Puzzles with numbers and things

[snookersfun]

Is it possible to make 35?

no, it has to be a multiple of 3....

[davis_greatest]

no, it has to be a multiple of 3....

I should really have made the scoring a bit simpler by dividing all the points by 3! Originally I was thinking of giving different numbers of points from those in the question. Never mind... continue as it is.

Scores of 30 received from snookersfun and Monique so far...

[Monique]

R170 Trophy woe closing ...

Snookersfun and Davis_Greatest rightly claimed that the snookatom can't be built! Congratulations again!
Here is the explanation hidden just in case someone wants to give it a last try

Suppose that it can be buit. Then count the pairs (ball, edge linked to that ball) in two different ways.
If you start from the balls you count 43 pairs as each ball carries a number of edges equal to its value and the cue ball was given a value of 1.
Starting from the edges and assuming you have n edges (n is of course an integer number), you get 2*n as each edge links 2 balls.
No way 43=2*n

[davis_greatest]

Take 9 basketballs shrunk to the size of ping pong balls and position them wherever you want on the bed of a snooker table enlarged to the size of a football pitch.

For any line of 3 basketballs shrunk to the size of ping pong balls in ANY direction, score 3 points.

For any line of 4 basketballs in ANY direction, score 6 points.

For any line of 5 or more basketballs in ANY direction, score 9 points.

Only the longest line counts. For example, a line of 9 balls in a line OOOOOOOOO scores 9 points (as it has 5 or more balls) - you cannot also count lines of 4 balls or 3 balls within that same line!


Post here whatever scores you can find - preferably as high as possible!
You don't have to find the highest theoretically possible - in fact, I don't know what it is!

From the answers I've had, there seems to be no advance on a score of 30 (which may well be the best possible, and received from both snookersfun and Monique) - so let's see the pictures on this thread...

[Monique]

Here a picture ... for the 30 bid

[IMG][/IMG]

[snookersfun]

my picture:

[snookersfun]

and Round 173:

I found another cross-number:
© Copyright 2000, Jim Loy


When solved, each little square will contain a one digit number. None of the multiple digit numbers (like 1-across) begins with a zero. The same digit may occur more than once in a multiple digit number.


Clues:

Across:

1. 11-across squared.
4. 3 more than 14-down.
5. 2 times 15-across.
7. 12-down + 5-across.
9. 2-down - 15-across.
10. 5 more than 9-across.
11. 10 less than 8-down.
13. 1-down + 3-down.
15. 1/2 of 5-across.
16. 4-down - 1-across.

Down:

1. 13-across - 3-down.
2. 11-across + 3-down.
3. 30 more than 5-across.
4. 2-down squared.
6. 1-across - 16-across.
7. 10-across rearranged.
8. 3 times 4-across.
12. 2 times 4-across.
14. 20 more than 3-down.
15. 3 more than 15-across.

[snookersfun]

update: d_g solved this one already despite taking the long route and programming it into excel

[snookersfun]

update: d_g solved this one already despite taking the long route and programming it into excel

...at this point I am happy to announce, that Robert602 joined us again and solved it perfectly as well, using pen and paper only!

[snookersfun]

...at this point I am happy to announce, that Robert602 joined us again and solved it perfectly as well, using pen and paper only!

...and Monique solved it as well yesterday. It seems with some help from excel following d_g's glorious lead

I'll leave it open for a little while longer... a little hint: 11 across, can be pretty much fixed down to 6 possibilities early on...

[snookersfun]

Round 174: one more of those... the excel geaks will be happy
But it can easily be done by hand...


Across:

1. 8D + 3x6A.
3. 2D + 10.
4. 8D + 1.
6. 3 x 1D.
8. 2D + 2.
9. 3A x 1D + 1.

Down:

1. 4A - 2.
2. 8D - 1.
3. 9A - 4A - 6.
5. (4A+4) x (3A-2).
7. 2D + 3A.
8. 1D + 1.

[snookersfun]

yes, d_g and Monique have solved this one already in record times! Well done!

[Monique]

Round 175 - Charlie in Brussels

I didn't spot any TSF member at the Belgian Open but Charlie was there definitely! When I saw (or better heard) him he was helding a press conference in front of a bunch of journalists (the cheeky chimp!).
The guy from TheMoon just asked "What was the total number of points scored in this tournament's centuries?"
"It was actually remarkable!" answered Charlie " There was a square number of centuries, performed by a square number of players, each of them achieveing a square number of centuries! Even better, all centuries got a different score and these scores formed a suite of consecutive numbers. The lower and the higher centuries were actually "monochromatic" (15 reds + 15 times the same colour) total clearances. Also one of the centuries's score was equal to twice the number of centuries times the number of 'century' players". Needless to say, the guys looked nonplussed...
"Yes but..." insisted the repesentative of ZooComics "What was the total number of points scored in this tournament's centuries?"
"Well" said Charlie "None of the opponents of the 'century' players were able to pot a single ball!".
"Great" said the guy from WorldOfTheNews, in a high pinched voice, "but what was the total number of points scored in this tournament's centuries?"
Too late ... Charlie was gone. A fresh load of bananas was served at the buffet...

Well, finally, what was the total number of points scored in the Belgian Open's centuries?"

Charlie appears courtesy of D_G. Thanks!

[Monique]

And Snookersfun solved it already! What can I say ... well done!

[Monique]

And D_G also. Congratulations
It remains open ... it's pretty straightforward