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rocdocmac

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Posts posted by rocdocmac

  8. What Question Must Be Asked?

    in New Logic/Math Puzzles

    Posted

    Spoiler

    What will happen if either Carroll or Kurt has 4 or 6 coins in their cell when Kleene's question on the "total  = 15" is NO? One can still arrive at a total 9 or 21. Carroll and Kurt don't know that Kleene has 5.

    One can possibly figure out a question to pose wrt the "total number of coins", but what if the OP stated "number of coins in each cell"? Then the answer is obviously 9 or 21 since no one would be able to figure out how to get to a total of 15 (four cases). So I think that, in such an instance, one should merely ask ... "Does the sum of the coins have an integer square root?" YES would mean a total of 9, NO would mean a total of 21.

    Sorry for side tracking!

     

  10. Cubicle Stack #2

    in New Logic/Math Puzzles

    Posted · Edited by rocdocmac
    Moved parts to spoiler

    @Thalia ...

    Done some counting, but still a lot to do! Nine combinations of 4-cubelet removal still to be done.

    Spoiler

    I found 190 so far, whereas you got 191, i.e. if we counted the same groups.

    Here's my list for checking on your side (*incomplete) ...  

     image.png.0e15536912b433fdefc021f994290cec.png

     

  11. 3 Quickies

    in New Logic/Math Puzzles

    Posted · Edited by rocdocmac
    Explanation refined

    Spoiler

    #1 goes to moderator @bonanova!

    7/12 = 0.58333 ... (whether marbles/counters/balls/whatever were put in the bag).

     

    Another way to calculate the answer using conditional probability (Thanks to Christian B):

    The probabilities that we have (1,2,3,4) white counters in the bag are

    (1/8,3/8,3/8,1/8)

    Given that we have (1,2,3,4) white counters in the bag, the probabilities that we draw two white counters are

    (0,1/6,1/2,1)

    The overall probability P(2W) to draw two white counters is given by

    P(2W) = 1/8*0+3/8*1/6+3/8*1/2+1/8*1 = 3/8

    The probability that all three counters are white results in

    P(3W) = 1/8*0+3/8*0+3/8*1/4+1/8*1 = 7/32

    Thus, the conditional probability P(3W|2W) is given by

    P(3W|2W) = P(3W∧2W)/P(2W) = P(3W)/P(2W) = 7/12 = 0.5833

     

     

     

     

     

  14. 3 Quickies

    in New Logic/Math Puzzles

    Posted · Edited by rocdocmac
    Expanded response

    Spoiler

    (1) There's more to it than meets the eye! Forget the 10 minutes .... take your time! Think of W1, W2, W3, W4, B1, B2, etc.

    (2) Does anyone wish to do some calculations? Again, don't worry about the 10 minutes.

    (3) The decimal point is not missing ... there isn't one!

     

  15. 3 Quickies

    in New Logic/Math Puzzles

    Posted

    10 minutes each!

    1.      A bag holds four counters.  One of them is white.  Each of the others is either black or white at equal chance.  One randomly draws out two counters, and discover they are both white. If one then randomly draws a third counter, what is the chance that it is white?

     

    2.      I’m about to play the music songs on Side 1 of a standard LP (diameter = 292 cm) that contains six tracks. The recorded surface area is 487.6 dm2 (including the “gaps” between each song) and the average distance between each annulus (i.e. distance between each “groove” along the LP radius) is 197 µm. The total time that it will take to play this side right from the beginning of the first song to exactly the end of the last (6th) song is 18:04 minutes. If the “needle” (stylus) is poised 12.5 mm directly above the beginning of the first song and 2 mm away from the edge of the record, how far does the “needle” travel to the point where the last song just ended?

     

    3.      What comes next?

    6 1 3 1 4 _

     

  16. I’m not a gardener

    in New Word Riddles

    Posted · Edited by rocdocmac
    some additonal words added

    Molly said: "Glass/Silicon (which is in the Carbon group!)"

    Spoiler

    Silicon ("silicium") is element #14 and glass is silicon dioxide (quartz or sand).

    But do we need an occupation here? Glassblower? Doesn't fit the description too well!

    And a lipidary/lipidist/lipidarist is already out, too.

     

  18. Cubicle Stack #2

    in New Logic/Math Puzzles

    Posted · Edited by rocdocmac
    Addendum to images

    Plotting the sequence 1, 4, 22, 139, ... as a scatter diagram, one can fit a 3rd degree polynomial curve with equation y = 14x^3 - 34.5x^2 + 23.5x + 1 (x = 0, 1, 2, 3).  Using this graph as an estimator, one can guess that the number of shapes will be in the region of 439 if four are removed. Therefore the guestimate for 500 is not far out!

    T25 appears to be 9 (see attached image) and I agree that T26 is 1. The roots indicate the distance between the two cubelet-middle points in the grid (x = y = z = 1).

    Thus, would the sequence end with ..., 9, 1, 0 (the 0 indicating no shape at all)?

    Will determine T24 a bit later!

    Remove 25.jpg

  19. Cubicle Stack #2

    in New Logic/Math Puzzles

    Posted

    2 hours ago, Thalia said:

    When you remove a cubelet, the kind you remove matters because it changes the shape. When you take away 26, it doesn't matter which one is left. There may be 4 different labels but in the end, when it comes to the remaining shape, all you see is a cube.

    Another thing. For T27, there is one possibility. But given the phrasing of the original question, does nothing count as a shape?

    Thanks Thalia,

    Interesting twist in the tale, for sure!

    Yes, surely 26 cubelets removed leaves a single cubelet, but where was the remaining one positionally located/"fixed" before removal of the other 26? Should be one specific remnant shape for each combination and, therefore,  four variants/labels/possibilties.

    "Nothing" or "no shape" may be regarded as a "blank" stacked cube.

    (1) What's your suggestion for the end of the sequence?

    (2) What do our other BrainDenners think?

    (3) How many distinct shapes can one expect for a 4-cubelet removal? Just a ball-park figure, perhaps!

     

     

     

  21. Cubicle Stack #2

    in New Logic/Math Puzzles

    Posted · Edited by rocdocmac

    37 minutes ago, rocdocmac said:
    Spoiler

    Tried to update the explanation, but time ran out!

    Series starts with T0 = 1, T1 = 4, T2 = 22, T3 = 139, ... and ends with its reversal, i.e. ... T24 = 139, T25 = 22, T26 = 4, T27 = 1.

    Only one possibility for T27, which is zero cubelets left.

     

     

  22. Cubicle Stack #2

    in New Logic/Math Puzzles

    Posted · Edited by rocdocmac
    Expanded response

    @Thalia ...
     

    Spoiler

    I actually noticed something in this line yesterday!

    T26<>1: If 26 cubelets are removed, one would be left with either X, M (= F), C or E, i.e. T26 = 4 (not "1"!).

    We now know that T3 = 139.

    Thus T0, T1, T2, T3, ... = 1, 4, 22, 139, ...

    The sequence should therefore end with

    ... T24, T25, T26, T27 = ... 139, 22, 4, 0

    Maximum values would be for T13 = T14

     

  24. Cubicle Stack #2

    in New Logic/Math Puzzles

    Posted

    Spoiler

    I agree with 24 for CEM, but we're still not on par with CEE and EEE. The total for 3-cubelet removal (say T3 = 13?) is surely somewhere between 135 and 140.

    I wonder whether an equation exists that one could use to predict the next number (4-cubelet removal or T4) and higher T-numbers in the sequence. Such a sequence should start with:

    T0, T1, T2, T3,  ..., (i.e. 1, 4, 22, 13?, ..., with T0 = 1, indicating a solid cube resulting from 0-cubelet removal),

    and end with:

    ..., T23, T24, T25, T26, T27 (i.e. 13?, 22, 4, 1, 0 with T27 = 0, which is equivalent to the removal of all 27 cubelets leaving nothing.

    The maximum value would be for T13.

     

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