Puzzles with numbers and things

[April madness]

I thought the cheapest way would be to clean them himself?

[berolina]

I thought the cheapest way would be to clean them himself?

No April, cheapest way is not to clean them at all!

[April madness]

No April, cheapest way is not to clean them at all!

Right, berolina, but since Scrooge was really considering the cleaning, the next cheapest option is what I said (at least I think so)

[davis_greatest]

And this true story round will close in 20 minutes...

This, I guarantee you, is a (99%) true story. I have only changed the odd, very minor detail (if you think you know which detail I have changed, you can guess it, and may earn some more fame, but that is not the point of this question)...

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

Last night, I played snooker with my apes. Usually, there are four of us: me, my pet orang-utan Oliver, my pet gorilla Gordon and, of course, my pet chimpanzee Charlie. Occasionally, one cannot make it, and last night was one of those nights.

When there are three of us playing, we play one against two (a bit like doubles, but the "one" on his own also plays the shots for his imaginary doubles partner). Then we rotate, so that we each play one frame alone.

Last night, with my opening visit, I played a very nice indeed long red, with a lot of left hand side to avoid the cue ball cannoning into reds after hitting the top cushion (as the black was tied up), thus finishing nicely on the blue. Potting the blue at pace enabled a full ball cannon onto the pink, to split the reds, one of which I then potted, followed by a pink... but I digress....

Depending on the standard, and the amount of safety played, sometimes we have a bit more time and play a 4th frame (sometimes with a smaller number of reds, such as 10). And so it was last night. For the 4th frame, all three of us play against each other, i.e. three "teams" of one, so we each come to the table every 3rd visit. This entirely ruins the tactical aspect, but it's still fun enough at the end of the evening...

If we counted the frames, a fair way to score it might be to award one frame to each person / ape for each frame won, except for frames won by a pair, in which case each member of the pair would be awarded half a frame.

Scoring the frames that way last night would have meant that everyone scored, we all got different scores, and I, of course, got the highest.

What was my outcome (win/loss) for each frame?

[rambon]

Well I think that you won your game of singles against your two friends. Then when paired with the gorilla you beat the orang-utang, and when paired with the Orang-utang, you beat the gorilla.

In the final frame you'd had one too many beers and lost to the gorilla.

As a result, you scored 2 points, the gorilla 1.5 and the Orang-utang wass last with 0.5 points.

Do I win a banana?

[davis_greatest]

Well I think that you won your game of singles against your two friends. Then when paired with the gorilla you beat the orang-utang, and when paired with the Orang-utang, you beat the gorilla.

In the final frame you'd had one too many beers and lost to the gorilla.

As a result, you scored 2 points, the gorilla 1.5 and the Orang-utang wass last with 0.5 points.

Do I win a banana?

Absolutely correct, except that I had only a coke and no beers. The reason that I lost the final frame was that, despite having come back from a near impossible position and back into contention on the final colours, I played a poor positional shot in potting the brown after winning a long safety battle, and was then unable to clear up and finish the job with a whitewash.

You may share the banana with snookersfun and abextra, who also answered this correctly.

[davis_greatest]

Roun 122 - I potted round balls and I potted snowballs

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

And all this in one visit! What was my break?

[Robert602]

I'll say (80)?

[snookersfun]

not 100% convinced of this one, but 80 works
but wouldn't 82 work as well?

[chasmmi]

ok i dont how how to do it so here is my final guess 849072

Please excuse the edits was trying to find the invisible colour.

[davis_greatest]

ok i dont how how to do it so here is my final guess 849072

Please excuse the edits was trying to find the invisible colour.

849072 is not correct, but robert602's and snookersfun's answers are!

[chasmmi]

my real guess can be found by dividing that number by 147 and then squarerooting that answer

[davis_greatest]

my real guess can be found by dividing that number by 147 and then squarerooting that answer

Ahhhhhh. Well then.... still no! Unless I have not worked out the question properly (which is not inconceivable as it was invented on the train while standing with no paper) and you can show that your answer is also possible?

[chasmmi]

I assumed the plurals meants more than one of everything in each sentence so suggested the break went:

Red Blue Red Pink Red Blue Red Pink Red Green Red Black Red Green Red Yellow Red Green Red Black Red Green Red Yellow Red Yellow Red Green yellow Green

but i could be wrong.

(If that is invisible then thanks Snookersfun. )

[davis_greatest]

Oh, I see. But doesn't your way involve starting potting the final colours after only 14 reds have gone down?