Puzzles with numbers and things

[davis_greatest]

Thanks for the answers to round 376.

And, on round 377, two perfect bids in very quickly from abextra and Monique! Very good Hopefully no one else will peek at those bids before bidding himself / herself!

[abextra]

R376

....

So Talia is still right.

I agree.

4x1 ... 2,3,4,5,6
3x1 ... 6x1
2x1 ... 7

[Monique]

R376

p1 black
p2 2 reds (more balls)
p3 3 reds (more balls, more points)
p4 4 reds (more balls, more points)
p5 yellow, green, brown, blue, pink (more balls, more points)
p6 6 reds (more balls)

and back to p1 (more points)

So Talia is still right.

[Monique]

R377 12 :snooker:

[abextra]

Round 377 - Clockwatching

This time, Talia Mabb herself ...

What is the greatest number of reds that she can place into any one pocket?

Bid here on the thread, and please post your answer in hidden text, like this (deleting the space after the two open square brackets):


Oh, and one more thing – no peeking at others’ bids!

I certainly can't do this - I mean "no peeking"... ...
12x1 ..... 7+6
2+3 ...... 3x1
5 .......... 4

[davis_greatest]

Congratulations on the answers put up to round 375!

Round 376 has also been solved quickly by moglet, abextra, Monique and snookersfun. Congratulations, again! Please put your answers up on the thread!

Another, then…

Round 377 - Clockwatching

This time, Talia Mabb herself puts all the snooker balls (the 15 reds, and six colours, but not the white) into the snooker table's six pockets. Again, she has arranged it so that she can walk around the table clockwise, as many times as she wants, and each pocket contains either balls of a greater value* than the previous pocket, or a greater number of balls (or both).

*the value of a ball is its usual value in snooker - 1 point for each of the 15 reds, 2 for the yellow,..., up to 7 for the black


What is the greatest number of reds that she can place into any one pocket?

Bid here on the thread, and please post your answer in hidden text, like this (deleting the space after the two open square brackets):

([ COLOR=#f1f1f1]
Enter break here, e.g. 147! [ /COLOR])

Oh, and one more thing – no peeking at others’ bids!

[snookersfun]

oh, snookersfun is late putting up all her stuff

Congratulations on round 374 - still some answers to go up, I think. (All the same, of course, as there is only one answer )

Meanwhile, I have had 4 correct answers to round 375 (out of the 797,856 possible solutions!) .... please put them up on the thread... and if anyone else has one, please put it up.

Lol, I can only say that I totally agree with Monique and Moglet - and I'm sure I will agree with Snookersfun too, if she should post her answer!

yeah, all the same and therefore simply agree as well

Round 375 update - yes, Talia was right! As well as those solutions above from Monique and abextra, different solutions were received by Private Message from snookersfun and moglet - hopefully also to be be put on the thread!

ok, this is mine, a beautiful one I think:

5,4,3.. 7,6

6x1.... 3x1

4x1.... 2, 2x1

[abextra]

Thank you for your answers, Monique and Moglet!

[moglet]

375

Can't remember which variation I had, maybe No. 35732

black
yellow and green
brown with 2 reds
blue with 3 reds
pink with 4 reds
6 reds
loop

[davis_greatest]

Round 375 update - yes, Talia was right! As well as those solutions above from Monique and abextra, different solutions were received by Private Message from snookersfun and moglet - hopefully also to be be put on the thread!

Round 375 - Clockwatching

Charlie and Gordon have just finished a frame of snooker, with the beautiful Talia Mabb refereeing once more. All the balls (except the white) are now in one of the snooker table's six pockets. Charlie says "Gordon - look at this, as I walk around the table clockwise, as many times as I want, each pocket contains either balls of a greater value* than the previous pocket, or a greater number of balls (or both)."

"I doubt that very much," says Gordon. "Must be impossible."

"No, I don't think it is," points out Miss Mabb.

*the value of a ball is its usual value in snooker - 1 point for each of the 15 reds, 2 for the yellow,..., up to 7 for the black


Who is right? Either prove that it can be done, or prove that it is impossible. Answers by Private Message please.

---------------------------------

Round 376 - More Talia Clockwatching

"No, no, no!" exclaims Gordon, looking at the answers received from TSF-ers to round 375. "The reason I said it [see round 375] was impossible, was that I had plainly seen that at least four pockets didn't contain any colours at all!"

"So?" asked Talia. "That still doesn't make it impossible!"


Who is right now? Either prove that it can still be done, or prove that it is impossible. Answers by Private Message please.

[Monique]

R375 Talia is right ...

pocket 1 : 4 reds - value 4
pocket 2 : 5 reds - value 5 (higher value and more balls)
pocket 3 : yellow and 4 reds - value 6 (higher value)
pocket 4 : black and 1 red - value 8 (higher value)
pocket 5 : pink and green - value 9 (higher value)
pocket 6 : brown, blue and 1 red - value 10 (higher value and more balls)

and back to pocket 1 - more balls.

[abextra]

Congratulations on round 374 - still some answers to go up, I think. (All the same, of course, as there is only one answer )

Lol, I can only say that I totally agree with Monique and Moglet - and I'm sure I will agree with Snookersfun too, if she should post her answer!

Once the first two or three have gone up, I think that Gordon might be waiting to jump in with the next round...

Sounds great, I like him jumping!

[abextra]

Round 375 - If the balls were pocketed like this (for example)

7,2 ..... 5x1
5x1 ..... 6,3
5,4 ..... 5x1

then Charlie could walk around the table endlessly and there would always be either more balls or balls for greater value in the next pocket. R383 - 15

[davis_greatest]

Congratulations on round 374 - still some answers to go up, I think. (All the same, of course, as there is only one answer )

Meanwhile, I have had 4 correct answers to round 375 (out of the 797,856 possible solutions!) .... please put them up on the thread... and if anyone else has one, please put it up.

Once the first two or three have gone up, I think that Gordon might be waiting to jump in with the next round...

[moglet]

4 frames, breaks 72,85,87,122 (∆=50)
Yellow numbers 16,9,4,1
4 yellows with 4 browns in third frame
Totals from each colour:
Blacks 7x6, 42
Pinks 6x7, 42
Greens, 3x14, 42
Blues, 5x12, 60
Browns, 4x15, 60
Yellows, 2x30, 60

Balls per break
1st all yellow clearance
2nd 8 yellows, 1 green, 6 browns, final colours
3rd 3 yellows, 9 greens, 3 browns, final colours
4th 2 browns, 8 blues, 3 pinks, 2 blacks, final colours