Puzzles with numbers and things

[abextra]

Congratulations on correct answers put up to round 357 by moglet and snookersfun; and to abextra who had answered by Private Message within minutes of the question going up - abextra, please put yours up too!

They are, indeed, all the same as each other (and as mine ), except for reflections / rotations... :snooker:

Here it is... was very lucky this time, this solution came to me itself.

[davis_greatest]

Round 359 - Square of balls

For his recent Chimpmas present, I gave Charlie a nice new snooker cue. The previous year, I gave him an empty square box whose interior is the width of 17 snooker balls and whose height is the same as one snooker ball.

Last January, Charlie entered our local Chimpsnook league for snooking apes. In the league, every ape plays every other one once, and is awarded 2 snooker balls for a win and one for a draw. The 4 apes with the most balls at the end qualify for the semi-final play-offs, due to take place this weekend.

During the season, Charlie put all the balls he was awarded into his box, lined up in rows and columns, and is delighted that he managed to fill it (with a 17x17 square) in time for the play-offs. The reason he is delighted is that he knew, before his first match, that he had to fill the box to be guaranteed of making the semi-final and so now knows that he must have qualified.

How many apes played in Charlie's Chimpsnook league?

Answers initially by Private Message please...

[davis_greatest]

Update on round 359 - 4 answers have come in:

one is close (but not quite there), one has the correct number but the explanation / reasoning is a bit dodgy (), and the other two also have the correct number and need to send an explanation.

Perhaps it would be easier to come back to round 359, after we first turn the question around with:

Round 360 More Chimpsnook

If there are c chimps in the Chimpsnook league (as played in round 359), how many points / balls does a chimp need to be guaranteed of finishing in the top t chimps (irrespective of how the other chimps do) after all the round robin matches?

Answers please (expressed in terms of c and t) directly on the thread.

[abextra]

Round 360 More Chimpsnook

If there are c chimps in the Chimpsnook league (as played in round 359), how many points / balls does a chimp need to be guaranteed of finishing in the top t chimps (irrespective of how the other chimps do) after all the round robin matches?

Answers please (expressed in terms of c and t) directly on the thread.

Maybe 2c-t-1?

[abextra]

or 2(c-t)+3?

I have to say that I'm totally confused - everything seemed so nice and clear and the answer I got was correct, but now when I started to explain it...

[snookersfun]

no, deleted mine already, didn't quite fit...

[Monique]

reasoning had lead me to 2*(c-t) +1 ... that's "close" but not correct. :snooker: thinking makes me think 2*(c-t) + 3 is indeed correct but don't see why

[snookersfun]

reasoning had lead me to 2*(c-t) +1 ... that's "close" but not correct. :snooker: thinking makes me think 2*(c-t) + 3 is indeed correct but don't see why

but it doesn't fit the top 2 case, does it?
so rather maybe 2(c-(t/2+1))+1= 2c-t-1, which abextra had all the time

[Monique]

So ... what about 2*(c-t) + t-1 that would fit ...

[Monique]

sorry ... collision. yes it's the same.

[abextra]

The draws are very confusing, aren't they?

[snookersfun]

maybe moglet can come up later with a 4th equal equation
night!!

[davis_greatest]

OK, round 360... yes, it is indeed 2c-t-1. We can prove it like this:

First, what is the highest score p that does NOT guarantee being in the top t places? This would be the highest score that at least the top t+1 players could all achieve.

These t+1 players play, in aggregate, (t+1)t/2 matches involving each other and (t+1)(c-t-1) matches not involving each other - adding these gives (t+1)(c-t/2-1) matches in all. Since each match gives 2 points, that is a maximum of (t+1)(2c-t-2) total points available to those t+1 players - so, dividing this by t+1 we see that the highest score p that the top t+1 players can all achieve is p=2c-t-2.

We must check (and can easily see) that it is possible for the top t+1 players each to score p. For example, they could draw all their matches against each other (giving each player t points) and win all their matches against the other c-t-1 players (giving a further 2(c-t-1)) = 2c-2t-2 points). (Note this is not the only way.) They would then each score t+2c-2t-2 = 2c-t-2.

Now, if any player scores at least p+1 = 2c-t-1, then at least one of the top t+1 players must score less than p. This means that scoring p+1 is sufficient to guarantee being in the top t places. So 2c-t-1 is the correct answer to round 360.

For round 359, we just have to solve 2c-t-1 = 17x17 = 289, where t=4. This gives c=147 players.

Congratulations to moglet, abextra and snookersfun, who all gave this as an answer somehow () and to Monique who found an extra chimp from somewhere to get 148.

[moglet]

Why I hate round robins

A guaranteed outright winner would have to win ALL his games and get 2(c-1) points (he can't play himself! and would leave a possible 2(c-2)pts for a clear second place...and so on), if there were two tied and guaranteed for first place then they would both have 2(c-1)-1pts, and for a three way tie 2(c-1)-2pts so for a "t"way tie each would have 2(c-1)-t-1pts.

Simplest expression I can see is 2c-t-1

Charlie needs to guarantee he has sufficient points to join the top group if there is "t"way tie for the top place.

I think, I'm still not sure..... so lets have the lowdown from d-g

[davis_greatest]

Round 361 - Charlie's box

How many snooker balls can Charlie fit into his Chimpmas box of round 259?

Answers by private message please.