Originally Posted by snookersfun
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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
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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
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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
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Originally Posted by snookersfun View Post5 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)
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Originally Posted by snookersfun View Poststart 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)
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Originally Posted by R_Demarco View PostAs a break has to start with a red, it will always be divisible by 1.
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I am not a wizzzz kid, but I count 16 reds on your table.
ah, the free ball
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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:
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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)
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As a break has to start with a red, it will always be divisible by 1.
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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.
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Originally Posted by Monique View PostBecause He was on a 154 starting with a black ... scenarios to be found in the "maximum" discussion thread?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.
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