Puzzles with numbers and things

[Monique]

If I read that right, the number of balls would be (answer invisible) -> 588

What puzzle are you answering?
I reckon it's R400 and if it is ... no that's not it. Please send me a PM with the reasoning that lead you to that answer and I can explain better if there is any misunderstanding.

Thanks for trying! Keep on it.

[Monique]

Abextra now solved R400 perfectly. Well done!
I'll leave it open two more days...

Meanwhlie Snookersfun answered R401 also perfectly. Congratulations

[neptar]

Sorry, I read it wrong. Also, shout we be PM'ing you the answers?

[Monique]

Sorry, I read it wrong. Also, shout we be PM'ing you the answers?

Yes, please do. This is the best way not to spoil the fun while allowing to be a bit more "explicit" when needed.

[abextra]

On Monday ...
Barry the Baboon has got a great idea to promote his shop... he's organising a challenge for the visitors - they dont have to buy anything...

...I want to make this challenge much more difficult and have the visitors to try to rearrange some big square into 147 diffrerent smaller ones. Is that possible? How many balls do I need?"
"Ha!" answers Charlie "Yes it is possible. In fact it is possible for any number of squares you want ... and there are many, many solutions for each number"
Charlie smiles mischieviously and asks "How many [ATTACH]2500[/ATTACH] will you give me if I help you?"
"Eehrr, 10" says Barry.
"Pah!" exclaims Charlie "I'm not interested! Ask those TSFers "
The shop's door slams shut, and he's gone ...

Who can help Barry?

I would be happy with these 10 bananas but I have no idea how to help Barry! I only guess he needs more than million balls, that's all. Any hints?

[Monique]

Moglet now solved R401 alswell with some advanced math research! Well done Moglet.

There is a simple way to solve this though ...
So here is a hint:
If you want to bulid a big square that is "splittable" in n smaller ones, start with n-1 squares that you chose all different, then try to find a nth one so that the total sum is a square again.

[abextra]

Moglet now solved R401 alswell with some advanced math research! Well done Moglet.

There is a simple way to solve this though ...
So here is a hint:
If you want to bulid a big square that is "splittable" in n smaller ones, start with n-1 squares that you chose all different, then try to find a nth one so that the total sum is a square again.

Thank you, Monique!

I've tried that, no success so far... ... obviously I've misunderstood something again.
I'm glad Barry has got some ideas from Snookersfun and Moglet!

[Monique]

I think it's time to close R400 ... So if anyone has a solution to offer please post it in the thread. If nobody does by tomorrow late evening, Snookersfun, Moglet or Abextra, please publish yours. Thanks and well done again!
R401 still open for a little while

[Monique]

R402 ... little pranksters

Charlie, Gordon, Oliver and Gwenny just arrived in Barry's shop and there in the middle of the shop is a huuuge pile of balls.

"Aaaah!" exclaims Barry "Good your lot is there! I just recieved all those premium quality snooker ball from Belgium and I must clean and store them ... BUT ... I promised Rollie to play some snooker with him. You know those humans, they are so slow! I can't close the shop for the rest of the day and I don't want those balls to disappear! Could you stay and keep the shop open for me while I play?" And without waiting for the answer he's gone!

Now his four friends are a bit annoyed. They also had planned some game of snooker :snooker: However they don't want to let Barry down. So they decide that they will keep the shop in turn while the three others go and play.

Charlie is first to stay in the shop and he's soon bored. So he decides to split this big pile of balls into 147 smaller piles et finds out he has one remaining ball. He decides to keep this one for himself, hides one of the smaller piles in Barry's cellar and jumbles back the 146 remaining piles into one big pile. Shortly thereafter Oliver is back to take his turn...

As it happens the apes' mischievious minds all work the same way, and soon Oliver is bored and - guess what - starts splitting the big pile into 147 smaller piles finds out he has a remaining ball, keeps it for himself, hides one of the small piles in Barry's cellar and jumbles back the 146 remaining piles into one big pile.

As it happens when Gordon and Gwenny take their turn ... they do exactly the same!

What's the smaller possible size of the original big pile?
What's the size of each of the smaller piles hidden in the cellar?

[abextra]

I think it's time to close R400 ... So if anyone has a solution to offer please post it in the thread. If nobody does by tomorrow late evening, Snookersfun, Moglet or Abextra, please publish yours.

Rollie and Gwenny are playing snooker for fun, only they play an "ape" version of the game, meaning that they have a much bigger triangle of reds than usual ...
....
1. How many frames have they played?
2. How many reds do they have on the table?
3. Why is there no need for Gwenny to compute anything?

1. Rollie and Gwenny just have finished 14th frame - the score in this frame was 377 points and in total they were gathered 986 points.
2. There were 120 red balls on the table - so the maximum would be 987 points.
3. I better leave this question for someone else to answer.

[moglet]

1. Rollie and Gwenny just have finished 14th frame - the score in this frame was 377 points and in total they were gathered 986 points.
2. There were 120 red balls on the table - so the maximum would be 987 points.
3. I better leave this question for someone else to answer.

I had a silly solution to start with, trying to keep the numbers realistic, but alas, in the ape world there are no such restrictions,

1, Frame 14 (986 total of breaks)
2, 120 reds (987 max. break, frame 16)
3, Fibonacci numbers

3, was not enough so I offered this:

3, The series has this property amongst others: the sum of n numbers in the series is always one less than the (n+2)th number. If I've phrased it correctly.

Seemed to be what was looked for.

[moglet]

R402

Charlie is first to stay in the shop and he's soon bored. So he decides to split this big pile of balls into 147 smaller piles et finds out he has one remaining ball. He decides to keep this one for himself, hides one of the smaller piles in Barry's cellar and jumbles back the 146 remaining piles into one big pile.

Monique , can you confirm that the smaller piles that Charlie makes are of equal size?

[Monique]

Monique , can you confirm that the smaller piles that Charlie makes are of equal size?

Yes each ape makes 147 piles of equal size out of the big pile they have to "guard"

Also regarding the property of the Fibonacci numbers, it's quite easy to prove that sum( i=1 to n; Fi) = Fn+2

the series starts with 1,1,2,3,5,8, ... so it's obviously true for n=2,3,4
the idea is to prove that if it is true for a certain value of n then it is also true for n+1; this by recurrence will prove it's true for any value.

so: sum(i=1 to n+1; Fi) = sum(i=1 to n; Fi) + Fn+1
because we assume the property is true for n this gives us
sum(i=1 to n+1; Fi) = Fn+2 + Fn+1 and this is Fn+3 by definition of the Fibonacci numbers.
hence the property is true for n+1

[Monique]

Snookersfun and Moglet both solved knotty R402... well done!

[Monique]

And now Abextra has solved R402 perfectly aswell... Well done!

Solutions for R401 and R402 on the thread anyone?