Puzzles with numbers and things

[davis_greatest]

OK - it's gone 8pm, so anyone who has bid a number of balls can now post an explanation of how to distribute them. You should show how the balls are distributed among the apes, in a similar way to how the example was given in the question for 2 apes.

However, instead of saying yellow, green etc, you can just say 2, 3 or use a similar shorthand if convenient.

You need to show how to distribute the balls for 2 apes, then 3 apes, then 4, ... all the way to 10.

Since several of you have bid the same, you will have to be trusted not to read the explanations of anyone else before you post your own!

[davis_greatest]

The people who bid are abextra, chasmmi, Sarmu and snookersfun. You all bid 111, so let's hear it!

[snookersfun]

Although feeling rather silly for the initial blunder: here it is withened out (may I suggest huge point deductions for not being able to read/or count to 10)
we always need to use 2,3,4,5,6,7=27 points, which means that upwards of dividing by two the minimum number of points given to each ape is 9 (as 7 would leave the pink ball 'unpairable' while 8 would leave the black ball useless. Therefore:
2 apes: 27 odd, +3 points=30 (4,5,6-2,3,3,7) 15p each-6+1 balls
3 apes: 27p (2,7-3,6-4,5) 9p each-6 balls
4 apes: 27p+9p (2,7-3,6-4,5-e.g.2,7) 9p each-6+2 balls
5 apes: 27p+18p (2,7-3,6-4,5-e.g.2,7-2,7) 9p each-6+4 balls
6 apes: 27p+27p (2,7-3,6-4,5-e.g.2,7-2,7-2,7) 9p each-6+6 balls
7 apes: 27p+36p (2,7-3,6-4,5-e.g.2,7-2,7-2,7-2,7) 9p each-6+6+2 balls
8 apes: 27p+45p (2,7-3,6-4,5-e.g.2,7-2,7-2,7-2,7-2,7) 9p each-6+6+4 balls
9 apes: 27p+54p (2,7-3,6-4,5-e.g.2,7-2,7-2,7-2,7-2,7-2,7) 9p each-6+6+6 balls
and.... not to forget 10 apes: 27p+63p (2,7-3,6-4,5-e.g.2,7-2,7-2,7-2,7-2,7-2,7-2,7) 9p each-6+6+6+2 balls
all in all 6*18+3=111
)

[chasmmi]

I feel like I have done something wrong but isn't it just that for two people you need an extra green to create an even number to split in two so:

2+7+6
3+3+4+5

And then for the rest you can just give the apes either 2+7, 3+6 or 4+5 to make a value of nine.

Therefore for three you have to give one of each to account for all six balls
Then for four you repeat one set so say 2+7, 2+7, 3+6, 4+5
then for five you repeat two sets and so on that the number of balls disytributed is aways two per ape except in the case of two apes where you need seven.

so the total is 7+6+8+10+12+14+16+18+20 = 111

[abextra]

Two apes 1) 7+6+2=15
. . . . . . . 2) 3+3+4+5=15 - 7 balls

Three apes 1) 7+2=9
. . . . . . . . 2) 6+3=9
. . . . . . . . 3) 4+5=9 - 6 balls

Four apes 1) 7+2=9
. . . . . . . 2) 6+3=9
. . . . . . . 3) 4+5=9
. . . . . . . 4) 7+2=9 - 8 balls

Five apes 1) 7+2=9
. . . . . . . 2) 6+3=9
. . . . . . . 3) 4+5=9
. . . . . . . 4) 7+2=9
. . . . . . . 5) 6+3=9 - 10 balls

Six apes 1) 7+2=9
. . . . . . 2) 6+3=9
. . . . . . 3) 4+5=9
. . . . . . 4) 7+2=9
. . . . . . 5) 6+3=9
. . . . . . 6) 4+5=9 - 12 balls

etc.

For every next ape there must be two more balls (value 9 points), so for
seven apes - 14 balls
eight apes - 16 balls
nine apes - 18 balls
ten apes - 20 balls

and in total there will be 111 balls.

[Sarmu]

Yea pretty much what abextra said, I find that the minimum scores each ape can get is 9 which requires at least 2 balls each, so it adds up to 111

[davis_greatest]

A point each to abextra, snookersfun, chasmmi and Sarmu!

SO HERE IS THE SCOREBOARD AFTER ROUND 85

snookersfun………..………….…..42½
abextra..............................24
davis_greatest....................19
Vidas.................................12½
chasmmi..............................9
elvaago...............................8
Sarmu.................................8
robert602.............................6
The Statman…………………. …...…5
Semih_Sayginer.....................2½
austrian_girl and her dad.........2½
April Madness........................1

[davis_greatest]

Round 86 - They call him Angles

Oliver, my pet orang-utan, is playing snooker with Angles McBum, on a perfectly rectangular snooker table with a playing area exactly 6 feet wide by 12 feet long. (The Statman, I know that this is slightly larger than the official playing area allowed, but that is the size of this snooker table and there is nothing I can do about it. Throughout this question, also, ignore the width of the balls - i.e. assume that the balls have zero diameter.)

Oliver and Angles are down to the final pink and black and Oliver has trapped Angles in a snooker, with the cue ball at least 5 feet away from the object ball (the pink).

The cue ball happens to be touching a cushion (not the top cushion) and is lying a whole number of feet (at least one foot) from the nearest corner pocket (or pockets).

The pink is not touching a cushion, but happens to lie a whole number of feet (at least one foot) from each cushion. This means that if you measured from the pink to the nearest point of each of the 4 cushions in turn, each time it would measure a whole number of feet exactly. (We are treating the table as having 4 cushions, one at each end and one at each side - ignore the fact that the middle pockets break up the cushions on each side.)

Angles is known for his superb knowledge of the angles on a snooker table and is extremely good at getting out of snookers. Furthermore, he is adept at using sidespin to such an extent that he can make the cue ball rebound off a cushion at whatever angle he chooses - the cue ball does not need to rebound at the same angle as it hit the cushion. (He can do this even when playing from tight on the cushion rail.)

Angles decides to play a one cushion escape - not off the top cushion though (and not off the cushion that the cue ball was already touching, as this would be impossible). Once he has chosen the cushion off which to rebound, he uses sidespin if necessary in such a way as to make the cue ball travel the shortest total distance possible, using that cushion, before the cue ball hits the pink. (This distance includes the travel to the cushion plus the travel from the cushion until striking the pink - i.e. add the two together.)

Note: this does not necessarily mean that the escape that Angles chooses makes the white travel the shortest possible distance of all possible escapes - there might be a shorter escape off another cushion. What it means is that, once the cushion for the escape has been chosen, Angles chooses the point of contact on that cushion in such a way as to make the cue ball travel the shortest distance.

Angles plays the shot perfectly, as expected, and strikes the pink full ball. It also just happens that the total distance that the cue ball travels in this escape, until striking the pink, is again exactly a whole number of feet (but not a multiple of 5 feet, so not 5, 10, 15 etc).

If Angles had chosen a different route and instead played the one cushion escape off the top cushion with no sidespin (so the cue ball rebounds at the same angle as it strikes the cushion), how far would the cue ball have travelled from being struck until hitting the pink?

Answers by Private Message please. Intial Deadline will be 23:00 GMT on Thursday 21 December.

[davis_greatest]

Ooooops! (round 86 update)

There was a bit of a problem in round 86, as originally worded, in that it was impossible! Sorry! The questions are supposed to be challenging, but not impossible. I have now added some wording in blue above, which should mean the question is now possible to solve.

The new wording in blue is "(but not a multiple of 5 feet, so not 5, 10, 15 etc)".

[davis_greatest]

Round 87 - Think inside the box

Just to confuse you all further, I will also ask round 87, to run at the same time as round 86. It’s a nice little puzzle (not devised by my apes – but nice all the same)...

Fill in digits (0-9) in the boxes below, to make a 10-digit number. (You can use any of the digits 0 to 9, but don’t need to use them all, and may use the same digit more than once.)

The digit in the box on the left should equal the number of zeroes in the 10-digit number you write. The digit in the next box should equal the total number of “1”s, etc, …. and the digit in the right-hand box should equal the total number of “9”s.

Again, answers by Private Message please! Initial Deadline of 23:00 GMT on Thursday 21 December (to be extended if I deem necessary).

[davis_greatest]

We've already had an early answer to round 87 from someone who will become a new entrant on the scoreboard once this round closes...

[davis_greatest]

A clue to Round 86

I am going to give a hint to round 86 - and it is this: the mention of sidespin, and Angles McBum being able to make the cue ball rebound off the cushion at any angle, was a deliberate red herring!

If you want to make a cue ball travel the shortest total distance, in travelling onto the cushion and then to an object ball, this will always be achieved by using no sidespin - so that the cue ball rebounds off the cushion at the same angle as it hits it. (We are ignoring "slide off the cushion" here, and assuming idealized cushions where the ball rebounds off at the same angle if no spin is used.)

As an example, suppose that you want to get out of a snooker and need to hit a pink that is (say) 3 feet from the cushion. See my brilliant picture below, which is the first time I have ever tried to draw anything using Paint. Good, eh?

You can hit the pink by visualising an imaginary "reflection" pink ball 3 feet behind the cushion, and aiming for that. The distance that the cue ball would travel, if it were able to travel through the cushion to the imaginary reflection pink ball, is the same as it will travel when rebounding off the cushion to hit the real pink ball. Since the shortest distance between two points (on a plane) is a straight line, the shortest total distance will be achieved by aiming straight for the imaginary ball.

It is the same principle as looking in a mirror - light travels the shortest distance between two points and rebounds off the mirror at the same angle as it hits it. The reflection of an object 3 feet in front of the mirror will appear 3 feet behind the mirror.

[snookersfun]

round 88: more numbers

I’ll put up another question, round 88:
You should try to arrange the numbers 1,2,3,….n (two of each) in such a way, that there is one other number between two ones, two other numbers between two 2s, three other numbers between two threes, etc.

Example: for n=4 i.e. using 1,2,3,4 twice each: 4,1,3,1,2,4,3,2
Try to get as high a sequence as possible in this way. Bids for number of n reached (so ½ the digits in the sequence) can be put up on the thread until Sunday 24.12. 12:00pm GMT. After that I’ll ask for the actual sequences.

[davis_greatest]

Where you say one digit between the ones, two digits between twos etc, do you mean digits or numbers? What happens if we get up to 10 or more so the numbers are 2 digits long?

I'll start with an opening bid of 37.









Just kidding I've only got 4, so far, which is the same as in the question!

[Snooker Rocks!]

Hmm. I've been working at this for a while now

I wonder if there is one possible sequence for any set of numbers