Puzzles with numbers and things

[davis_greatest]

And now snookersfun has matched Sarmu and found the highest possible break!

I've also had a very good submission from austrian_girl - high, but not the maximum possible

[davis_greatest]

By the way, I have only had one acceptable answer to this - submitted by Sarmu a mere 3 hours after I posted the question! Sarmu's solution was also the highest break possible! (I have also had at least two invalid answers.)

Answers by Private Message please by the Initial Deadline of 18:00 GMT on Sunday 3 December

I've just had a very good submission from snookersfun to Ape Break Madness - not as high a break as Sarmu's though!

[davis_greatest]

I'd better quickly award davis_greatest a point for The Statman's anagram round 71, before The Statman tries to scrap that excellent round!

snookersfun's geometry question is round 72.

SO HERE IS THE SCOREBOARD AFTER ROUND 71 BUT BEFORE ROUND 68, APE BREAK MADNESS AND BEFORE SNOOKERSFUN'S TRIANGLES OF ROUND 72

snookersfun……………………….…..32½
abextra...............................16
davis_greatest.....................14½
Vidas..................................12½
robert602.............................6
elvaago...............................6
chasmmi..............................5½
The Statman……………………..……5
Semih_Sayginer.....................2½
April Madness.......................1

[elvaago]

Well, you got my reply. :-)

[snookersfun]

whatever you can come up with

[elvaago]

Do you want a number of a formula, expressing the angle in F with other angles?

[snookersfun]

As we are all happily back to puzzling, I have some geometry here for you.
Deadline Saturday noon GMT. I would like the answers by PM to reduce the mayhem on here a bit

Attached is a figure, containing a parallelogram ABCD. Also given is that the two angles on C are of same size and ED=AD.

Question: how big is the angle on F (with explanation of course)

[elvaago]

It's still interesting, even if you mathematify it! ;-)

[davis_greatest]

Did you know that if you take any two digit number and then reverse it, the difference is always divisable by 9?

32 - 23 = 9
91 - 19 = 72 = 8 x 9

It always works. Try it!

If the first digit is x and the second digit is y, then the number is 10x + y

Reverse the digits, and you get 10y + x

The difference is then (10x + y) - (10y + x) = 9(x-y)

which of course is divisible by 9!

[The Statman]

Sorry - regarding what previous question?

The one about the ape who had seven balls.

Oh, I thought PM stood for post meridiem. So I may answer questions in the morning, as long as it is by Private Message?

Ah, I see.

[elvaago]

Did you know that if you take any two digit number and then reverse it, the difference is always divisable by 9?

32 - 23 = 9
91 - 19 = 72 = 8 x 9

It always works. Try it!

Did you know that you can easily square numbers close to one hundred?

For x < 100
x ^ 2 = x - (100-x) + (100 - x)^2 (where , incast of (100 - x)^2 < 10 you add the zero before it.
For x > 100
x ^ 2 = x + (x - 100) + (x - 100) ^ 2, same rule applies in case of 1, 4 or 9 where you write 01, 04 and 09.

I.e.
97 ^ 2 = 9409
94 ^ 2 = 8836
102 ^ 2 = 10404
107 ^ 2 = 11449.

[davis_greatest]

In any case, the TWELVE plus ONE and ELEVEN plus TWO is the answer I was looking for.

Regarding previous question, I DID reply by PM. I was on the BBC board until it closed at noon, so by definition it was afternoon by the time I answered here.

Sorry - regarding what previous question?

[davis_greatest]

Reminder of round 69 deadline

By the way, I have only had one acceptable answer to this - submitted by Sarmu a mere 3 hours after I posted the question! Sarmu's solution was also the highest break possible! (I have also had at least two invalid answers.)

Answers by Private Message please by the Initial Deadline of 18:00 GMT on Sunday 3 December

Would you like another Ape Break round?

I'm sure you are familiar with the rules by now. You need to find the highest break you can (without using a free ball).

This time, it's a bit different. It's like normal snooker, still with 15 reds, but now there are 4 extra colours. You have to pot red, colour, red, colour etc, just like in normal snooker, but then pot the 10 colours in order (instead of the usual six).

The colours are:

yellow: 2 points (if you can't read that, it says yellow: 2 points)
green: 3 points
brown: 4 points
blue: 5 points
pink: 6 points
black: 7 points
orange: 8 points
silver: 9 points
olive: 10 points
purple: 11 points


There are now 10 pockets (one extra pocket added on each edge of the table). From top left, going clockwise, the pockets are:

purple, brown, orange, green, pink, silver, blue, olive, yellow, black

So the table looks a bit like this - I've put the value of the colour of each pocket to help you see, in case you are colour blind.

11------4------8
!......................!
!......................!
!......................!
7.....................3
!......................!
!......................!
!......................!
2.....................6
!......................!
!......................!
!......................!
10------5------9


The rules are

a) Once a colour has been potted, the same colour cannot be potted following the next red, nor following the red after that. (Once the 15th red and colour have been potted, this rule no longer applies - the final colours may and must be potted in the usual order of yellow, green, brown, blue, pink, black, orange, silver, olive, purple, regardless of the colours potted with the final reds.)

Example 1: Red Brown Red Yellow Red Blue Red Brown IS allowed
BUT
Example 2: Red Brown Red Yellow Red Brown IS NOT

Example 3: For the 14th and 15th reds,
Red Black Red Yellow Yellow Green Brown Blue Pink Black Orange Silver Olive Purple IS allowed


b) Whenever a colour has been potted into a corner pocket, the following colour cannot be potted into any pocket that lies along the same edge - that means it may not be on the same side (left or right) of the table, nor at the same end (top or bottom).

Example A: after potting a colour into the orange pocket, it would not be permissible to pot the next colour into the orange, green, pink or silver pockets (same side), nor into the purple or brown pockets (same end).

c) Whenever a colour has been potted into a pocket that is not a corner pocket, the following colour must be potted into a corner pocket, but must not lie on the same edge.

Example B: after potting a colour into the yellow pocket, the following colour can only be potted into the orange or silver pockets.

Example C: after potting a colour into the blue pocket, the following colour can only be potted into the orange or purple pockets.

d) Rules b) and c) apply even when down to the final 10 colours after all the reds have gone.

e) None of these rules apply to reds. It makes no difference into which pockets reds are potted.

f) And this is the important bit: potting a colour into a pocket of the same colour as the ball (e.g. pink into pink pocket) scores double points (in this example 2 x 6 = 12).


As ever, your question is: what is the highest break (ignoring free balls) you can make?

You don't need to say the highest theoretically possible - you just need to send me the highest break that YOU can find by Private Message by the Initial Deadline of 18:00 GMT on Sunday 3 December.


If you have any questions, please ask them on the thread.


You should explain how you get your break. For example, you might say:

Red
Green into Yellow pocket (or whatever)
Red
Pink into Orange pocket (or whatever)
....
...

and after all 15 reds and colours...

Yellow into Yellow pocket (or whatever)
Green into Silver pocket (or whatever)
Brown into Black pocket (or whatever)
....

[The Statman]

Sorry, The Statman, I've forgotten my own rule!... Should we be posting answers here or sending them to you by Private Message? I assumed as no deadline was given, it was the former.

Indeed.

Until it was mentioned earlier today I did not even know that rule had come into force! So I apologise for forgetting to implement the Rule I knew nothing about!

I think I shall just scrap this question and leave it for someone who has more experience in posting questions on Internet forums.

[elvaago]

In that case, I believe davis greatest's answer to be the lowest you can score if you remain truthful to the question being the two numbers before the = sign have to be different.
Otherwise it's as simple as
10 + 1 = 11
1 + 10 = 11
And I don't want points for this one. DG deserves it!