wu :: forums « wu :: forums - Two cubes with Total Volume 17 » Welcome, Guest. Please Login or Register. May 18th, 2022, 9:27pm RIDDLES SITE WRITE MATH! Home Help Search Members Login Register
 wu :: forums    riddles    hard (Moderators: Grimbal, Eigenray, SMQ, towr, ThudnBlunder, Icarus, william wu)    Two cubes with Total Volume 17 « Previous topic | Next topic »
 Pages: 1 2 Reply Notify of replies Send Topic Print
 Author Topic: Two cubes with Total Volume 17  (Read 4169 times)
Barukh
Uberpuzzler

Gender:
Posts: 2276
 Two cubes with Total Volume 17   « on: Jul 6th, 2007, 5:47am » Quote Modify

Two equal cubes with side length 2 units have total volume of 16 cubic units.

Can you find the exact sizes of 2 cubes that have total size of 17 cubic units?
 « Last Edit: Jul 6th, 2007, 5:48am by Barukh » IP Logged
SMQ
wu::riddles Moderator
Uberpuzzler

Gender:
Posts: 2084
 Re: Two cubes with Total Volume 17   « Reply #1 on: Jul 6th, 2007, 6:07am » Quote Modify

Huh?

Is the cube root of 17/2 not an exact size?  Is there some further restriction?  Is there some significance I'm missing to your using "length" and "volume" in the first line, but "size" (apparently to mean two different things) in the second?

--SMQ
 IP Logged

--SMQ

towr
wu::riddles Moderator
Uberpuzzler

Some people are average, some are just mean.

Gender:
Posts: 13730
 Re: Two cubes with Total Volume 17   « Reply #2 on: Jul 6th, 2007, 6:40am » Quote Modify

Do the two cubes that total 17 have to be the same size? Must the side lengths be integer? Can they be negative?
 IP Logged

Wikipedia, Google, Mathworld, Integer sequence DB
Barukh
Uberpuzzler

Gender:
Posts: 2276
 Re: Two cubes with Total Volume 17   « Reply #3 on: Jul 6th, 2007, 7:45am » Quote Modify

Well, well... I should state it more carefully.

What was meant is: the side lengths should be positive integers or rationals.

Sorry for inconveniece.
 IP Logged
Sir Col
Uberpuzzler

impudens simia et macrologus profundus fabulae

Gender:
Posts: 1825
 Re: Two cubes with Total Volume 17   « Reply #4 on: Jul 7th, 2007, 3:48pm » Quote Modify

Duh! [/edit]

I've not solved it, butand I've made a littleno progress...

Ignore fumbling nonsense below[/edit]
::
(a/c)3 + (b/d)3 = 17
a3 + b3 = 17(bd)3 = 17e3
(a+b)(a2-ab+b2) = 17e3

As 17 is prime, a+b=17 or a2-ab+b2 = 17

If a+b=17 then a2-ab+b2 = e3.
(a+b)2 = a2+2ab+b2 = 289 => a2+b2 = 289-2ab

So 289-3ab = e3 => 289-e3 = 3ab.

Only 289-64=225 and 289-1=288 are divisible by 3 and with 3ab=225 or 288 it can be verified that there are no rational solutions in a.

So if a solution exists then a2-ab+b2 = 17 and a+b = e3.
(a+b)2 = a2+b2+2ab = e6
a2+b2 = e6-2ab
So e6-3ab = 17 or e6-17 = 3ab.

As 17==2 mod 3 we are looking for e==2 mod 3.

But that's as far as I can get!
::
 « Last Edit: Jul 8th, 2007, 2:07am by Sir Col » IP Logged

mathschallenge.net / projecteuler.net
Obob
Senior Riddler

Gender:
Posts: 489
 Re: Two cubes with Total Volume 17   « Reply #5 on: Jul 7th, 2007, 3:56pm » Quote Modify

on Jul 7th, 2007, 3:48pm, Sir Col wrote:
 As 17 is prime, a+b=17 or a2-ab+b2 = 17 ::

Doesn't it only follow that 17 divides a+b or 17 divides a2-ab+b2?
 IP Logged
FiBsTeR
Senior Riddler

Gender:
Posts: 581
 Re: Two cubes with Total Volume 17   « Reply #6 on: Jul 7th, 2007, 7:34pm » Quote Modify

on Jul 7th, 2007, 3:48pm, Sir Col wrote:
 (a/c)3 + (b/d)3 = 17 a3 + b3 = 17(bd)3

Can you explain how you got from the first line to the second line?

EDIT: I assumed that maybe you meant a3 + b3 = 17(cd)3, but even then I still don't follow how you can arrive at this, as I get a3d3 + b3c3 = 17(cd)3
 « Last Edit: Jul 7th, 2007, 7:36pm by FiBsTeR » IP Logged
SWF
Uberpuzzler

Posts: 879
 Re: Two cubes with Total Volume 17   « Reply #7 on: Jul 7th, 2007, 9:37pm » Quote Modify

We need to find a3+b3=17c3 with a,b, and c integers. Then a/c and b/c are the sizes of the two cubes. Assume a and b are relatively prime since can multiply a,b,c by any integer for other solutions.

If you look at the values of integers cubed mod 17, it turns out that the form of and and b must be  a=17*A+n and b=17*B-n where A and B are integers and n is an integer from 1 through 16.  The original equation then becomes:

c3=(A+B)( 289*(A2-AB+B2) + 51*n*(A-B) +3*n2)

That doesn't look much better, but it did eliminate many possibilities.
 IP Logged
towr
wu::riddles Moderator
Uberpuzzler

Some people are average, some are just mean.

Gender:
Posts: 13730
 Re: Two cubes with Total Volume 17   « Reply #8 on: Jul 8th, 2007, 10:48am » Quote Modify

Using a3 +b3 =17c3, I think c has to be a multiple of 7
 IP Logged

Wikipedia, Google, Mathworld, Integer sequence DB
Barukh
Uberpuzzler

Gender:
Posts: 2276
 Re: Two cubes with Total Volume 17   « Reply #9 on: Jul 8th, 2007, 10:59am » Quote Modify

on Jul 8th, 2007, 10:48am, towr wrote:
 Using a3 +b3 =17c3, I think c has to be a multiple of 7

Why?
 IP Logged
towr
wu::riddles Moderator
Uberpuzzler

Some people are average, some are just mean.

Gender:
Posts: 13730
 Re: Two cubes with Total Volume 17   « Reply #10 on: Jul 8th, 2007, 11:13am » Quote Modify

on Jul 8th, 2007, 10:59am, Barukh wrote:
 Why?
Why not?

I tried sums of two cubes modulo 7 that are also 17 times a cube mod 7. And the latter turns out to 0 mod 7, hence a multiple of 7.
 « Last Edit: Jul 8th, 2007, 11:41am by towr » IP Logged

Wikipedia, Google, Mathworld, Integer sequence DB
towr
wu::riddles Moderator
Uberpuzzler

Some people are average, some are just mean.

Gender:
Posts: 13730
 Re: Two cubes with Total Volume 17   « Reply #11 on: Jul 9th, 2007, 2:41am » Quote Modify

Maybe it would work to combine the results when you use results modulo primes.
For example with p=19 we have only 7 of the possible 19 sums of cubes (mod 19) that can be 17 times a cube (mod 19). With
337  it's  113,  for 2953  it's  985. It seems to be about a third each time.
Now if these results were somewhat independant, you could quickly narrow down our a,b,c.
 IP Logged

Wikipedia, Google, Mathworld, Integer sequence DB
towr
wu::riddles Moderator
Uberpuzzler

Some people are average, some are just mean.

Gender:
Posts: 13730
 Re: Two cubes with Total Volume 17   « Reply #12 on: Jul 9th, 2007, 4:28am » Quote Modify

Seems the maths required to deduce an answer is a bit beyond what I know: http://www.math.niu.edu/~rusin/known-math/94/cubes.sum
 IP Logged

Wikipedia, Google, Mathworld, Integer sequence DB
Barukh
Uberpuzzler

Gender:
Posts: 2276
 Re: Two cubes with Total Volume 17   « Reply #13 on: Jul 9th, 2007, 5:18am » Quote Modify

on Jul 9th, 2007, 4:28am, towr wrote:
 Seems the maths required to deduce an answer is a bit beyond what I know

I don't think so.

Now, when the answer is known: how difficult it is to find the solution with negative sizes?
 IP Logged
revenge
Junior Member

Posts: 61
 Re: Two cubes with Total Volume 17   « Reply #14 on: Jul 9th, 2007, 3:42pm » Quote Modify

Would this help?
 IP Logged
SWF
Uberpuzzler

Posts: 879
 Re: Two cubes with Total Volume 17   « Reply #15 on: Jul 9th, 2007, 9:36pm » Quote Modify

Looks like people have managed to look up the answer already, but let me give the progress I have made so far. Continuing from where I left off last time, make the substitution A=x+e and B=x-e, and the equation becomes:

c3=2x*(289*x2+3*q2) where q=17*e+n

The problem is reduced to finding integers x and q that give a cube for this formula, and e is found from floor(q/17), and n=mod(q,17). I gave up on an elegant way to solve the equation, but did find some solutions.

Solution 1: From the formula, c must be a multiple of 2, and from towr's observation also a multiple of 7. Try for the simpliest possible solution:  c=2*7=14. A simple thing to try is x=1, and fortunately solving for q gives an integer: q=19. So e=1 and n=2. A=x+e=2, B=x-e=0, a=17*A+n=36, b=17*B-n=-2.  A solution is a=36, b=-2, c=-14, or can factor out the 2 for a=18, b=-1, c=7. Unfortunately this has a negative number.

Solution 2: Another easy thing to try is x a perfect cube: x=d3 and c must be of the form c=14*d*f, and

(14*f)3=2*(289*d6+3*q2)

Try various values of f and d, and solve for q until an integer q is found. A simple computer program quickly finds d=19 and f=307 as a solution, and it gives x=6859, q=93277, a=209880, b=23326, c=81662. A common factor of 2 can be removed giving: a=104940, b=11663, c=40831. I could find no other positive solution except multiples of this.
 « Last Edit: Jul 9th, 2007, 9:38pm by SWF » IP Logged
towr
wu::riddles Moderator
Uberpuzzler

Some people are average, some are just mean.

Gender:
Posts: 13730
 Re: Two cubes with Total Volume 17   « Reply #16 on: Jul 10th, 2007, 12:40am » Quote Modify

on Jul 9th, 2007, 9:36pm, SWF wrote:
 I could find no other positive solution except multiples of this.
From as much as I understood from the link I found, once you have one solution, you can, somehow, find infinitely many others (alternating negative and positive, and not merely multiples).
 IP Logged

Wikipedia, Google, Mathworld, Integer sequence DB
Barukh
Uberpuzzler

Gender:
Posts: 2276
 Re: Two cubes with Total Volume 17   « Reply #17 on: Jul 10th, 2007, 12:40am » Quote Modify

Well done, SWF!

Your negative solution is what I meant in my previous post by negative sizes. Fortunately, it's easy to find in this particular case (but not in general).

Once you've got this solution, an infinite number of others may be generated by a simple transformation formula. Can you find it? As a reference, we've got 2 solutions:

a1 = 18, b1 = -1, c1 = 7.
a2 = 104940, b2 = 11663, c2 = 40831.

What's the relation between the two?
 IP Logged
Hippo
Uberpuzzler

Gender:
Posts: 919
 Re: Two cubes with Total Volume 17   « Reply #18 on: Jul 10th, 2007, 3:49am » Quote Modify

The answer to the last question was given in already mentioned link ... the squaring in elliptic curves ... btw: this connection with elliptic curves was unknown to me.
 IP Logged
towr
wu::riddles Moderator
Uberpuzzler

Some people are average, some are just mean.

Gender:
Posts: 13730
 Re: Two cubes with Total Volume 17   « Reply #19 on: Jul 10th, 2007, 4:16am » Quote Modify

on Jul 10th, 2007, 3:49am, Hippo wrote:
 The answer to the last question was given in already mentioned link ... the squaring in elliptic curves ...
But what does that mean, that's what I'm wondering.

An actual procedure to find the next triple can be found halfway down on http://www.math.niu.edu/~rusin/known-math/95/numthy.cub
 IP Logged

Wikipedia, Google, Mathworld, Integer sequence DB
Hippo
Uberpuzzler

Gender:
Posts: 919
 Re: Two cubes with Total Volume 17   « Reply #20 on: Jul 10th, 2007, 7:28am » Quote Modify

For description of the group operations you can also google the description of Lenstra's Elliptic Curve Factorizacion Algorithm.

But may be there is a difference in the curve shape ... so I should think about it a little bit ...

... I can see I know very little about such curves ... if I have read it well in wikipedia ... this was the base for the proof of the Last Fermat Theorem ...
 « Last Edit: Jul 10th, 2007, 4:21pm by Hippo » IP Logged
Barukh
Uberpuzzler

Gender:
Posts: 2276
 Re: Two cubes with Total Volume 17   « Reply #21 on: Jul 10th, 2007, 8:37am » Quote Modify

Quote:
 An actual procedure to find the next triple can be found halfway down on http://www.math.niu.edu/~rusin/known-math/95/numthy.cub

Exactly! Looks like Dave knows everything...

on Jul 10th, 2007, 4:16am, towr wrote:
 But what does that mean, that's what I'm wondering.

Maybe, it's the same as the above? I will need to find out...
 IP Logged
SWF
Uberpuzzler

Posts: 879
 Re: Two cubes with Total Volume 17   « Reply #22 on: Jul 10th, 2007, 6:27pm » Quote Modify

No wonder I couldn't find any more positive solutions, they get big fast! The formulas given in towr's link seem to have an error. I think it should say that new values of a,b,c can be generated from those of a previous step with:

anew= a*(a3+2*b3)
bnew= -b*(2a3+b3)
cnew= c*(a3-b3)

In the x and q of the formula I posted earlier, the next x of a solution is found from the previous x and q with the simple formula, xnew=x*q3. The next q is qnew=(3*(17x)4 + 6*(17xq)2 - q4)/8. Values of a and b in terms of x and q are:  a=(17x+q)/2   b=(17x-q)/2.

 « Last Edit: Jul 15th, 2007, 11:48am by SWF » IP Logged
Hippo
Uberpuzzler

Gender:
Posts: 919
 Re: Two cubes with Total Volume 17   « Reply #23 on: Jul 25th, 2007, 2:43am » Quote Modify

I don't know how to continue ... using elliptic curves requires other curve than x^3+y^3=n.

One can choose v=(x+y),w=xy and got nice elliptic curve, but rational v,w solution seems do not help finding rational solution for x,y.

So I look forward for anyone's explanation of the citations.
SWF: Can you give more details of your computations? (may be subresults?)
 IP Logged
srn437
Newbie

the dark lord rises again....

Posts: 1
 Re: Two cubes with Total Volume 17   « Reply #24 on: Aug 29th, 2007, 10:05am » Quote Modify

17 expressed as two cubes. Can the sides have irrational or negative length. If no to both, it's impossible.
 IP Logged
 Pages: 1 2 Reply Notify of replies Send Topic Print

 Forum Jump: ----------------------------- riddles -----------------------------  - easy   - medium => hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »