Puzzles with numbers and things

[davis_greatest]

Round 171 hints

I've had a couple of answers to round 171 - both quite close but no one quite there. So, a couple of clues:

First, you need to work out how many chimps are at the party, to result in 171 hugs!

Second, there is a clue in the fact that these are (tango) dances. If you were at a party, which people (or apes, depending on your taste) would you consider dancing with? And is there any chance that you and the person you are dancing with might both have done the tango with the same "third" person? If not, why not?

Enough clues for now!

[davis_greatest]

Don't forget to do Monique's Round 170 Trophy woe (post 1805) above - as she says, it's not difficult!

And, once you've done that...

Round 171 - 171 Hugs at the Dancing Chimpanzee Tango Party

Charlie rolled in at 7 o'clock this morning and woke me up to tell me about all the fun he had just had at the Dancing Chimpanzee Tango Party.

At the party were Charlie and many of his chimpanzee friends, and they had a tango dancing competition. As the chimpanzees hadn't seen each other for some time (since the previous chimpanzee party - I forget which round), at the start of the party each chimpanzee had a hug with every other chimpanzee (i.e. one hug between each pair of chimps).

I don't know why, but apparently Charlie counted these hugs - and said there were 171 of them. But I digress - the real purpose of the party was the tango dancing.

Chimpanzee Tango Dancing is a bit like Baboon Fanta Dancing, except that each dance is done by two chimps instead of two baboons, and it is a bit more sexy. Anyway, every chimp did the tango with every other chimp except - and here's the funny thing - for some strange reason, no two chimps would agree to dance with each other if there were any other chimp at all at the party with whom they had both previously danced that night!

For example, if Charlie and Chrissa were about to dance together but discovered that there was another chimp who had already danced with both of them (say, Charlie had danced with Chump, and Chrissa had also danced with Chump), then Charlie and Chrissa would not dance together! No matter how good their moves!

Anyway, it all turned out OK and there were still plenty of dances, despite the chimps having that funny rule concerning with whom they would dance. In fact, it turned out that the maximum possible number of dances took place.

And - the most important thing of all: Charlie won the prize for the best dancing! All that time watching me finally paid off.

How many tangos took place last night at the Dancing Chimpanzee Tango Party?

[davis_greatest]

Round 167 answer

Gordon: 6 balls - all colors except the yellow + 1 red - value 26
Charlie: 7 balls - yellow + 6 reds - value 8
Oliver: 9 balls - 8 reds + cue ball - value 8

... is the correct answer! Well done Monique and snookersfun!

So....

...If Oliver has 7 reds and the white, 7 points, and Charlie has 6 of the remaining reds and the yellow, 8 points, then Gordon would have the last two reds and the five remaining colours, 27 points?

see how close you were, icantplay! I told you you were nearly there! You just had to pass one of those reds from Gordon to Oliver, so Gordon had one red and Oliver eight...

[Monique]

R170 Trophy woe

Was short-lived! Congratulations to Davis_Greatest and Snookerfun who solved it already. It remains open ... it is not difficult!

[Monique]

Round 167 as requested ...

Gordon: 6 balls - all colors except the yellow + 1 red - value 26
Charlie: 7 balls - yellow + 6 reds - value 8
Oliver: 9 balls - 8 reds + cue ball - value 8

[davis_greatest]

Thanks Monique. And the answer to round 167?

This evening, I went to play snooker. However, it was a rather boring affair and I struggled to make any sizeable breaks, as all the balls were missing. Eventually, I found them - my three apes were playing with them and had shared them out.

They had, among them, the complete set of 22 balls, and each ape was adding up the value of his balls (1 for a red, 2 for a yellow, 3 for a green etc, with no value for the cue ball).

Although my pet gorilla Gordon had the fewest balls (and Oliver, my little pet orang-utan had the most), the value of Gordon's balls was at least 3 times the value of my pet chimpanzee Charlie's. Nevertheless, the value of Charlie's balls was at least as great as Oliver’s.

What balls did each ape have?

[Monique]

Round 168

Gordon: Yellow + 4 reds; 5 balls; 6 pts
Oliver: Brown + 3 reds; 4 balls; 7 pts
Charlie: everything else; 13 balls; 29 pts

Charlie had over 3 times as many balls as Oliver
yes 13> 3*4

and the value of Charlie's balls was over twice that of the balls of Oliver and Gordon combined! Cheeky Charlie!
yes 29> 2*(6+7)

(Mind you, Gordon's balls were of less value than Oliver's, but don't tell Gordon that.)
yes 6<7

The ape who ended up with the yellow ball had one red more than the ape with the brown ball,
yes 4=3+1

whereas the number of balls held by the ape with the black ball was less than 3 times the number held by the ape playing with the yellow ball.
yes 13 < 3*5

I do remember that the value of Gordon's balls was either greater or less than the number of balls he held, but I don't recall which.
well value=6 > number=5

[davis_greatest]

Rounds 167 and 168.... correct answers from Monique and snookersfun - congratulations again! Someone please post on this thread the answers!

Monique - round 170 - hehehehe. I'll pager / PM you my answer...

[davis_greatest]

I'll bid 92 points:

...O
OOOO
.OOO
OOOO
...O

2 × 10 = 20
6 × ..6 = 36
12 × 3 = 36

Congratulations, The Statman (and the others who then got the same arrangement)! This is the best arrangement possible - a Star of David formation with the 13 balls.

If it were not for the fact that you are already in the Puzzles with numbers and things Hall of Frame, you would be entering it now!

[Monique]

R170 Trophy woe

Every tournament needs a trophy. The Belgian Open in no exception ... After many rounds of discussion the "Trophy Committee" finaly opted for the "Snookatom", a design clearly inspired by the Atomium (thus a Belgo_Belgian affair ...)

The snookatom is built of the 22 snooker balls (6 colours, 15 reds, 1 cue ball) and a number of edges. 2 balls are linked by no more than one edge. An edge always links exactly two balls.

O---O OK; O===O NOK; O---- NOK

Each ball is the vertex of exactly the number of edges corresponding to its value (ex: pink is at the extremity of 6 edges). The cue ball has been given a value of 1.

Now, while the players are lining up for the Snookesnoops Candy Fest, the "Trophy Committee" is facing a crisis. Apparantly the Snookatom designer has fled to Brazil ... However, in a bout of honnesty, he's sent his unfinished Snookatom to the Committee by mail. The box with the Snookatom spare parts is now on the table, and the honourable members of the Committe try to figure out how to assemble the thing!

There is just one lady who keeps telling everyone that the Snookatom is not "feasable". Well, women are useless with maths and technique, so nobody actually listen to her. But tension is building ...

Before anything really unfortunate happens, the TSF help desk is asked:

Either to explain how the Snookatom can be built
Or to actually prove it can't be done

[Monique]

You've lost me. There are no more triangles. There are only 2 triangles of 10.

If you were to count one 10-triangle for each of the six vertices of the star, you would be counting each triangle 3 times - for each triangle has 3 corners!

I've lost myself (not you) indeed! Too many point of views, but the same triangles.
But I still think there is no better bid. If you look at the two 'ten' triangles - from any perspective (or vertex) - they intersect and have, and must have, a minimum 7 elements in common. Here, only the 3 vertexes from each triangle are not "reused". I don't see how you could possibly have seven elements in common, including one vertex, or more than seven elements in common.

[davis_greatest]

In fact if you look at this properly - I mean if you use real circles - you see that there are more triangles than counted by Statsman as this is a "David Star" with a "six" symmetry ...

So you have 6 "Ten triangle" for example, one for each vertex of the star, and this is a much higher bid!

You've lost me. There are no more triangles. There are only 2 triangles of 10.

If you were to count one 10-triangle for each of the six vertices of the star, you would be counting each triangle 3 times - for each triangle has 3 corners!

[Lee Vilenski]

Wow, so Thats well over 100 yeah?

[Monique]

A fine submission, The Statman! In fact, this was the pattern I had in mind originally when I asked you to use 13 balls. My instinct would be that this is the best possible - although I haven't thought about it yet to see if this can be proved, and would be delighted to see anyone better it!

In fact if you look at this properly - I mean if you use real circles - you see that there are more triangles than counted by Statsman as this is a "David Star" with a "six" symmetry ...

So you have 6 "Ten triangle" for example, one for each vertex of the star, and this is a much higher bid!


I share D_G feeling that this is the best achievable ...

[davis_greatest]

Er....

---0
--000
-00000
00000
Er...

Thats 10x2
3x8
6x5

20+ 24 + 30 = 74... Not a bad effort!

ok, changed, doesn't change the outcome.

Um.... haven't you now just moved one ball from one row to another from your original drawing - there are still 14 balls there!