Puzzles with numbers and things

[dantuck_7]

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

[davis_greatest]

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).

[R_Demarco]

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

[snookersfun]

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)

[snookersfun]

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)

[PaulTheSoave]

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

ah, the free ball

[davis_greatest]

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.

[davis_greatest]

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.

[davis_greatest]

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....

[PaulTheSoave]

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

[snookersfun]

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

[snookersfun]

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

[PaulTheSoave]

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

[PaulTheSoave]

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

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:

[snookersfun]

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

No, see post 2812, R. 310, and maximum discussion thread....