Brain Teaser No 1

Tennis trilogy - Part 3

Can you see a neat link between part 1 and part 2 results? Can you find, and prove, a formula for the number of matches in a tennis tournament with any number of contestants?

Cheers

Don

In any tournament everybody but the overall winner must lose a match. Every match has just one loser. So the number of matches is the number of losers, or the number of participants less one.

Didn't really answer the question, sorry....

Paul

Didn't really answer the question, sorry....

errr.....I think you just did

Cheers

Don

It appears that Paul has answered both parts of the trilogy in one fell swoop.

Contestants less one in any knockout competition.

Matthew

I came across this problem in a book I am reading at the moment (Dr Riemann's Zeros). The equation:
a^3+b^3=22c^3
has a solution for integer values of a,b,c. What is it? Fairly obviously it is possible to take a brute computational force approach to this, but is there another way to easily get a solution?
John

If it is any help, a and b are prime numbers and c is even.
John

....obviously it is possible to take a brute computational force ......

presumably this is reference to "Google's" brute force searching powers..........

Cheers

Don

A way of finding (more) solutions to this is to take a simple solution to the equation, express it as a rational point on an elliptic curve and then use this to find the coordinates of the next solution and so on. Unfortunately the only simple solution I can see is (1,-1,0) which is no good.

I guess you are asking for a proof in the absence of any "seed" solution ?

presumably this is reference to "Google's" brute force searching powers........

No, I have a solution to this equation which I got by writing some code in java.
John

John,

That would appear to be the sledge hammer brute force method.

Matthew

Cracked nuts,

17299, 25469, 9954
34598, 50938, 19908
51897, 76407, 29862

I guess an 'etc' might be appropriate. I haven't looked deeper than a max for c of 100000 in case there's another family of solutions.

Paul

Well, my poor PC has looked as deep as,

605465^3 + 891415^3 = 22(348390^3)

Which is obviously one of the family.

Paul

17299, 25469, 9954

Well done Paul, that was the solution I found. As you point out for each a, b, c, na, nb, nc is also a solution, but I believe (but can't prove) this is the only set.

I coulnd't find any mathematical tricks to get at a solution without a computer, I assume that was the case for everyone else as well?
John

What is the smallest integer n that is the sum of two different integral cubes? That is, what is the smallest n for integers a, b, c, d such that:
a^3 + b^3 = n
c^3 + d^3 = n
John

quote:

---

Originally posted by John Channing:
I coulnd't find any mathematical tricks to get at a solution without a computer, I assume that was the case for everyone else as well?

---

I certainly couldn't find a paper and pencil solution. A little web searching revealed these sort of problems are called Diophantine equations but no real hints at an elegant solution. It has made me think I should pick up a text on number theory though.

Is your new problem doable without a computer?

cheers,

Dan

What is the smallest integer n that is the sum of two different integral cubes? That is, what is the smallest n for integers a, b, c, d such that:

1729?

Cheers

Don

Is your new problem doable without a computer?

Yes (assuming I have got the right answer of course)

Cheers

Don

a^3 + b^3 = 22c^3........

can anybody enlighten us on the significance of the "22"

are there other 'solution families' if the "22" is replaced by (say) 21....or 19.....or.....

of course we know that there are NO solutions when "22" is replaced by "1"

Cheers

Don

quote:

---

Originally posted by Don Atkinson:
_What is the smallest integer n that is the sum of two different integral cubes? That is, what is the smallest n for integers a, b, c, d such that:_

1729?

Cheers

Don

---

Otherwise known as the "Hardy-Ramanujan Number".

quote:

---

Originally posted by Don Atkinson:
a^3 + b^3 = 22c^3........

can anybody enlighten us on the significance of the "22"

are there other 'solution families' if the "22" is replaced by (say) 21....or 19.....or.....

of course we know that there are NO solutions when "22" is replaced by "1"

Cheers

Don

---

Otherwise known as "Fermat's last theorem".

Minky,

Otherwise known as the "Hardy-Ramanujan Number".

i'm almost afraid to ask who, or what, Hardy-Ramanujan is (are ?)

Femant.......well, I think the margins of this post are slightly too small to explain who he was.....

Cheers

Don

Otherwise known as the "Hardy-Ramanujan Number".

.....and there is still an opportunity for someone to excel by providing numbers for a, b, c and d........

Cheers

Don

quote:

---

are there other 'solution families' if the "22" is replaced by (say) 21....or 19.....or.....

---

It took about 8 hours of CPU time to search up to a or b = 1000000, only minutes for 100000. So more searching is possible. It seems a bit crude though.

I'll set it going tonight....

Paul

Is your new problem doable without a computer?

Yes (assuming I have got the right answer of course)

Well....I've now re-checked all the smallish cube numbers and I'm confident there is nothing smaller than 1729 that fits the bill.

No doubt there are bigger numbers....but they will take a computer IMHO....over to Paul...or John

Cheers

Don

quote:

---

Originally posted by Don Atkinson:
No doubt there are bigger numbers....but they will take a computer IMHO

---

Here are a few bigger ones:
99 29 92 60 994688
98 35 92 59 984067
98 24 89 63 955016
97 20 96 33 920673
97 47 90 66 1016496
...
51 12 43 38 134379
48 6 45 27 110808
48 4 40 36 110656
40 12 33 31 65728
39 17 36 26 64232
36 3 30 27 46683
34 9 33 16 40033
34 2 33 15 39312
32 4 30 18 32832
27 10 24 19 20683
24 2 20 18 13832
16 2 15 9 4104
12 1 10 9 1729

Obviously if (a,b),(c,d) is a solution multiples are also solutions.

cheers,

Dan

p.s. what's the largest 4-tuple that's not a multiple that someone can compute?

[This message was edited by Dan M on FRIDAY 21 November 2003 at 00:49.]