Puzzles with numbers and things

[snookersfun]

Elvaago, there are 'high number' calculator applets out there, but as you stated, I doubt we need them...
They are fun to play around with though

[elvaago]

When numbers get this big, you're not supposed to really use them anymore, except as a point of reference. But I'm a little too busy at work today to really work on this problem right now. Maybe tonight or tomorrow.

[snookersfun]

OK, just as a first thought (and totally ignoring the craziness of that number), how about not a lot of the original red remain on the table (only those at 'square positions' (do you want a number?) and the last ball (cue ball) would be red now on table 1.

[davis_greatest]

OK, just as a first thought (and totally ignoring the craziness of that number), how about not a lot of the original red remain on the table (only those at 'square positions' (do you want a number?) and the last ball (cue ball) would be red now on table 1.

Getting warm.

Yes, I want a number! (Doesn't need a computer.)

I need an explanation for why the last ball will be whichever colour you think it will be.

[elvaago]

The last ball will be red because it will be switched an odd number of times due to the fact that the total number of balls is odd. (147etc * 147etc-1)/2
7 x 6 = 42, not dividable by four, therefore when divided by two it's always odd.

I got that far. I have no idea whatsoever as to how many reds are left.

[davis_greatest]

The last ball will be red because it will be switched an odd number of times due to the fact that the total number of balls is odd. (147etc * 147etc-1)/2
7 x 6 = 42, not dividable by four, therefore when divided by two it's always odd.

I got that far. I have no idea whatsoever as to how many reds are left.

Hmmm... not really... actually the number of reds in each triangle ends in 78, not half of 42. You've also ignored the fact that there are 8 triangles. And being odd doesn't make any difference, anyway. But nice idea

[elvaago]

See, too big for my head. I'll wait this one out. :-)

[snookersfun]

Getting warm.

Yes, I want a number! (Doesn't need a computer.)

I need an explanation for why the last ball will be whichever colour you think it will be.

I just had another inspiration
number of reds left is 2 * your 147147..... number
and the last number is switched, as it is not a square (that I will explain after the match hopefully (n,n+1 business?)

[davis_greatest]

I just had another inspiration
number of reds left is 2 * your 147147..... number

Yes!

and the last number is switched

No

, as it is not a square (that I will explain after the match hopefully (n,n+1 business?)

No

[snookersfun]

Yes!


No


No

So, it is a square, well, so I'll be... but the first part was good, wasn't it?
Anyway number of red balls (n-1) are with x=147....147


(n-1)= 8*x*(#+1)/2 = 4*x(x+1)
the white ball (which I forgot in my square calculations) is then in position n= 4*x(x+1)+1= 4x^2+4x+1= (2x+1)^2 a perfect square!!!

[snookersfun]

some mess up, sorry again:
So, it is a square, well, so I'll be... but the first part was good, wasn't it?
Anyway number of red balls (n-1) are with x=147....147


(n-1)= 8*x*(x+1)/2 = 4*x(x+1)
the white ball (which I forgot in my square calculations) is then in position n= 4*x(x+1)+1= 4x^2+4x+1= (2x+1)^2 a perfect square!!!

[davis_greatest]

Yes, that's pretty much it! So can you tie it up with your final answer to both parts of the question?

That includes explaining what the significance of square numbers has to do with the number of switches.

[snookersfun]

Yes, that's pretty much it! So can you tie it up with your final answer to both parts of the question?

That includes explaining what the significance of square numbers has to do with the number of switches.

So, now I have to work hard again....

first one should notice, that final colour changes do not occur, when a ball is in a position, which is a perfect square.
Colours are switched, for the number itself and possible number of factors of the number (which are even for all numbers (e.g. 24=2*12, 3*8, 4*6) except perfect squares as one factor will appear twice naturally). Thus uneven number of switches alltogether result in colour change, whereas even number of switches (all squares) result in no colour change.

then back to the previous:
number of red balls (n-1) are with x=147....147
(n-1)= 8*x*(x+1)/2 = 4*x(x+1)

the white ball (which I forgot in my square calculations) is then in position n= 4*x(x+1)+1= 4x^2+4x+1= (2x+1)^2 a perfect square!!!

therefore, there are (2x+1)-1 = 2x square numbers before that last number (in red ball positions)

I really hope, that this is clear.

[davis_greatest]

Congratulations snookersfun

Yes, that will do! Well done.

So the answer to the question is that the last ball on Table 1 finishes up white, as it started (having been switched an even number of times), and the number of reds left on Table 1 is twice that big 147-digit number, i.e. there are

29429429429429429429429429429429429429429429429429 42942942942942942942942942942942942942942942942942 94294294294294294294294294294294294294294294294 reds left!

Every other ball on Table 1 - which is almost all of them (!) - finishes white!

I thought up this question in bed last night while happening to think about triangular numbers


HERE IS THE SCOREBOARD AFTER ROUND 39:

snookersfun……………………….…..17½
Vidas……………………………………….8½
abextra……………………………..…...5½
robert602…………………………………5
davis_greatest…………………..……5
elvaago...............................2
Semih_Sayginer.....................½

(some rounds may be worth more than one point)
(especially ones won by davis_greatest)

[snookersfun]

I thought up this question in bed last night while happening to think about triangular numbers

thanks, good puzzle, but.... what some people do in bed....