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  • Originally Posted by abextra View Post
    Hi, Snookersfun!

    What does Charlie mean by " discard any fractional portion"? Should I try to spell RED HERRING?
    Originally Posted by tallguy View Post
    Hi abextra, It means you ignore the remainder. For eg. 7/2 = 3 if you discard any fractional portion.

    The spelling of RED HERRING is correct.
    and despite the red herring R. 349 now also solved by abextra and d_g.
    That means we can pretty much close that round. Please somebody put up his/her solution!

    Comment


    • Mon solved the tilted balance as well, even with a slightly banged up head Congrats!

      and with that to:

      R. 351 break off- distances

      This time around the apes play a proper game of snooker, or almost, as they are experimenting with different numbers of lines in the original triangle of reds. In one of those efforts involving seven lines of reds, Gordon smashes the pack (Quinten Hann like), potting two reds right away in the process, and leaving the other reds in the following positions on the table.
      26 balls-1.bmp
      Charlie, being his regular clever self, takes a glance and exclaims: Wait, before you play on, look how neatly you scattered those balls. As I see it, you could organise all balls into pairs, so that each pair has discrete and distinct distances between them. Starting from a pair of touching balls, continuing with the next pairs exactly one, two, etc. balls apart. In addition no ball is positioned with his center on the imaginary line between a given couple.

      Can you assign the pairs?

      Comment


      • A breakdown of R349, Gwennie would have gone straight to the last paragraph!

        If we take the four scores as a,b,c and d and do the algebraic calculations as instructed (a full stop denotes multiplication):

        (a.b.a.c.a.d).(b.a.b.c.b.d).(c.a.c.b.c.d).(d.a.d.b .d.c)

        collect terms:

        a^6.b^6.c^6.d^6

        then divide by the product of the four scores:

        a^6.b^6.c^6.d^6/a.b.c.d = a^5.b^5.c^5.d^5

        No remainder, red herring, as was the number and distribution of the four scores, we just needed to know they were positive and greater than 1.
        We are asked for the remainder after subtracting the four score product and division by 10, anything to the left of the decimal point is not relevant.
        If we let:

        n = a.b.c.d

        Then we have;

        (n^5-n)/10

        It is a property of the fifth powers that the final digit of n (a whole number) is always the final digit of its fifth power, so, there will never be any “remainder” after division by ten.
        Last edited by moglet; 30 December 2008, 12:01 PM.

        Comment


        • Originally Posted by moglet View Post
          A breakdown of R349, Gwennie would have gone straight to the last paragraph!

          If we take the four scores as a,b,c and d and do the algebraic calculations...
          Thank you, Moglet! The explaining is always the hardest part for me.

          Comment


          • Originally Posted by abextra View Post
            Thank you, Moglet! The explaining is always the hardest part for me.
            yes, thank you indeed, moglet! Very nice and complete explanation (not my strong point either)

            meanwhile I have a first entrant for R.351. tallguy solved it and also learnt to attach pictures in PMs. Congrats

            Comment


            • Originally Posted by snookersfun View Post
              R. 351 break off- distances

              This time...

              ...you could organise all balls into pairs, so that each pair has discrete and distinct distances between them. Starting from a pair of touching balls, continuing with the next pairs exactly one, two, etc. balls apart. In addition no ball is positioned with his center on the imaginary line between a given couple.

              Can you assign the pairs?
              Well... are the imaginary lines between balls always horizontal or vertical? I was hoping so, but...

              Comment


              • Originally Posted by abextra View Post
                Well... are the imaginary lines between balls always horizontal or vertical? I was hoping so, but...
                see, erm, no, another red herring that grid But they are straight lines.

                Comment


                • more solutions in now:

                  moglet and d_g on the distances and abextra on tilting. Congratulations everybody.

                  goes 'sheltering' again now...

                  Comment


                  • Round 352 - Take Five

                    OK, a quick one from here... :snooker: ....complete the five missing numbers in this sequence...

                    18916, ?????, 53233, ?????, 84956, 57646, 57273, ?????, 88996, ?????, 10511, 21131, 20122, 12512, 91341, ?????
                    "If anybody can knock these three balls in, this man can."
                    David Taylor, 11 January 1982, as Steve Davis prepared to pot the blue, in making the first 147 break on television.

                    Comment


                    • Breaking News

                      First congratulations for solving round 352 go to snookersfun - well done

                      Edit - and Monique now joining suit!
                      Last edited by davis_greatest; 31 December 2008, 08:10 PM. Reason: Monique added
                      "If anybody can knock these three balls in, this man can."
                      David Taylor, 11 January 1982, as Steve Davis prepared to pot the blue, in making the first 147 break on television.

                      Comment


                      • Happy new year to all on TSF.

                        Next answer (or request for hints if needed, although the clue is in the question :snooker to round 352 on the thread please.

                        Round 353 New Year Number Magic

                        Using as many as you like of the numbers 1,5,5,6,8,9 (but not using any more than once), and any combination of addition, subtraction, multiplication and/or division, get as close as you can to today's target, which is 2009.

                        Again, answers please directly on the thread. Quickest / closest wins!
                        "If anybody can knock these three balls in, this man can."
                        David Taylor, 11 January 1982, as Steve Davis prepared to pot the blue, in making the first 147 break on television.

                        Comment


                        • let's launch the race ... 5*5*8*(9+1) + 6 = 2006
                          Proud winner of the 2008 Bahrain Championship Lucky Dip
                          http://ronnieosullivan.tv/forum/index.php

                          Comment


                          • Needing one more "1" to get the target - 5*5*81-6-9=2010

                            Comment


                            • Is that 81 allowed? the way I understand "numbers" not "digits" i supposed not ... D_G?
                              Proud winner of the 2008 Bahrain Championship Lucky Dip
                              http://ronnieosullivan.tv/forum/index.php

                              Comment


                              • Originally Posted by davis_greatest View Post
                                ... :snooker: ....complete the five missing numbers in this sequence...

                                18916, ?????, 53233, ?????, 84956, 57646, 57273, ?????, 88996, ?????, 10511, 21131, 20122, 12512, 91341, ?????
                                ...

                                Next answer (or request for hints if needed, although the clue is in the question :snooker to round 352 on the thread please.

                                I guess one of the missing numbers is 80818?

                                The :snooker:was VERY helpful!!!

                                Comment

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