Puzzles with numbers and things

[davis_greatest]

If g,c,o are the scores of Gordon, Charlie and Oliver
we have:
g+c = a^2
o+g=(a+1)^2
c+o=(a+2)^2

Congratulations, Monique!

To find the points, you can simplify it a bit and take out the trial and error in finding n if you write

g+c = (n-1)²
o+g= n²
c+o= (n+1)²

and g+c+o = m²

If you add up the first three equations, you get
2(g+c+o) = 3n² +2, which must equal 2m² from the 4th equation.

You can quickly see that n must be even, so writing n=2k, we can simplify
3n² +2=2m² to
6k² +1 = m². (*)

You can either solve this in integers by trial and error or recognise it as a form of "Pell's equation" ( http://en.wikipedia.org/wiki/Pell's_equation )

The easiest way to solve this sort of equation is to produce "continued fractions" that converge to the square root of 6 - there is an algorithm to do this. The fractions are:

2/1, 5/2, 22/9, 49/20, 218/89, 485/198, 2158/881 ...

which give possible values for m/k.

You need to stop when you find one that satisfies equation (*) - actually, all the even ones do (the 2nd, 4th, 6th). The 2nd is no good as it gives 0 points for one of the apes - but 49/20 works. So k=20; n = 2k = 40 and m=49.

Then from the first 3 equations it is easy to arrange them to get:

g = 2k² - 4k = 720
c = 2k² +1 = 801
o = 2k² + 4k = 880

[Monique]

If g,c,o are the scores of Gordon, Charlie and Oliver
we have:
g+c = a^2
o+g=(a+1)^2
c+o=(a+2)^2
and 2*(g+c+o)=3*a^2+6*a+5=2*(a+n)^2 or a^2+(6-4*n)*a+(5-n^2)=0 (E)
where a and n>2 are integers.

Solving E we find a discriminant D=4*(6*n^2-12*n+4) and D must be a square.
The first possible value for n=10 which gives a=39

Solving the system above we find
o=880
c=801
g=720

which are reasonable values for snookers scores. A bigger n ... would establish some records...

Because the ape who has the lowest score is not the one who won the fewest match Gordon must have won one match, and at least 5 frames. Charlie has won more frames ... but can only have won a maximum of 6.
So the match scores must be:
o vs c 4-3
o vs g 4-1
g vs c 4-3

[davis_greatest]

I think sweet, but we'd have to ask Barry, who felt very full by the time he had finished cleaning it off.

Meanwhile, first to give correct answers were snookersfun and abextra - congratulations. Next answer with explanation please on the thread.

[April madness]

was that sweet marmalade or more like sweet-and-sour?

[davis_greatest]

Round 267 - Round Robbing

Last night, Gordon, Charlie, and Oliver played a mini snooker tournament. The format was very simple - each ape played a best-of-7 match (i.e. first to 4 frames) against each of the other 2 apes, so 3 matches were played in all.

The matches were very high quality, and none culminated in a whitewash.

At the end, in order to try to determine a winner, we looked at the points and frame scores. The apes also totalled how many points they had each scored over all the frames in their matches. Then, during the night, they broke into Barry The Baboon's Ball Shop and each ape stole / borrowed a number of balls equal to the number of points the ape had scored.

It was all a bit of fun, as they were intending to cover the balls in marmalade and put them back in their friend Barry's shop this morning before he had got up.

The ape who had won the most matches was also the one who had the most marmalade-covered balls. The ape who won the fewest frames had taken the fewest balls but was not the ape who had won the fewest matches.

Before they put the balls back in Barry's shop, they tried to stick them together in a nice pattern with the marmalade. First, Gordon and Charlie noticed that if they combined their balls, they could stick them in a nice solid square formation. However, Oliver felt very left out and cried, so Charlie took his own balls away and Oliver and Gordon instead put their balls together - again, they found that they could form a perfect solid square, this time one row larger than the previous square.

This was no good, as now Charlie was left out - so, instead, Charlie and Oliver combined their balls, and again they formed a perfect square! And the square was now even bigger still - one row more than before!

The apes thought long and hard about what to do, so no one would be left out. So, Gordon added his balls, and they found that with all three apes' balls there, they could again make a perfect square! Bigger than ever! And the marmalade held it together nicely.

So, they carried the square of balls back to Barry's shop, and it formed a nice marmalade-covered surprise for him this morning!

What was the frame score in each of the 3 matches, and how many points did each ape amass in total?

Answers initially by Private Message please

[davis_greatest]

Very nice, snookersfun! Congratulations to you and Monique! You can also do this without using remainders - Monique did it following the outline below.

With the same notation as you had (p black balls and x red balls), we have
7p+x = a², p+x = b²;

subtracting the second from the first gives
6p = a²- b² = (a-b)(a+b).

a-b and a+b must both be odd or both be even; since 6p is even they can't both be odd so must both be even. Hence p is even.

The only even prime is 2, so p=2, 6p=12, and so a-b=2, a+b=6. This leads to b=2, so x=2.

Thus there are p+x = 2+2 = 4 balls.

[snookersfun]

Yes, congratulations, Monique! 4 balls it is!

Monique also sent a proof by private message that 4 must be the greatest number of balls. Does anyone else want to put up a proof here?

pondered, as my proof is far from elegant, but what the heck:

7p+x = a^2 and p+x = b^2

squares being mod8: 0,1,4 and primes (except the 2) 1,3,5,7, while 7xprime reverses those mods into 7,5,3,1. Therefore there is no possible x (mod 0-7) to be added to prime and 7xprime to both result in a square. Thus prime can only be 2 and x mod8, 2 or 6. Difference between the squares being 12, only x=2 works actually

[davis_greatest]

R265 : 4 ... 2 blacks and 2 reds.

Yes, congratulations, Monique! 4 balls it is!

Monique also sent a proof by private message that 4 must be the greatest number of balls. Does anyone else want to put up a proof here?

[davis_greatest]

Abextra you are the champion at these!

So, this took a bit longer than expected to come up, but blame Fu and Greene, who carried on for ever...
but on the bright side, I now have the 273 as well (wohee), so following are both breaks in hidden text:

Congratulations, snookersfun, for finding these two maximum possible breaks, for rounds 263 and 264 - and congratulations also to abextra, who had sent them by private message!

[davis_greatest]

Here it is in hidden text (maybe somebody is still working on it ).

black in black (21) - green in green (9) - blue in blue (15) - pink in pink (18) -
black in blue (12) - blue in pink (11) - brown in blue (9) - pink in pink (18) -
blue in blue (15) - green in green (9) - black in black (21) - brown in brown (12) -
pink in black (13) - green in green (9) - black in black (21) = 213 points

15 red balls = 15 points

yellow in brown (6), green in black (10), brown in brown (12), blue in black (12),
pink in brown (10), black in black (21) = 71 points

in total 213+15+71=299 points

Congratulations, abextra, for finding the 299 maximum break for round 261!

[PaulTheSoave]

Unbelievable! I surrender!

Sorry, the Zohar was the copyright aspect!

Apologies

[snookersfun]

cueimages10 036.jpg
the Zohar having nothing to do with it...

What are you reading Paul

[PaulTheSoave]

Here is a small puzzle for the cleverclogs. I have listed the elements of a fomula, but mixed them up a bit. Show me the formula in graphic display. Well, for this one the deadline is 31st december 2007. Answers on here. hehehehe .

ps. dont worry, will be the only one I will post!

R3 R3
R1 R2
OH
LiAlH4 then H+
Et2O
R3 R3
R1 R2
OH
R3 R3
R1 R2
OH
R3 R3
R1 R2
R4
Et2O
1) 1eq Ti(Oi-Pr)4
2) 1.2eq DET-(+)
3) 0.65eq t-BuOOH
MS, CH2Cl2
-20oC, 1h
R4MgI, 10% CuI
ee > 98%
1 2
3

Copyright. Zohar

[snookersfun]

R265 : 4 ... 2 blacks and 2 reds.

proof by mod8 again, right?

[snookersfun]

Welcome back, Snookersfun!

I've found 273 for R263 and 300 for R264 , will send them by PM, so feel free to post your breaks here!

Abextra you are the champion at these!

So, this took a bit longer than expected to come up, but blame Fu and Greene, who carried on for ever...
but on the bright side, I now have the 273 as well (wohee), so following are both breaks in hidden text:

263:

7 in 7, 4 in 4, 6 in 7, 7 in 4,
5 in 7, 4 in 4, 7 in 7, 6 in 4,
5 in 7, 4 in 4, 7 in 7, 6 in 4,
5 in 7, 4 in 4, 7 in 7
which should add up to 192

15 reds and

2 in 4, 3 in 7, 4 in 4, 5 in 7, 6 in 4, 7 in 7 = 66

sum: 192+15+66=273

300:

7 in 7, 4 in 4, 6 in 7, 3 in 3,
7 in 7, 4 in 4, 6 in 7, 3 in 3,
7 in 7, 4 in 4, 2 in 2, 6 in 6, 5 in 5, 3 in 3, 7 in 7
which should add up to 215

15 reds and

2 in 4, 3 in 7, 4 in 4, 5 in 7, 6 in 4, 7 in 7 = 70

sum: 215+15+70=300