedron?


  5. Spoiler

    I'm happy with the first set of 12 possibilities (parallel faces) and the number of combinations (6561). Watch out, however, that there are repetitions in the second lot with the crossovers, e.g. 41238567 is essentially the same as 67412385. Also, the entry "21658347" is invalid!


    On 4/11/2018 at 11:25 AM, rocdocmac said:

    Sorry for the "icosagon"  in the title! Should read isocagon.

    Should have said "isocagon" in the OP is wrong - icosagon in the title needs no change. Double dyslexia!!

  6. Sorry for the "icosagon"  in the title! Should read isocagon. 'Twas a dyslexic moment!

    16 hours ago, CaptainEd said:

    tetrahedron: success only if no two ants arrive at same cell

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    AD: BE, FC
    AE: BF
    BA: FE
    CA: DF
    CD: BE
    only 6/81 cases succeed, by my count



    So far both correct (a) 1/524288 and (b) 2/27


  7. (a)    If an ant is placed on each vertex of a regular isocagon, what is the probability that there will be no encounter between any two ants if all ants start moving simultaneously and randomly (either clockwise or anti-clockwise) along the edge to the next vertex without changing direction?

    What are the chances of no encounter if there are …

    (b)   Four ants, each placed on the vertices of a tetrahedron?

    (c)    Eight ants, starting at the corners of a cube?

  8. 7 hours ago, ThunderCloud said:

    Hmmm, I'll go for the seemingly obvious…

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    8 ?


    I tend to agree with 8. If the problem is exponential and not linear, how come "9" is right in the middle between 6 and 12? After all, 12*1 = 12, 6*2 = 6, and 8*1.5 = 12, whereas 9*1.5 = 13.5!





    Sorry ... nobody said 9! Didn't read carefully!!!


    10 minutes ago, rocdocmac said:

    I tend to agree with 8. If the problem is exponential and not linear, how come "9" is right in the middle between 6 and 12? After all, 12*1 = 12, 6*2 = 6, and 8*1.5 = 12, whereas 9*1.5 = 13.5!



  9. On 3/23/2018 at 12:56 PM, harey said:

    I do not get what the 12.5 mm are there for. Do you mean after the begin of the first song?

    The needle goes down 12.5 mm to the very beginning of the first song, i.e. before the first song actually starts.


    Pay attention to both Bonanova's remarks on 2 February 2018. Otherwise you'll go off track!


  10. If a piece of rectangular paper is folded once only and a single straight cut is made in the following ways,

    how many pieces of paper will there be after the cut paper is unfolded in place?


    (1) A cut perpendicular to the (last) fold


    (2) A cut parallel to the (last) fold


    (3) A diagonal cut


    (4) A diagonal cut in the opposite direction


    (5) Both (1) and (2)


    (6) Both (3) and (4)


    (7) Both (5) and (6)



    The answers for cuts (1) to (7) are 2, 3, 3, 3, 6, 7, and 14, respectively.


    It is very difficult (rather practically impossible) to fold a piece of paper more than 6 or 7 times.


    But, suppose we have a sheet of paper of infinite size and of negligible thickness, what would the

    outcome be if such a chunk of paper is folded 21 times and then cut accordingly to the mentioned seven ways?

    Each new fold from the start of folding is at right angles to the previous one.


    The images below show the different cuts to be made. The single fold in the given example

    (or the final fold when multiple folds are involved) appears on the right-hand side (||) of each drawing.