Puzzles with numbers and things

[davis_greatest]

Good - let's have all bids please... even if someone has already bid lower.

[Sarmu]

Yea i worked out to be 111 this morning, but didnt' bother to post since someone worked out only 90

[chasmmi]

Ah it is 111 balls I thought it was up to 9 monkeys not 10

[chasmmi]

And there was me trying to seeing if any player had ever ranked 3 3 1 2 4 x

[davis_greatest]

OK - you can all have till a deadline of 20:00 GMT tomorrow (Tuesday 19 December) to keep bidding.

Any time from 20:00 GMT tomorrow, please all post here a detailed explanation of how you distributed the number of balls that you had bid before the deadline (even if someone else has bid lower). Then I'll decide how to distribute points.

If you won't be able to explain after 20:00 GMT tomorrow how you distributed the balls, then you can send me your explanation by Private Message any time beforehand, and I will then post it on the thread for you after 20:00 GMT.

If I think of anything else to ask tomorrow morning before I go off for a whole day of playing snooker (), I will also post it here.

Oh yes - another reminder - no editing of any posts that include an answer please. Edited posts will be void.

You are organising a game at Oliver's party. 27 of the guests have gone home, having had enough cake, so there are 10 apes left to play.

You need to take some snooker colours - as many as you want, but no reds, and there must be at least one yellow, one green, one brown, one blue, one pink and one black (so there can be more than one of each).

Then you must share the balls out between just two apes (Oliver and Gordon), so each has balls of the same total value.

For example, you might have one yellow, two greens, one brown, one blue, one pink and one black, and split them as follows:

Oliver: one yellow, one pink, one black - total value 2+6+7 = 15
Gordon: two greens, one brown, one blue - total value 3+3+4+5 = 15.


Then, you take some colours again (not necessarily the same ones as before), and must share them among 3 apes, so each has balls of the same total value.

Then again, but sharing among 4 apes, then 5, 6, 7, 8, 9 and finally among 10 apes. Each time, you take a new set of balls. So, in total, you must distribute 9 sets of balls.


You need to try to do this using as few balls, in total, as possible. Make your bids here, of the fewest total number of balls you think you can use (e.g. just bid "100" or whatever).

Once there has been enough time for the bids to come in, you will then be asked which balls you chose for each round and how they were distributed. But you should not give that information yet.

[davis_greatest]

Round 83.

I'm going to award half a point to davis_greatest because he is the only one who got 4 questions correct, answer + explanation.

Excellent.

5) This one proved difficult for everyone. The answer is this: Count the number of lines in the letters in the alphabet. A has 3, B has 3, C has 1, D has 2, E has 4, so the answer is 3 for F has 3 lines.

Even Charlie says he wouldn't have got that in a billion years.

[elvaago]

Round 83.

I'm going to award half a point to davis_greatest because he is the only one who got 4 questions correct, answer + explanation.

Below you will find all the answers and explanations.

1) The answer of course is 24. The explanation? red, black, red, black, red, black. ONLY ONE PERSON GOT THIS EXPLANATION! This is a snooker forum! :-D
2) The answer is 36. It's a series of squares.
3) The answer is 13. A+B=C, B+C=D, etc.
4) The answer is 65. A^2 + B = C, B^2 + C = D, etc.
5) This one proved difficult for everyone. The answer is this: Count the number of lines in the letters in the alphabet. A has 3, B has 3, C has 1, D has 2, E has 4, so the answer is 3 for F has 3 lines.

[abextra]

Oh wait, it's worse, I need 111 balls!

[abextra]

I can't manage with less than 110 balls.

[chasmmi]

ok now i have 90, the end

[davis_greatest]

Incidentally, you may still bid even if someone else has already bid lower than you can manage.

I'll still probably give at least a half point for decent, not necessarily winning, bids - and it could also be that someone has bid incorrectly.

[chasmmi]

dammit i can do 91 too, struck down by unfortunate idiocy

[snookersfun]

91, if I didn't calculate wrong just now

[chasmmi]

I'll bid 92 balls

[davis_greatest]

I'll put chasmmi's questions here and my answers, so everyone gets the same information.

"A few questions:

Can the total value cahnge from set to set or does it have to be say 15 every single time?"

Yes, the total value can change every time. Each time you need to take a new set of balls (the only requirement is that each time only colours are allowed, and there must be at least one of each colour).


"Is there anything to stop me just giving say all six colours to every ape every time?"

Nothing to stop you doing that. It probably wouldn't result in the fewest balls though.