Puzzles with numbers and things

[davis_greatest]

I think they can finish in 1h 15 minutes, three pairs can play at the same time?

OK - can you please demonstrate how the pairings could go to get them all done in 1hr 15 minutes? E.g. suppose the players are A, B, C, D, E and F...

[abextra]

Here you are

A-F , B-E , C-D
A-E , B-D , C-F
A-D , B-C , F-E
A-C , B-F , D-E
A-B , C-E , D-F

[davis_greatest]

Here you are

A-F , B-E , C-D
A-E , B-D , C-F
A-D , B-C , F-E
A-C , B-F , D-E
A-B , C-E , D-F

Yes, that works perfectly. So that proves that 1hr and 15 minutes is possible, and it is easy to see that it cannot be done any quicker than that, so that proves that 1hr 15 minutes is the minimum time possible.

Any ideas on solving part b), finding the "groups of 3"?

[snookersfun]

I see, I totally clubbered this one, didn't I.

[abextra]

You're doing great, snookersfun! I've got no ideas about part b), have you?

[abextra]

Why can't they all play with strawberry jam? It would be so much tastier and less painful!

[Vidas]

Why can't they all play with strawberry jam? It would be so much tastier and less painful!

they will grow fat from so much strawberry jam - some physical activity should be there as well!

Some ideas about part b):

Player (A) has two options for each opponent and five opponents
(to smack or give pot of jam).
so (at least) 3 of his partners(say B,C,D) got the same(say smacking).
if any two of B,C,D smacked each other -
they and A are the group,
if not, (B,C,D) are the group(pots of jam between all).

P.S. I can't make out - can there be exactly one such group? I always get at least two...

[davis_greatest]

Pot Smack - The Morning After: solved!

they will grow fat from so much strawberry jam - some physical activity should be there as well!

Some ideas about part b):

Player (A) has two options for each opponent and five opponents
(to smack or give pot of jam).
so (at least) 3 of his partners(say B,C,D) got the same(say smacking).
if any two of B,C,D smacked each other -
they and A are the group,
if not, (B,C,D) are the group(pots of jam between all).

P.S. I can't make out - can there be exactly one such group? I always get at least two...

Nicely done, Vidas! Well done! This indeed proves that there must be at least one group of at least 3 players, who all mutually gave each other jam or all mutually smacked each other's faces. In fact, this particular one is a well-known problem.

You said you always get at least two groups - that's right - there will always be at least two such groups (although they may be overlapping, i.e. there may be a player who is in both groups).

I suppose I had better think of another question! ...

[davis_greatest]

Question 11 - Pot Smack Marathon

Encouraged by the incredible success of my Pot Smack tournament, I've managed to get the BBC to secure uninterrupted coverage of my Pot Smack Marathon, which runs over the whole autumn.

The Pot Smack Marathon consists of the 3 most popular players on the circuit: Maun Shirty, Damon Grott and Ebony Petdon. Each player has his own table, and at the beginning of the autumn, the host John Vertigo says "Pot as many balls as you can." They then spend the whole autumn potting balls as fast as they can, an activity ideally suited to the speed machine Petdon in particular, while Vertigo puts the balls back and keeps count.

At the end of the autumn, each player is given a number of points - one point for every ball he potted. It comes as no surprise to me to see that Petdon potted the most, while Grott, who struggled a little in keeping position, potted the fewest.

I am, however, surprised by how similar the scores are, and even more surprised to notice that Shirty's score is exactly halfway between Petdon's and Grott's. Even more astonishingly, the number of balls that Shirty potted is a perfect square, in stark contrast to the shape of his face, which is a perfect circle.

Now comes the fun part, which all the audience has been eagerly awaiting. Each player's score is cubed, to give his Cubescore. Petdon, whose Cubescore is of course more than Shirty's, then gives (with the help of the audience) Shirty one smack for every point difference in their respective Cubescores. The use of Cubescores is, of course, an element of genius on my part, since it allows for far more smacks to be given. The sponsors pay me handsomely for this.

Then Shirty, whose Cubescore is greater than Grott's, gives Grott (with much help from the audience) one smack for every point difference in their respective Cubescores.

The game seems slightly unfair because Shirty, who potted more balls than Grott, actually gets smacked more times than Grott. However, none of the audience complains about this, and the sponsors appear delighted and sign me up to organise the competition again next year.


I tell my friend about the whole happy event. "How many more smacks than Grott do you think Shirty got?" I ask her.

"I don't know," she says "147 million?"

"Not a bad guess," I reply. In fact, Shirty received precisely 148,862,166 more smacks than Grott."


So.... how many balls did each player pot during the Pot Smack Marathon?

[snookersfun]

lol
that was worth reading. I got to work today... not sure I get to this... will be stuck in my mind though now... thanks

OK, didn't leave me alone and I reached something, not finished as I have to pick a kid from his first day at school

anyway under here hidden (hopefully):

one has eventually devide 24810361 by a perfect sqaure (# of pots by the middle guy- and that # I still have to find- no time now) to get the sqaure of the difference in pots between either one and the middle one.

[snookersfun]

Oh, the last part was easy :
blunder corrected now...

Grott 24.810.360
Shirty 24.810.361
Petdon 24.810.362

unless there are more solutions (kind of unthinkable looking at it), I just stopped after the first possibility...

Solution (you will ask, won't you):
one can elegantly (always looks so easy after breaking one's head for an hour) write the story up in the following equation:

(x+a)^3 -2x^3 + (x-a)^3 = 148.862.166 (x being Shirty's pots and a being the difference in pots)

this will then resolve into:
6a^2x = 148.862.166 or a^2x = 24.810.361

bearing in mind, that x has to be a perfect square, one is just reduced to the problem of finding two squares multiplied yielding that # ('a' being rather small). The above # already is a perfect square, thus a=1 and x= 24.810.361 .

[davis_greatest]

Almost there, snookersfun.

However, although there are exactly two sets of numbers that are theoretically possible (and you have given one), I had decided when I wrote this that the 24 million+ pots answer that you have given would not be allowed. This is because it would not be feasible to pot so many balls during the autumn. Even if they played 24 hours a day every day, that would need over 3 balls to be knocked in every second, which is faster than even Petdon can play.

So you need to find the other answer - which is smaller than this!

[snookersfun]

just came back to the computer
a=17, therefore x=85849
Grott 85832
Shirty 85849
Petdon 85866

makes more sense (ca. 300 pots/per hour for just 3 h play through the season?)

[davis_greatest]

Well done, snookersfun!

Bingo! That is the correct answer.

I'll write out your solution and fill in some of the gaps:

Suppose Shirty makes x pots. We are told x is a perfect square. Suppose x=j^2

Shirty's score is the average of Petdon's and Grott's. So for some "a", Petdon made x+a pots and
Grott made x-a pots.

Petdon's Cubescore is (x+a)^3, Shirty's is x^3 and Grott's is (x-a)^3

Petdon gives Shirty [(x+a)^3 - x^3] smacks.
Shirty gives Grott [x^3 - (x-a)^3] smacks.

Shirty receives [(x+a)^3 - x^3] - [x^3 - (x-a)^3] more smacks than Grott,

i.e. 6x.a^2 = 6 (j^2) (a^2), which I told you equals 148,862,166

So (j^2)(a^2) = 148,862,166 / 6 = 24,810,361
giving j.a = 4,981

The only factorisations of 4981 are:

4,981 = 1 x 4981, or
4,981 = 17 x 293 (snce 17 and 293 are both prime numbers)

Since Grotty made j^2 - a pots, we know that j^2 > a.
Therefore the only possibilities are:

a=1, j=4981, or
a=17, j=293

The first one gives over 24 million pots, which is unreasonable, so

a=17, j=293, and Shirty made x=293^2 = 85,849 pots.
Petdon made 17 more and Grott 17 fewer.

[snookersfun]

precise and clear as always