Puzzles with numbers and things

[chasmmi]

So if I learn to count and add red green???

[davis_greatest]

So if I learn to count and add red green???

That should do it!

[berolina]

...
Round 120 - You wait all day for an ape, and then they come in threes

Last night, I played snooker with my apes...

Do you know about this?
http://dilbertblog.typepad.com/the_d...e_mysteri.html

[snookersfun]

Do you know about this?
http://dilbertblog.typepad.com/the_d...e_mysteri.html

yikes didn't get the relevance though!

...but back to the question, I had another row in my answer:
how about if you add in a blue, minus a green? I think that would also qualify, wouldn't it?

[berolina]

yikes didn't get the relevance though!

Oh, I came across that article today and thought it was funny and what with davis_greatest and him playing with apes...well well well

[davis_greatest]

yikes didn't get the relevance though!

...but back to the question, I had another row in my answer:
how about if you add in a blue, minus a green? I think that would also qualify, wouldn't it?

Let's see the whole break.

[The Statman]

Oliver broke off, nothing went in, so it was my shot. I potted red balls and I potted blue balls and I potted pink balls (at least as many blue balls as pink balls and at least as many red balls as blue balls and at least as many pink balls as blue balls).

...2 reds, pink, 2 reds, blue, 2 reds...

Then I potted red balls and I potted black balls and I potted yellow balls and I potted green balls (at least as many green balls as yellow balls and black balls combined).

...black, 3 reds, yellow, 2 reds, green, 2 reds, green...

Then I potted yellow balls and I potted red balls and I potted green balls (at least as many green balls as red balls).

...Red, green, red, yellow.

Break was 46.

This assumes that 'I potted red balls and blue balls [etc.]' doesn't necessarily stipulate that all balls are potted in the plural (and that multiple reds in one shot are allowed).

[snookersfun]

something like this:Red Blue Red Pink Red Blue Red Pink Red Blue Red Green Red Black Red Green Red Yellow Red Green Red Black Red Green Red Yellow Red Yellow Red Green yellow Green

[The Statman]

something like this:

That fails on the first criterion, though, doesn't it? At least as many blues as pinks AND at least as many pinks as blues – suggests number of blues and pinks must be equal.

[snookersfun]

blimey, I totally overread that third statement. So then, only that one possibility it seems.

[The Statman]

I think there are at least 2 possible answers.

For the first segment you can have:
Four reds, two blues, two pinks (total 26)

For the next segment you can have:
8 reds plus two blacks, two yellows, four greens — plus the following (thirteenth) red. (total 65)

The for the final segment you can have:
yellow, red, green, red, green; followed by the yellow (which satisfies the condition by itself) and the green (which still satisfies the condition).

Total 77 (up to final yellow) or 80 (to final green).

[It is by putting the 13th red at the end of the second segment, rather than at the beginning of the third segment, that means the green condition in the third segment is satisfied without the final green having been played.]

[snookersfun]

right! I had the reds in different segments as well. Didn't follow it up after finding the 80 though...

[The Statman]

I've just edited message 1462.

[davis_greatest]

This assumes that 'I potted red balls and blue balls [etc.]' doesn't necessarily stipulate that all balls are potted in the plural (and that multiple reds in one shot are allowed).

Balls, Statman, balls. It does mean the plural.

[davis_greatest]

I think there are at least 2 possible answers.

For the first segment you can have:
Four reds, two blues, two pinks (total 26)

For the next segment you can have:
8 reds plus two blacks, two yellows, four greens — plus the following (thirteenth) red. (total 65)

The for the final segment you can have:
yellow, red, green, red, green; followed by the yellow (which satisfies the condition by itself) and the green (which still satisfies the condition).

Total 77 (up to final yellow) or 80 (to final green).

[It is by putting the 13th red at the end of the second segment, rather than at the beginning of the third segment, that means the green condition in the third segment is satisfied without the final green having been played.]

Congratulations The Statman! This (the 77 one), I think, is the only other answer apart from the more obvious 80 one (although I haven't given it a huge amount of thought since noticing that I had not ruled out the possibility of a segment ending in a red, and someone might just find another!).