Puzzles with numbers and things

[davis_greatest]

Round 234 - Touching Pineapples

This is just like round 231 (to save you reading it all), except for the bit in magenta...

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Take an ordinary set of snooker balls. Place the cue ball on your nose, and the remaining balls on the bed of the table wherever you like.

Whenever there is a straight line of (2 or more) touching balls, in any direction, every ball in that line scores a number of points equal to the number of balls in the line.

For example, if there is a line of 7 touching balls, each of those 7 balls scores 7 points.

If a ball lies in more than one line, it scores points for every line it is in. For example, if a ball is in a line of 7 balls and also is in a different line of 6 balls, that ball scores 13 points, like the top left ball in the example below. (Those balls are supposed to touch, even though they don't appear to.)

Eg.

ooooooo
o
o
o .... ooooooooo
o
o

You are not allowed to place all 21 balls in just one line! Apart from that, anything goes.

Your final score is just the number of points scored by your highest-scoring ball (excluding the cue ball, which doesn't score, but must remain nasally balanced while you post your bid). You will then be able to claim from my apes, when they return, a number of FREE pineapples equal to your final score.

Post here the number of pineapples you are claiming. Once we have some decent claims, I will then ask you to explain where you placed your balls.

[abextra]

I'm afraid you will get nothing because of the one separate ball that scores 0 plums...

[davis_greatest]

I think 11 is possible...

Yes, it is! Let's see the arrangement.

So you can put 20 in a line ... and one alone (just one ball is not a line )?
for 20 plums?

The ball alone would be the lowest scoring ball (0 points) and so I'm afraid you'd get no plums for that (to add insult to no-plums, you wouldn't even get any bananas either, unfortunately. It would be rather a fruitless arrangement).

[abextra]

21 pineapples... not very hungry now...

[Monique]

Lets start with 23 pinapples.

arrangement for 11 plums?
O
O
O
O
O
O
O
O
O
O
OOOOOOOOOOO

[abextra]

Yeah, they could also be in the shape of "+" or "x" or any two lines wich have one mutual (or dear, sorry for my English! ) ball.

[davis_greatest]

Lets start with 23 pinapples.

arrangement for 11 plums?
O
O
O
O
O
O
O
O
O
O
OOOOOOOOOOO

Yes, this arrangement will get you the 11 plums! (Or anything similar, such as a cross.)

Great opening bids from abextra and Monique for 21 and 23 pineapples.

I'm sure you'll both soon outbid yourselves though on those pineapples

[davis_greatest]

Yeah, they could also be in the shape of "+" or "x" or any two lines wich have one mutual (or dear, sorry for my English! ) ball.

.....yes!

[Monique]

Not now though ... Nigth all, sweet dreams ...

[Monique]

A very last bid ... 25!

[davis_greatest]

A very last bid ... 25!

Yes, very nice! And I believe that's the greatest available number of pineapples! We'll see if anyone wants to match it and then invite an explanation of the arrangement.

[snookersfun]

as I can't fathom this 25 (21 and 23 pineapples more than enough for tummy aches)

moving on to R235: triangular (oh, inspiration! triangle blue chalk on hexagonal part of snooker table

HappyY.gif

arrange the triangle chalk (possible arrangements of 4, 3, 2, 1 pieces on right of pic) on table. Chalk pieces (unless part of an arrangement) are not allowed to touch (even at a single point). Again number clues on sides of the hexagon point to number of chalk pieces in respective line.

P.S. the wavy thing is a part of table

Answers to me initially...

[davis_greatest]

R234 ... Monique (or anyone else!), please explain the arrangement for 25 pineapples.

Hint for anyone trying to fnd such an arrangement... think of a star...

[Monique]

Build a star with 5 "legs" of four balls each around a central ball. This way all the 21 balls are taken into account and the central ball "weights" 5. You cannot do better because you can only put 6 balls maximum "around" one central ball. But if you do that you have only three lines, not 5, because opposite legs are aligned. So the central bal weights only 3 and the total is then 23.

D_G please revise the english ... Not sure it is understandable

[davis_greatest]

Build a star with 5 "legs" of four balls each around a central ball. This way all the 21 balls are taken into account and the central ball "weights" 5. You cannot do better because you can only put 6 balls maximum "around" one central ball. But if you do that you have only three lines, not 5, because opposite legs are aligned. So the central bal weights only 3 and the total is then 23.

D_G please revise the english ... Not sure it is understandable

Perfectly understandable... and correct, of course! Congratulations on the 25 pineapples!
The central ball, of course, scores 25, as it appears in 5 lines of 5 balls.