Puzzles with numbers and things

[Monique]

more ...

[abextra]

The Belgian Open Management Team is already preparing the next Belgian Open...

What does Charlie have on his mind?

Maybe something with 494 balls? Does Charlie need more or less?

[davis_greatest]

R189 - nice pictures from snookersfun... and picture from Monique abextra, please post yours here too...

[Monique]

Round 190 b - Preparing the new season ...

The Belgian Open Management Team is already preparing the next Belgian Open. They definitely don't want to repeat the snookatom disaster. Therefore they decide to send an e-mail to the most distinguished mathematicians on TSF with the following specification:

The new trophy, named snookedron, must

- have the shape of a polyhedron
- each of the faces of this polyhedron is a regular polygone
- each face has a different color
- each face is made of snooker balls
- each edge must be at least 36 cm long

we will put the following material at your disposal:

- 127 black balls
- 130 pink balls
- 147 yellow balls
- 132 blue balls
- 28 brown balls
- 36 red balls
- 29 green balls
- 27 white balls

Please use as many balls as possible.

The next day they receive the following answer

"Dear Sirs,
May I advise you to look for the missing snookatom cue ball?
Your devoted Charlie."

What does Charlie have on his mind?

[Monique]

Round 190 a - Mr Grott unlucky again ...

This is a "wink" puzzle without numbers ... so no excuse!

The Belgian Open "Trophy Committe" finally managed to assemble the snookatom, discarding the cue ball. It was decided that the cue ball would be offered by Miss Tallia Mabb to the runner-up, with as many kisses as points scored in the highest break on the tournament! Well Mr Grott is the runner-up and to say that he is a bit excited is an understatement!

In due time, the winner takes the snookatom from the hands of the officials and flees home with it immediately, gutted. If he had the choice, he would have opted for the runner-up price, in particular as HE made the highest break!
Las! When the great time comes for Grott the snookatom cue ball has disappeared and Miss Tallia with it. Alledgedly she's still looking frantically for the *** ball ...that remains undiscoverable.
Gossip tells that the precious cue ball was stolen by a fan of the winner who got it signed by his idol and uses it now as an avatar on a certain forum ...

Where is the snookatom cue ball now?

[snookersfun]

[Monique]

R 189b ( and R189a ipso facto)

[davis_greatest]

Round 189 and 189b - had some nice pictures from snookersfun, abextra, and Monique!

Please put them up on the thread...

[davis_greatest]

Round 189 - Time for a quickie

Place 16 snooker balls in 8 lines of four. (They can't simply be in a 4 by 4 grid pattern - that's too easy.)

... and also round 189, part (b): place 16 snooker balls, anywhere you like, to make 12 lines of four (as snookersfun has already managed).

[davis_greatest]

Yes, eight only. The balls shouldn't be touching.

(Certainly, more than 8 would be possible. It was late, I posted it while a bit tired - and then realised that it was a very silly question indeed, as you could just put 16 balls in a 4x4 square! So I nearly deleted it, but decided to leave it and say that you can't do the obvious square...)

[snookersfun]

Round 189 - Time for a quickie

Place 16 snooker balls in 8 lines of four. (They can't simply be in a 4 by 4 grid pattern - that's too easy.)

let me understand that... you want 8 only? Do they need to be touching or anything else, I might have missed? I'll PM you 11 lines now, I am sure one could do more...

[davis_greatest]

Round 189 - Time for a quickie

Place 16 snooker balls in 8 lines of four. (They can't simply be in a 4 by 4 grid pattern - that's too easy.)

[davis_greatest]

From the sequence of hexagonal numbers 7,19,37,61....

Does anyone remember how to expand :

n = 1 + SUM(6n) {n=1:n}

to f(n) = 1+ 3(n+1)(n)..

And after that, sum the sequence again to get to

n = (n+1)^3 -1.


Snookersfun?

What are you trying to do? Do you mean
SUM (from k=1 to k=n) of {3k(k+1) + 1} to solve the previous problem?

That's a pretty long way of doing it, but you can. It equals:

3n(n+1)(2n+1)/6 + 3n(n+1)/2 + n
= 3n(n+1)(2n+1+3) / 6 + n
= n(n+1)(n+2) + n
= n³ + 3n² + 3n
= (n+1)³ -1

[dantuck_7]

From the sequence of hexagonal numbers 7,19,37,61....

Does anyone remember how to expand :

n = 1 + SUM(6n) {n=1:n}

to f(n) = 1+ 3(n+1)(n)..

And after that, sum the sequence again to get to

n = (n+1)^3 -1.


Snookersfun?

[davis_greatest]

Correct answers to R188 so far from abextra, dantuck_7 and Monique .... well done. Round still open...

Round 188 - Congratulations then to abextra, dantuck_7, Monique and now snookersfun for finding that Oliver had potted 999,999,999 balls. No wonder he became sleepy!

After n-1 times of potting the balls, Oliver had potted n³-1 balls.
Put n=1000, and you see that after 999 times, he had potted 1000³-1 = 999,999,999 balls.

The first time, he potted 7 balls (one central ball, with 6 balls around it);
the next time there were 19 balls (12 extra ones);
then 37 balls (18 extra ones)....
... each time the number of extra balls goes up by six, because the shape is hexagonal.

These numbers are the differences between successive cubes (this can be shown with algebra or geometrically):
2³-1³ = 8=1 = 7
3³-2³ = 27 -8 = 19
4³-3³ = 64 - 27 = 37
etc.

so adding them all up gives 1000³-1.