Puzzles with numbers and things

[davis_greatest]

I think it is perhaps time to close round 368. Congratulations to abextra, snookersfun, moglet and Monique, who all found the maximum break of 123 with the aid of a free ball!

Round 369 will remain open for another day.

Round 370 - Gorillooker Cue Patrol

Gordon is playing a game of Gorillooker against Oliver - just like normal snooker except that the triangular pack of reds (while still an equilateral triangle) need not necessarily have 5 rows.

Gordon makes a neat total clearance (potting a colour after every red, and no free ball). The funny thing that Oliver noticed was that, at all times during Gordon's break, the total points that Gordon had scored from any one colour (up to the pink) was always at least as great as the total points he had scored from all higher valued colours combined.

(For example - the total points he had scored from potting browns was always at least as great as the total points he had scored from potting blues + pinks + blacks.)

Gordon had made the triangle of reds at the start of the frame the smallest possible that would allow him to achieve this feat.

What was Gordon's break?

Once again, please post your answer in hidden text, like this (deleting the space after the two open square brackets):

([ COLOR=#f1f1f1]
Enter break here, e.g. 147! [ /COLOR])

Just put up the value of the break for now - later you will be asked to explain why it is so!

[Monique]

Although I'm no certain I got the explanation correctly here is a bid


with 136 reds a break of 509 :snooker:

I supposed that at any given moment in time during the break (including just after last black sinked in) the points gained on colour "n-1" where superior to the cumulated points gained from colour n and above.

[snookersfun]

I for one don’t seem to understand at all, as it seems to be possible to get various breaks, but here are two options

a black anywhere in the break before the colours: 264 (hopefully counted right)
Otherwise 30?

[snookersfun]

and in case that 'at all times during Gordon's break' means something like 'overall'
444:snooker:

[moglet]

R370

According to my interpretation:

264 from an 11 row pack

[moglet]

R370

Aside from the written conditions and the size of the triangle can we have confirmation that all other snooker regulations are as normal and it is a standard full size table?

My other interpretation at the moment depends on the above.

364 from 15 row triangle, this is not right I didn't total the yellows properly, number of balls not their value, so it should be 444 from 15 row triangle

[davis_greatest]

Thank you for the various answers / attempts to, and questions on, round 370, here and by private message.

It seems to have caused rather more confusion among a number of entrants than I had anticipated, so to be fair to all and ensure that everyone has the same clarification, I shall post a few summary points here arising from some of the queries received and/or submitted solutions:

1) "at least as great as" means "greater than or equal to". So, for instance, 6 is at least as great as 5, and 5 is also at least as great as 5.

2) "at all times during Gordon's break" means at every point in time. So it includes the start of the break, the end of the break, and every time in between, including before, during and after every stroke.

3) "the total points that Gordon had scored from any one colour (up to the pink) …" - this must apply for each and every colour up to the pink! And it must apply at all times (see above)!

4) Nowhere is it specified that all colours or any particular given colour must be potted before the final colours sequence.

5) "at all times......was always at least as great as the total points he had scored" refers to… "the total points he had scored up to that point in time" - not the total points that he might go on to score in future.

6) All snooker regulations are normal, except that:
(a) there may be a number other than 5 rows of reds;
(b) the table is large enough to fit however many reds are required - in the event that that should mean that it must be larger than normal, or the balls smaller than normal, then so be it!

[moglet]

Thank you in return for these clarifications, I, for one, am completely exhausted now and this will have to wait until I am restored by some sleep.

[abextra]

Thank you for the explanations.

[davis_greatest]

Round 370 update

I supposed that at any given moment in time during the break (including just after last black sinked in) the points gained on colour "n-1" where superior to the cumulated points gained from colour n and above.

Monique - remember the wording was "at least as great as", not "greater than" - so do you want to rethink your answer?

Other than that, there are various answers on the thread - the last hidden answer from snookersfun and the last (edited) hidden answer from moglet are both correct - congratulations!

Apologies if I have missed any other answers that were too well hidden! Let me know if so?

Shall we close this round then and move on? Let's say, any more answers to round 370 to go straight up on the thread please without using hidden text.... preferably with a simple explanation of how to solve it (to save me from putting one up ).

[davis_greatest]

While round 370 stays open a little longer, here are the next two rounds, appearing together, for the weekend. (Answers by Private Message please - so no one can peek at hidden answers )


Round 371 Respot that!

Gordon and Charlie have been playing a few shots on the respotted black in their final frame decider.

From experience, I know that on every shot in such situations, Charlie attempts a spectacular double, never fouls, and has the same constant likelihood of success.

The same can be said of Gordon, except that he attempts trebles and so is less successful.

Charlie's chance of potting the black on any shot is the same as his opponent's chance of missing.

The same, of course, can be said of Gordon.

As Charlie addresses the cue ball for his next shot, I quickly calculate that he is 6 times as likely as Gordon to win. Unfortunately for him, he misses - although I am confident that the noise of my abacus did not distract him - and so Gordon now approaches the table.

Who is now more likely to be victorious - and what is that ape's chance now of winning?



===============================

…. Meanwhile, on the adjacent table….

Round 372 - Psychic Oliver?

While I was keeping an eye on Gordon and Charlie’s final frame respotted black shoot-out, I was also having a frame with little Oliver this evening. We have a custom that whoever is not at the table should keep score.

Well, Oliver broke off, left me a long red on, which I potted, followed by a colour. As soon as he had seen those first two shots of mine, Oliver popped out for a banana and left me to continue my break, and arrived back just in time to see me pot the final black!

“Oooh," he said. "Another total clearance?”

“Of course!" I replied "15 reds, each with a colour, and then the final 6 colours – again!”

“How many did you score this time?” asked Oliver, who really should have been watching.

“Well, you were supposed to be scoring, not eating bananas," I replied. “But, put it this way - for each of the six colours, the total points I scored from that colour equalled the total points I scored from one of the other colours.”

“Wow!” said Oliver, apparently less excited than I had thought he would be by this news.

“And, what’s more,” I continued, “From the 4th colour onwards (which happened to be a yellow), I never potted the same colour as the previous one, nor as the one before that. “

And that, it turned out, was enough information for Oliver to tell me every single ball I had potted during my break, including the order!

What was my break (total, and also listing every ball in order potted)?


Answers to both rounds by Private Message please.

Any questions or requests for clarification, on the thread please.

[Monique]

R 370 ...

Looking at the situation at the end of the break:

- at least one black has been potted, for 7, so at least 2 pinks have been potted
- because that yields at least 19, we must have at least 4 blues, adding 20 points
- we now need at lest 10 brows, adding another 40 points...

going on we find that we needs at least 1 black, 3 pinks, 4 blues, 10 browns, 27 greens and 80 yellows. This yields a minimum of 118 pots with "reds". Therefore the smallest triangle must have 15 rows (and 120 reds).
Now if we add 2 reds with yellows to "consume" the 120 reds we reach a break of 444
However if were allowed to start with a free ball we could also add a green after those 2 yellows for 448 :snooker:

[davis_greatest]

Thank you Monique (note, 2 pinks, not 3, in the final paragraph - and free balls were expressly forbidden). It was indeed, a break of 444 - with a triangle with 15 rows and 120 reds!

The way to solve it was indeed to start at the highest value colour and work your way down (knowing that there must always be at least 1 black in a total clearance!), finding the minimum number of each colour required.

So, to complete the answer above:

1 black, giving 7 points - total so far: 7 points
So need 2 pinks (to be >=7 points), giving 12 points - total so far: 19 points
So need 4 blues (to be >=19 points), giving 20 points - total so far: 39 points
So need 10 browns (to be >=39 points), giving 40 points - total so far: 79 points
So need 27 greens (to be >=79 points), giving 81 points - total so far: 160 points
So need 80 yellows (to be >=160 points), giving 160 points - total so far: 320 points

That has given 1+2+4+10+27+80 = 124 colours (being 118 with reds plus the 6 colours). The next triangular number after 118 is 120, corresponding to 15 rows - and the problem only works if those extra 2 reds are also potted with yellows, giving 82 yellows potted in all.

To the 320 points above, add 120 for the reds, and 2 for the 2 extra yellows, and we get 444.

Congratulations to all who submitted the 444 break - snookersfun, moglet and Monique!

[abextra]

…. Meanwhile, on the adjacent table….

Round 372 - Psychic Oliver?

...
And that, it turned out, was enough information for Oliver to tell me every single ball I had potted during my break, including the order!

What was my break (total, and also listing every ball in order potted)?

Any questions or requests for clarification, on the thread please.

It is not enough information for me - obviously I'm missing something... I see more than one possible break and order!

[davis_greatest]

R371 update - one correct answer so far - it is still open, and anyone who has solved it correctly will be named here when the round closes!

R372 update - I think a few are struggling slightly with R372, having found a few breaks, and not knowing which one to choose. There is, however, only one possible answer to this problem!

So, in order to help see how to address R372, here is a mini (much easier) round 373. Once you have answered it, it should help you with R372. It is open to everyone except the one successful solver (edit - now 2 successful solvers) so far of R372.... answer on the thread, please.

Mini Round 373 -

Charlie, Gordon and Oliver know that my box contains 3 red balls and 2 pink balls, and each ape takes one ball, which he looks at (unseen by the others). Gordon then secretly shows Oliver his (Gordon's) ball, at which point Oliver announces that he knows which ball Charlie has too.

What colour ball does each ape have?