Puzzles with numbers and things

[davis_greatest]

Anyone who has an advance on 9 reds for round 378, please put the arrangement up here, so we can close that round.

Snookersfun’s round 380 is still open.

But a quick one also for a Sunday evening:

Round 381 - Snap happy

The world’s top 16 ranked snooker players are lined up in a queue (in some order) for their group photo with Barry Bailey the Baboon. Barry sends the first one in the queue to stand next to the snooker table, but sends the next one back, to the back of the queue.

The next player in line is sent to the snooker table, and then the next two are sent to the back of the queue! And Barry sends the next player to the snooker table, and then the next three to the back of the queue. And so on, sending one player to the snooker table, and the next four to the back, and one player to the table, the next five to the back, etc… until every player has been sent to the snooker table.

Fortunately, the players have now arrived at the snooker table in order of their ranking – 1 to 16 – ready for Barry to take the photo.

In which order were they initially standing in the queue?

Answers initially by private message please.

[snookersfun]

update: I have two correct answers already by d_g and moglet. Well done!

[snookersfun]

I just found another good one (should probably credit some javaguru for those):

R.380 more primes in 6-digit numbers

This time Gordon tells Charlie he's thinking of a 6-digit number. All its digits are different. The digital sum of the number is equal the number formed by the last 2 digits in the number. Moreover the sum of the first 2 digits is the same as the sum of the last 2 digits.
If one takes the sum of the number, the number formed by rotating its digits one to the left, the number formed by rotating digits one to the right, the number formed by rotating digits pair-wise to the right, the number ‘pair-wise rotated’ to the left, and the number with the first three and last three digits swapped, one can observe that the first and last digits of this sum match the last two digits of the number, in some order.
Gordon concludes: ‘If each of the three numbers formed by the digit pairs in the number is prime, then what is my number?’
Charlie has a little ‘think’ and comes up with which number?

Answers by PM please

[davis_greatest]

Congratulations to moglet and snookersfun who have sent their answers for [hidden number] of reds to round 378 by PM - and to abextra for her 9 reds above

[abextra]

Anyone else who has 8 reds or higher, please put the the arrangement up on the thread.

With 9 reds (hope it works):

2+7
2x1+3 - less points
6+4 - less balls
9x1 - less points
5 - less balls
4x1 - less points

back to 2+7 - less balls. sorry for peeking

[davis_greatest]

Round 379 update - congratulations to snookersfun, moglet and Monique who have collectively found the 14 ways this can be done. Well done

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Round 378 update - snookersfun and moglet have also posted the greatest possible number of reds that can be put in the pocket diagonally opposite to the yellow - congratulations! Monique, your bid is a nice opening one but can go higher.

snookersfun and moglet - please PM me your arrangements for round 378. Anyone else who has 8 reds or higher, please put the the arrangement up on the thread.

[Monique]

R379 another proposal

7,1 - 2 and 4x1 - 5,4 - 6x1 - 6,3 - 5x1 :snooker:

[moglet]

R378
11
R379
All variations on abextra's original as long as the colour pairs sum to at least 8 work in both directions, i.e. 26,5reds,35,5reds,47,5reds loop

[Monique]

R378
8 for now :snooker:

[snookersfun]

adding the rest
2,7.....5x1
5x1.....3,6
4,5.....5x1
basically 4 of those (3,6 and 4,5 switching places and same using pairs 3,5 and 4,6)
or
2,6....5x1
5x1...4,7
3,5...5x1
again 4 of those (also 3,7 and 5,4)

2 of: (switching 7,1 and 5,3 pockets)
2,4x1...5,3
4,6......5x1
5x1....7,1

2 of: (switching 7,1 and 6,3 pockets)
2,4x1...6,3
4,5......5x1
5x1....7,1

2 of: (switching 7,1 and 4,5 pockets)
2,4x1...4,5
6,3......5x1
5x1....7,1

[snookersfun]

Round 379 - Anticlocknotwatching

Again, Talia has placed the same balls into the pockets, and now when she walks around the table anticlockwise, each pocket contains either balls valued at least 3 points greater than the previous pocket, or at least 3 more balls (or both).

The yellow is still placed in the yellow pocket.

Find as many ways as you can, and put up any that you find - this time do show the arrangement of balls in each pocket.

2,7.....5x1 or 2,6....5x1 and several permutations
5x1.....3,6----5x1...4,7
4,5.....5x1----3,5...5x1

[snookersfun]

Round 378 - Anticlockwatching

Talia Mabb has again put the snooker balls (the 15 reds, and six colours, but not the white) into the snooker table's six pockets, but in a very different way from before. She has arranged them so that she can walk around the table anticlockwise (counterclockwise, for anyone who learnt US English), as many times as she wants, and each pocket contains either balls of a lower value* than the previous pocket, or a smaller number of balls (or both).

The yellow is placed in the yellow pocket.


What is the greatest number of reds that she can place in the pocket diagonally opposite to the yellow?

Oh, and once again, try not to peek at others’ hidden answers!

lucky for me nothing to peek at yet 11

[davis_greatest]

Congratulations again to those who have put up the correct answers on the thread to round 377 - abextra, Monique, snookersfun and moglet!

I think I've had enough of these, so now for something completely different. In fact, two rounds, completely different from anything in recent times.

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Round 378 - Anticlockwatching

Talia Mabb has again put the snooker balls (the 15 reds, and six colours, but not the white) into the snooker table's six pockets, but in a very different way from before. She has arranged them so that she can walk around the table anticlockwise (counterclockwise, for anyone who learnt US English), as many times as she wants, and each pocket contains either balls of a lower value* than the previous pocket, or a smaller number of balls (or both).

The yellow is placed in the yellow pocket.

*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 in the pocket diagonally opposite to the yellow? (When bidding, you can put up the number of reds only, but this time not the arrangement yet until asked later, please.)


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Round 379 - Anticlocknotwatching

Again, Talia has placed the same balls into the pockets, and now when she walks around the table anticlockwise, each pocket contains either balls valued at least 3 points greater than the previous pocket, or at least 3 more balls (or both).

The yellow is still placed in the yellow pocket.

Find as many ways as you can, and put up any that you find - this time do show the arrangement of balls in each pocket.

===========================

Answers to both rounds can go straight up on the thread, in hidden text please, like this (deleting the space after the two open square brackets):

([ COLOR=#f1f1f1]
Enter answer here! [ /COLOR])

Oh, and once again, try not to peek at others’ hidden answers!

[moglet]

R377
Can't find more than 12: 67,3 reds,4,5,23,12 reds

[snookersfun]

12

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

silly