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  • April madness
    replied
    ah, thanks for clarification.

    Leave a comment:


  • davis_greatest
    replied
    Originally Posted by April madness
    does that mean that the house itself is small, just the toilet is huge?
    Hehe No, the toilet is smaller than the house. If we'd used the whole house, we could have had 12,988,816 apes.

    Leave a comment:


  • April madness
    replied
    Originally Posted by davis_greatest
    We held the party in the toilet.
    does that mean that the house itself is small, just the toilet is huge?

    Leave a comment:


  • dantuck_7
    replied
    http://www.thesnookerforum.com/photo....php?n=240&w=o

    Graph regarding the earlier thread.

    Dan.

    Leave a comment:


  • davis_greatest
    replied
    Originally Posted by davis_greatest

    Round 116 - Break it up

    In this post http://www.thesnookerforum.com/showp...ount_1356.html, we listed the number of all possible snooker breaks (ignoring free balls, misses, or potting more than one red in one stroke). A break of red-yellow-red-black, say, was counted as different from red-black-red-yellow, as the balls were potted in a different order.

    What is the total number of different breaks? You must show a straightforward method for arriving at your answer and you are not allowed simply to add the 147 numbers shown in the list - although that would give you the right answer!
    I don't like leaving rounds open for too long, especially when other rounds have been asked since, so I'll close this one.

    Congratulations to snookersfun who correctly counted there being 4,043,590,867,366 possible breaks. In fact, the easiest way (at least, my way) to obtain this number is to see that there are 43/5*(6^15-1) + 21 possible breaks! (Explanation available on request.)

    Leave a comment:


  • davis_greatest
    replied
    Originally Posted by April madness
    you have a big house, DG, to have 3604 apes partying there
    We held the party in the toilet.

    Leave a comment:


  • April madness
    replied
    you have a big house, DG, to have 3604 apes partying there

    Leave a comment:


  • davis_greatest
    replied
    ... and that, dantuck_7, is the correct answer! Well done.

    The number of apes is, of course, simply the square root of the number of kisses.
    So 3,604 apes were invited to the party, being the square root of 12,988,816.


    Congratulations also to snookersfun who submitted the same answer by Private Message.


    So here, with new joiner dantuck_7, is the latest Puzzles with numbers and things Hall of Frame

    Oliver (my pet orang-utan)
    Gordon (my pet gorilla)
    Charlie (my pet chimpanzee)
    snookersfun
    abextra
    davis_greatest (Oliver's, Gordon's and Charlie's pet something)
    Vidas
    chasmmi
    elvaago
    robert602
    Sarmu
    The Statman
    austrian_girl
    austrian_girl's dad
    Semih_Sayginer
    Snooker Rocks!
    Ginger_Freak
    April Madness
    steveb72
    rambon
    Microsoft Excel
    dantuck_7


    I will also take this opportunity to remind the newer joiners to this thread, who may not have caught up with the previous 1401 posts, of the rule that posts with answers may not be edited. If you wish to change an answer, please post again.

    That rule does not, of course, apply to posts with questions - nor, indeed, to any posts at all made by davis_greatest.

    Leave a comment:


  • dantuck_7
    replied
    'z' chimps,'x' orang-utans, 'y' gorillas

    1) (x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2zy
    2) x^2+y^2=z^2

    0)3604^2 = 2xy+2xz+2yz + 2z(z-1) + 2z
    3604^2 = 2xy+2xz+2yz + 2z^2 - 2z + 2z

    Looking at 1)
    3604^2 = (x+y+z)^2 ((because x^2+y^2+z^2 = 2z^2))
    so x+y+z = 3604

    There we go.

    Dan.

    Leave a comment:


  • davis_greatest
    replied
    And for those who don't like counting breaks, here at the same time is ...

    Round 117 - Kiss me

    In order to celebrate the fact that Steve Davis is fewer than 100 days from taking his 7th World Title, I throw a little party for all my chimpanzee, orang-utan and gorilla friends. Actually, quite a big party.

    Before the fun begins, they all stand along the sides of a triangle, equally spaced, holding hands - the chimpanzees along one side, the orang-utans along another and the gorillas along the third. As it happens, the triangle is a right-angled one, with the chimpanzees along the longest side.

    When I blow my whistle, they all run around, greeting each other, and the fun begins. Every ape kisses every ape of a different species on the left cheek. (This excludes me.)

    The chimpanzees are especially friendly, and every chimpanzee also kisses each other chimpanzee, but on the right cheeks. I say "right cheeks", because the chimpanzees don't just get kissed by chimpanzees on the cheeks on their faces. It is, to borrow snooker parlance, a "double kiss" (one upstairs and one downstairs).

    I also give every chimpanzee a kiss on each ear.

    Once that's all over, some 12,988,816* kisses later, we get on with the games.

    How many apes (and I am not included!) came to the party?


    For the first 24 hours from the time of this message, answers by Private Message please - unless you are not already in the Puzzles with numbers and things Hall of Frame, in which case answer on this thread. After 24 hours, everyone should answer on this thread.



    *I just noticed that that number is exactly 995,000 fewer than the number of possible sets of numbers in the UK national lottery. That has no relevance whatsoever to the question - just noticed it, that's all...

    Leave a comment:


  • davis_greatest
    replied
    I think we await Snooker Rocks! to close round 115:

    http://www.thesnookerforum.com/showp...ount_1354.html

    The answer, in any case, to Snooker Rocks!'s question was 2, 2 and 9. Congratulations to snookersfun, davis_greatest and anyone else who may have solved it - perhaps Snooker Rocks! can confirm.



    So...

    Round 116 - Break it up

    In this post http://www.thesnookerforum.com/showp...ount_1356.html, we listed the number of all possible snooker breaks (ignoring free balls, misses, or potting more than one red in one stroke). A break of red-yellow-red-black, say, was counted as different from red-black-red-yellow, as the balls were potted in a different order.

    What is the total number of different breaks? You must show a straightforward method for arriving at your answer and you are not allowed simply to add the 147 numbers shown in the list - although that would give you the right answer!

    Leave a comment:


  • snookersfun
    replied
    Originally Posted by chasmmi
    I believe that is 543 and counting if you are interested.
    chasmmi, that is grand as well. Believe me, I do understand you...

    Leave a comment:


  • davis_greatest
    replied
    Originally Posted by snookersfun
    Oh, how nice! Thousands of tiny numbers!
    OK, and now that you got that out of your system, what next?
    We are impressed
    Next this thread will go back to having puzzles with numbers and things.

    I will post the next puzzle tonight. No idea what yet, but I'll think of something before I go to bed...

    Leave a comment:


  • snookersfun
    replied
    Oh, how nice! Thousands of tiny numbers!
    OK, and now that you got that out of your system, what next?
    We are impressed

    Leave a comment:


  • davis_greatest
    replied
    And here's a graph, showing that of all the breaks from 1 to 147, a break of 81 is the break with the most possible number of ways (8,993) of making it (where the same colours potted after reds, even if potted in a different order, are counted as the same way). (If the order matters and breaks with colours potted after reds in a different order are counted as different ways, then a break of 88 has the most ways: 133,835,290,090 of them.)

    Leave a comment:

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