Puzzles with numbers and things

[moglet]

Wow! I hope this doesn't start a trend, it'll take longer than solving the puzzle

[snookersfun]

...and as I had my doubts about much variety in this round as well, I subcontracted my artwork (so, all credits to little son)

snookerballs3.jpg

[moglet]

Round 357

I think the deadline for this round has expired:

[snookersfun]

update:

R. 358 took ja. (welcome back!), Mon and moglet no time at all! Well done
and Abextra had no problem either. Congrats
same goes for d_g now. Well done!

[snookersfun]

A slight dip in activity, so a good time to get an intermezzo in
R. 358 ape arithmetic
After their Wednesday night league session, Gwenny, Gordon, Charlie, Olli and (erm whatshisname) Ginger (?, the gibbon) recapitulate their high breaks. It turns out that none of them had managed a century break that evening and Charlie relates their respective high breaks this way:
Gwenny’s high break was ¾ of Olli’s
Charlie’s hb was 14 points higher than Olli’s
Ginger’s hb was 2/3 of Charlie’s
And Gordon actually got the average of all their scores

Please PM me the respective high breaks

[davis_greatest]

R357 - good to see that we some 3 bananas & no pineapple solutions, as Charlie doesn't have any pineapples.

Will give until midnight UK time tomorrow (c.28 hours from now) in case anyone might still be searching - and after that, all arrangements up on the thread please.

[moglet]

Round 357

I also bid 3 bananas and no pineapples

[moglet]

Aah! Monique, I see how you have annotated the answer, this is a pictorial version:

[Monique]

pairs in ascending order distances from 0 to 12 ...

13,16
8,12
25,26
15,20
11,17
14,23
10,21
4,18
9,24
1,2
3,22
5,6
7,19

phew I struggled with this one

[snookersfun]

Breaking news:

one more correct answer to R.351 (distances): Monique joined the crowd.
So congratulations to tallguy, moglet, d_g and Mon on solving the round.

Closing this round now, please somebody put up his/her result!

[snookersfun]

I'm still bidding 3 bananas
- and just can't believe we all got the same arrangement with 6 pineapples!!!

well done for seeing that so fast!
and not wondering we all got the same initially

and yes, have the 3 bananas as well now

[davis_greatest]

abextra has bid 3 bananas and no pineapples (and already sent the arrangement by PM) - so a total of just 3 fruit!

I won't ask anyone to beat it - but bid here if you can match it!

[abextra]

I'm still bidding 3 bananas
- and just can't believe we all got the same arrangement with 6 pineapples!!!

[davis_greatest]

Well, here's mine... totally different... lol!

Thanks abextra - well done too!

.....I hope your not going to tell us there is a solution for 2 bananas

Not quite, but:

Round 357 - Pineapple small print (continued from round 356)

As Charlie is about to pay for his new balls in round 356, it turns out that you and he hadn't read all the small print on the price tag in Barry's shop - as well as the bananas cost, Charlie may also have to pay with some pineapples.

Barry explains that in the triangle there are 4 pairs of adjacent rows - and since the triangle can be rotated 3 ways, there are 12 possible pairs of adjacent rows after allowing for rotation. Every time that Barry finds that the difference in points value between the two rows in one of those 12 pairs is different from the cost in bananas, Charlie will also have to pay a pineapple.

So in the example given by moglet, snookersfun and abextra, the cost would be 3 bananas and 6 pineapples = 9 fruit. Far too expensive.

(For example, the top two rows in moglet's and snookersfun's arrangement are worth 2 points and 4 points - the difference is 2 points - because this 2 is different from the banana cost of 3, those two rows contribute a cost of a pineapple.)

You need to help Charlie minimise the total number of fruit (bananas plus pineapples) he must pay to buy those 15 balls. Please bid here on the thread your total fruit number.

[abextra]

nice solution moglet!
wonder if there is another option at all

It seems there is no other way!