Puzzles with numbers and things

[davis_greatest]

Hehehe... I'm not so shy... I'll send it by PM to leave a chance to others who might be interested...

OK.... meanwhile, on checking messages, I see that snookersfun has answered it too. So, congratulations abextra and snookersfun. Next answer on the thread please!

[abextra]

Hehehe... I'm not so shy... I'll send it by PM to leave a chance to others who might be interested... Isn't it sad that most of our puzzlers are gone... I wonder why have they all quitted?

[davis_greatest]

abextra, don't be shy Place your winning bid here for all to see!

[davis_greatest]

Wow, 300th round! Big thanks to davis_greatest for running this thread and congratulations to everyone around here!

Back to these Big Pockets - sadly I didn't see the Doherty-Selby match, was it something special about the set of snooker balls before the end of the mid-session interval? If there were 15 reds and 6 colours, I hope I could get the Score over 100 000!

Oh, yes, normal set of snooker balls - 15 reds and 6 colours. It just so happened that the question was posted during the mid-session interval!

[abextra]

Wow, 300th round! Big thanks to davis_greatest for running this thread and congratulations to everyone around here!

Back to these Big Pockets - sadly I didn't see the Doherty-Selby match, was it something special about the set of snooker balls before the end of the mid-session interval? If there were 15 reds and 6 colours, I hope I could get the Score over 100 000! (Maybe 115 248 or so...)

[davis_greatest]

Round 300 update

OK, I've had a couple of bids in. One is too high, and the other, a bid of 11,000, has come from Monique.

In the light of the latest bid, which put balls in only 4 pockets, I think perhaps the question may need to be clarified: a snooker table has 6 pockets and you are intended to put at least one ball into every pocket. If you leave one or more pockets empty, that (or those) pocket(s) will have score 0, which will make your final Score 0.

So, make sure that the total number of balls in top pockets equals the total number of balls in bottom pockets, but don't leave the centre pockets empty!

[davis_greatest]

OK... still a few open rounds I see.

Here is a simpler one then, I think, which I hope shouldn't stay open long...

Round 300 - Big Pockets

Take the set of snooker balls (forget the white) from the Doherty-Selby match before the end of the mid-session interval, and place them in the pockets of the table wherever you like.

Each pocket, however, is only big enough to hold a maximum of 5 balls. As the table is a little uneven (causing one of Selby's slow shots across the nap apparently to drift), it is important to keep it balanced. So you must make sure that the total number of balls in the two top pockets (combined) is equal to the total number of balls you place in the two bottom pockets.

Each pocket receives a score equal to the value of the balls it contains. (E.g. if a pocket contains just a blue, a pink and a red, it scores 5+6+1 = 12.)

Now, your total Score is equal to the product that you get by multiplying the scores of each of the 6 pockets.

What is the highest Score you can get (and where do you place the balls)?

[snookersfun]

...and although I still have only Mon's answer to that previous round, here is the next one now (easier):

R.299(wow): Pentaminos

Very much like the Tetraminos, the tabulated words have to be arranged as a crossword in the following grid. All unused cells will make up the complete set of possible Pentaminos.
penta.bmp

[snookersfun]

update: Monique first in with an answer to R.298! Very well done! Always good with those difficult ones

[snookersfun]

R. 298: Differences

place the snooker balls (colours, value 2-7) into the white cells of the following grid, so that every row and column contains all different colours.
The two digits in the dark cells show the min. and max. difference in value between pairs of balls touching the dark cells (by edge or corner)
diff3.bmp
example shows a 1-4 digit case

[Monique]

I did it again ....

[davis_greatest]

I'm not sure I understand the scoring... ... anyway,

I bid 165 for Round 296 and 129 for Round 297...

(In Round 297 I got two equal values of a Line, black-brown-green and green-pink-blue both add up to 14, I assumed one of them is the highest and the other one is the second highest... )

here is my 130 (4 lines of 13)
and 165 was my highest for R.296 as well
[ATTACH]994[/ATTACH]

For R296

1_4
576
2_3

gives 5*(5+6+7) = 90
+ 3*(3+6+4) = 39
+ 2*(2+7+4) = 26
+ 1*(1+7+3) = 11
+ 1*(1+2+5) = 8

total 174

unless I misunderstood the rules ...

abextra and snookersfun - yes, very nice pictures... and these are valid bids . (I haven't thought whether higher is possible, but these look hard to beat.)

Monique - nice arrangement too, but unfortunately I don't think this one is possible, since the balls would not all be touching (remember they are round ).

[davis_greatest]

It seems Round 295 has left for me...

I think colouring a 1470 x 1470 "chessboard" with six colours shouldn't be a big problem for little Oliver.

The easiest way would be using some of the 10x10 "chessboard" pics posted above, instead of one square he only has to colour 147x147 squares (for example the yellow patch must be 147x294 squares).

I'm sure there are LOADS of different ways to do it, and I'm also pretty sure that he could colour every "chessboard" using six colours, if the number of rows is even and bigger than six... Sadly I am not clever enough to give a proper explanation... ... somebody please help!

Congratulations snookersfun, Monique and abextra! Round 295 is now closed. Yes, taking a 10x10 coloured chessboard, and dividing each square into a new 147x147 grid, will give the required 1470x1470 coloured chessboard.

[Monique]

For R296

1_4
576
2_3

gives 5*(5+6+7) = 90
+ 3*(3+6+4) = 39
+ 2*(2+7+4) = 26
+ 1*(1+7+3) = 11
+ 1*(1+2+5) = 8

total 174

unless I misunderstood the rules ...

[snookersfun]

here is my 130 (4 lines of 13)
and 165 was my highest for R.296 as well