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  • PaulTheSoave
    replied
    Does the free ball black not count as 1 then instead of 7?

    Originally Posted by snookersfun View Post
    aww, shucks...
    anyway, I know it is too late, but started something with free ball black last night in bed... (and I am sure one can work on that as well)
    so here is that:

    Leave a comment:


  • PaulTheSoave
    replied
    I put my alarm to receive my congratulations and 1st prize,

    Leave a comment:


  • snookersfun
    replied
    one more now (out of contention, just for the fun of it), think it is 97 now:

    black-7
    2 reds-9
    black-16
    red-17
    black-24
    red-25
    blue-30
    red-31
    black-38
    red-39
    black-46
    red-47
    blue-52
    red-53
    blue-58
    red-59
    black-66
    red-67
    pink-73
    red-74
    blue-79
    3reds-82
    pink-88
    red-89
    pink-95
    yellow-97

    Leave a comment:


  • snookersfun
    replied
    aww, shucks...
    anyway, I know it is too late, but started something with free ball black last night in bed... (and I am sure one can work on that as well)
    so here is that:
    black-7
    3 reds-10
    black-17
    red-18
    black-25
    red-26
    pink-32
    red-33
    pink-39
    2 reds-41
    pink-47
    red-48
    black-55
    3 reds-58
    black-65
    2 reds-67
    pink-73
    red-74
    black-81
    yellow-83
    green-86

    Leave a comment:


  • PaulTheSoave
    replied
    Its late, so this could be totally wrong, but have a try.

    5 reds = 5
    black = 12
    2 reds = 14
    black = 21
    red = 22
    black = 29
    2 reds = 31
    black = 38
    red = 39
    black = 46
    red =47
    pink = 53
    red =54
    blue = 59
    2 red = 61
    black = 68
    1 red = 69
    1 blue = 74

    free ball at start

    Leave a comment:


  • davis_greatest
    replied
    Originally Posted by snookersfun View Post
    5 reds=5
    black=12
    2 reds=14
    black=21
    red=22
    black=29
    2reds=31
    black=38
    red=39
    black=46
    red=47
    pink=53
    red=54
    black=61
    3reds=64
    rather:
    black=71
    yellow=73

    end? (1 free ball at beginning)
    and this one is about an hour and a half from winning the round....

    Leave a comment:


  • davis_greatest
    replied
    Originally Posted by snookersfun View Post
    start with several reds
    5 reds=5
    black=12
    2 reds=14
    black=21
    red=22
    black=29
    2reds=31
    black=38
    red=39
    black=46
    red=47
    pink=53
    red=54
    black=61
    3reds=64
    yellow=66
    yellow=68
    green=71

    end? (1 free ball at beginning)
    This one is disqualified, I'm afraid, as 66 is divisible by 22.

    Leave a comment:


  • davis_greatest
    replied
    Originally Posted by R_Demarco View Post
    As a break has to start with a red, it will always be divisible by 1.
    A break doesn't have to start with a red. It can start with a colour (either if no reds are left, or in the rare situation when a Foul And A Miss was called after an attempt on a colour) or with more than one red potted in a single stroke.

    Leave a comment:


  • PaulTheSoave
    replied
    I am not a wizzzz kid, but I count 16 reds on your table.

    ah, the free ball

    Leave a comment:


  • snookersfun
    replied
    5 reds=5
    black=12
    2 reds=14
    black=21
    red=22
    black=29
    2reds=31
    black=38
    red=39
    black=46
    red=47
    pink=53
    red=54
    black=61
    3reds=64
    rather:
    black=71
    yellow=73

    end? (1 free ball at beginning)

    Leave a comment:


  • snookersfun
    replied
    start with several reds
    5 reds=5
    black=12
    2 reds=14
    black=21
    red=22
    black=29
    2reds=31
    black=38
    red=39
    black=46
    red=47
    pink=53
    red=54
    black=61
    3reds=64
    yellow=66
    yellow=68
    green=71

    end? (1 free ball at beginning)

    Leave a comment:


  • R_Demarco
    replied
    As a break has to start with a red, it will always be divisible by 1.

    Leave a comment:


  • davis_greatest
    replied
    Round 311 - Indivisible Oliver

    Oliver has just made a break wearing his magic cloak. When he puts it on, both he and his breaks become Indivisible.

    An Indivisible break in snooker is a break in which, after each shot, the break is never divisible by the running break after any previous shot.

    For example, potting green and then brown would be an Indivisible break - since it goes 3, 7; and 7 is not divisible by 3.

    However, potting green, brown and then blue is not an Indivisible break - since it goes 3, 7, 12; and 12 is divisible by 3.

    Post here on the thread the greatest Indivisible Break you can find. You can post as many as you like, as long as each one you post is higher than the previous one you posted.

    When posting a break, you should state not only the total break, but the balls potted: e.g. "7 - green, brown".

    And hurry... this is all about speed. The round will close with the greatest Indivisible break posted here before midnight UK time tonight (or until the first answer appears, if there are no answers before midnight).
    Last edited by davis_greatest; 9 February 2008, 06:46 PM.

    Leave a comment:


  • dantuck_7
    replied
    On Round 309 I think it was the same as Moniques solution.

    Leave a comment:


  • davis_greatest
    replied
    Originally Posted by Monique View Post
    Because He was on a 154 starting with a black ... scenarios to be found in the "maximum" discussion thread?
    Yes! Congratulations, Monique - and very quick! Exactly so. He had been put back in by Charlie after a "Foul and a Miss", having missed a colour following a "free ball" red on his previous visit.

    Leave a comment:

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