Puzzles with numbers and things

[snookersfun]

I got the same numbers as the others, but used the n(n+1)(2n+1)/6 formula and worked out the totals for 50 orangs and 49 baboons then for 100 orangs and 99 baboons, then looked at interpolating between the two. Like Monique, I went straight to 74/73 as an educated guess. This didn't take long, the extra time - many hours, was taken up trying to find a simple and direct formula to give the answer, without much success. Snookersfun's analysis is brilliant.

awwww, thank you moglet!!
But snookersfun spent many more or less fruitless hours as well using the above formula

and with this to

Round 389: no stopping the madness-hexagonal snooker

Being a bit bored of the regular snooker table, the apes ordererd a new table from Barry the Baboon. Oliver sets it up the moment it arrives (barely challenged by it after hefting those last 49 tables around) and the apes all gape at it contently.

It looks quite amazing and different, it still has those 6 pockets, but as the table is now a regular hexagon, all pockets lie evenly distributed at the six corners of the hexagon. In clockwise order we have yellow, green, brown, blue, pink and black pocket. They wonder a bit if the templates for the pockets should yet be altered again, but decide on a test run of the table nonetheless.

Just setting up a practice run, they start with the cue ball right in front of the black pocket, at which point Charlie does some calculations and asks his friends the following:

Look, just imagine you wanted to reach either the yellow, green or brown pocket from here (let's say either by as many shots as needed or by lining up balls), but say you could move only along certain lines, which are those formed by connecting all corners of the hexagon with each other (including those along the cushions). How many different ways would there be to reach each of those pockets if starting from the black pocket. But just one more thing, at no point during each path would you want to move farther away from the destination pocket (i.e. at each node only directions which lead closer to the pocket can be chosen).

And now I only hope that this is halfway clear, should you have questions though, feel free to ask on the thread.
Solutions (paths/pocket) please by PM

[davis_greatest]

Thank you for the answers to round 387 - here is another possible method of solution, which takes out the need for 'guesswork'

1) Let there be b baboons.

2) Then there are b+1 orang-utans and 2b+1 apes in all.

3) Let the tallest orang-utan (who happens to be called Othello) have n rows in his square, and n² balls in all.

4) Then, excluding Othello, for every k from 1 up to b, there is exactly one orang with (n-k)² balls and one baboon with (n+k)² balls.
The difference between these is (n+k)² - (n-k)² = 4nk

5) Adding all these differences (4nk) for every k from 1 up to b gives:
4n b(b+1)/2 = 2n b(b+1)

6) This total must equal the number of balls of Othello because the orangs and baboons have the same number of balls in total.
So, n² = 2n b(b+1)
i.e. n = 2b(b+1) = 2b² + 2b

7) The number of rows in Oliver's square is 10731 = n - b = b(2b+1)
= number of baboons x number of apes.

8) So we need to find b, where b(2b+1) = 10731.
Observe that √(10731 / 2) is slightly greater than 73 and, indeed, 10731 = 73 x 147 = 73 x (2 x 73 + 1). So there are b=73 baboons.

9) Therefore, there are 147 apes, and the number of baboons x number of apes is the number of rows in Oliver’s square.

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

And congratulations to Monique, moglet and snookersfun who all correctly identified that the answer to round 388 is 'no, you can't' - because it is, of course, impossible to draw such lines without crossing.

[snookersfun]

update to R. 389:
d_g and moglet sent correct answers for all three destination pockets. Very well done!

[davis_greatest]

Round 389: no stopping the madness-hexagonal snooker

Solutions (paths/pocket) please by PM

update to R. 389:
d_g and moglet sent correct answers for all three destination pockets. Very well done!

By the way, I only gave the numbers of paths to each of those 3 pockets - I didn't list them all, as I assumed that wasn't required (and wouldn't have the patience ) Does "paths/pocket" mean you want to see every path, or just the number of paths to reach each pocket?

[davis_greatest]

Round 389 is still open, but that doesn’t stop the apes having a go on the table meanwhile for…

Round 390 – Mexican Hexagon

Oliver, Charlie, Gordon, Barry, Gwendoline and their new friend Señor Gibbon decide to have some fun playing with their new hexagonal table from the previous round.

Each ape stands at one pocket and, whenever the cue ball comes to him or her, s/he knocks it to any one of the three apes across the other side of the table, choosing randomly each time, with each of the three apes being equally likely. So, no ape ever knocks the ball along the cushion to either of the adjacent apes.

Exactly 5 seconds elapse between shots.

The start is delayed slightly because it turns out that Señor Gibbon is out placing bets on when Oliver will play his 147th stroke! However, he is about to arrive back and Oliver will start the game and play his first stroke at the exact moment of this post (which I expect will be 12:34pm UK time).

What is the expected UK time at which Oliver will play his 147th stroke?

Answers either by Private Message or – for extra credit of one banana – you can post your answer here on the thread if the time you post it is the same as your answer!

[snookersfun]

Answers either by Private Message or – for extra credit of one banana – you can post your answer here on the thread if the time you post it is the same as your answer![/COLOR]

by any chance now?

[davis_greatest]

by any chance now?

No, not 3pm

[davis_greatest]

Update to round 390 - correct answers in from snookerfun and moglet so far - well done! Edit - and Monique too!

While round 389 and 390 remain open, I will throw in a similar round which may or may not help with round 390:

Round 391 - Hex Mex part II

From the moment that Oliver (who is at the black pocket) plays his first stroke (see round 390 above), what is the expected (i.e. average) number of seconds until the ball first arrives:
(a) to the ape on Oliver's left, at the yellow pocket?
(b) to the ape at the green pocket?
(c) to the ape at the brown pocket?
(d) to the ape at the blue pocket?
(e) to the ape at the pink pocket?
(f) back to Oliver at the black pocket?

[davis_greatest]

Any answers to rounds 390 or 391 can go on the thread please...

[moglet]

Rs 390,391

Over time and on average each ape will play the ball the same number of times, so each ape will receive the ball 146 times and play it 146 times, Oliver has already played his first shot and will have to wait 146x6x5/60 = 73 minutes before he receives the ball for his 147th shot.

Following the same principle that each ape, on average, will play one shot after Oliver's initial stroke before Oliver gets it back and that there are six possible routes, visiting each pocket once, back to "base":

E.g. for the ape to the left of Oliver, for each of the possible routes it would take 2,2,3,3,4 or 4 shots to reach his pocket if these are averaged, sum/6 we get 3 shots, so 15 secs
For the ape opposite Oliver it would take either 1,1,3,3,5 or 5, sum/6 3shots. And so on until finally, for Oliver to get the ball back will take one of six possible routes, each of six shots, 36/6 or 30 secs.

If this is correct the "average" time to each pocket after Oliver's first stroke are all equal at 15 secs, and 30 to get back to Oliver.

[davis_greatest]

Thank you moglet. The first part is fine.

Round 390 is correct - since the ball can get (in time) from any pocket to any other, and by the symmetry of the arrangement, the ball spends an equal amount of time on average at each pocket. Hence it takes 6 shots (30 seconds, or half a minute) to return to each pocket. Hence 73 minutes for 146 returns to Oliver, and he played his 147th shot at 1.47pm

-------------------------

Round 391 - some hints then, as there have not been any correct answers so far:

By symmetry:

(a) the expected (average) time to reach Oliver from the yellow pocket is the same as the time to reach the yellow pocket from Oliver (by playing "backwards"); and also the same as the time to reach Oliver from the pink pocket, and also the same as the time to reach the pink pocket from Oliver. Call this expected time A and label the two pockets on each side of Oliver A.

(b) Similarly, label the pockets two along from Oliver (green and blue) B - and the expected time to reach Oliver is B.

(c) let the expected time to reach Oliver from the opposite pocket (brown) C.

Now, look at where the ball can go from each of pockets A, B or C.

For example, from A there is a ⅓ chance the next shot will be from the other A, and a ⅓ chance it will be from a B and a ⅓ chance it will be from C. The ball can't go directly to Oliver from A.

The shot takes 5 seconds, so:
A = 5 + ⅓A + ⅓B + ⅓C

Do something similar for pockets B and C and you quickly get an answer...

[snookersfun]

By the way, I only gave the numbers of paths to each of those 3 pockets - I didn't list them all, as I assumed that wasn't required (and wouldn't have the patience ) Does "paths/pocket" mean you want to see every path, or just the number of paths to reach each pocket?

oh, forgot about this, but the number of paths/pocket are sufficient.

Here maybe a hint on how to do the counting without having to list all possible paths systematically.

Doing it here for the easiest, the paths to yellow.
hex-yellow.GIF
In the above picture (all credit to moglet,... and can't seem to upload at the moment so leaving the small gif in, please click to see the bigger version) one can first draw some one way signs pointing into directions bringing one closer to the destination pocket. Now one can go 2 ways about this, either start at the black pocket or the destination one and label that 1 (for each one way to reach it). In the picture I used the method starting at the destination.

So, I labeled that 1. Now I am looking at the closest point next to it, point C (or rather one with only one directional pointer leading closer to the yellow pocket), which will then still be labeled 1 (as there is only one possibility to reach yellow from there). The next point a bit farther out, B, can move via C and directly to yellow, so I add 1+1=2, A can only continue via B, so keeps B's 2 as its label. The next point out is the starting point, and now I add all ways from here out to the nodes 1+2+2= 5 possible ways.

I hope this is clear.
Now for the green and the brown situations one would have to put different pointers accordingly, but using the same method should be able to calculate the number of paths within minutes.

[Monique]

Round 392 - Rollie didn't practice enough

Rollie O'Sunnyman isn't happy! He just lost all the frames he played against Charlie, Gordon and Oliver His own fault of course: he hasn't practiced much recently )

Now as a reward for winning their frames, each of the apes kept all the balls they had been playing with except the cue ball so that Barry the baboon is over the moon: Rollie, the loser, needed to buy a new set for each frame!

It's funny, says Oliver, we all have the same nice number of balls, maybe we could imagine a lovely set-up for Barry's shop with all these balls?

Yes, says Charlie, we could for instance build a number of cubes with them, all different, and arrange those cubes in ascending order, starting with just one ball. That would be nice as their "edges" would be one ball bigger each time ...

Oh, says Gordon, I can think of at least two other nice arrangements with those balls

Meanwhile a very depressed and skint Rollie reflects that with the cue balls he still has ... he can't even paint them to make it a complete set and start practising again

How many balls do Gordon, Charlie and Oliver have each?
Can you think of nice/remarkable ball arrangements other than the cubes proposed by Charlie?

[davis_greatest]

Round 391 update - only correct answer so far, via PM, is from snookersfun (maybe with a teeny bit of help ) ... a reminder that answers can go on the thread. A hint - apart from part (f), the answers are not multiples of 5 seconds.

[snookersfun]

update on R. 389:

Mon already yesterday solved the yellow and the brown paths. Well done! Slight problems with the green still

With that we should probably close this round and solutions for green and brown can go up on the thread now.

and here another short one (we might have done something similar, but none the less...)

R.393: separating

Charlie, Oliver and Gordon by now had many practise hours on their hexagonal table. They come to the conclusion that the final colour clearance is too easy as it is, with all the colours one after each other around the table; the colour spots actually lying on a smaller inner hexagon (points A,C,E,G,I,K on moglet's graphic from my previous round). Charlie's plan is to separate at least pink and black by not having them on neighbouring spots, and see if that will show a difference in play.

So, how many different ways are there to do that? Out of how many different ways to arrange all colours on those spots without those constrictions?