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# Find five numbers 1-10. sum of three each cubed = sum of two each cubed watch

1. Tried fitting it all in the title. To make it more clear:

Find five distinct whole numbers between 1 and 10 so that the cubes of three of them add up to the same number as the cubes of the other two.

Can someone tell me the technique to doing this please?

Thanks.
2. 3^3 x 4^3 x 5^3 = 216=1^3 x 6^3
- think i just lucked it! i cubed them all, was going to try different combos, and hit this straight off!
be interested to see a method, though.
3. (Original post by Hasufel)
3^3 x 4^3 x 5^3 = 216=1^3 x 6^3
It is the sum. Not the products :/

Try going for 3 smaller ones equal to 2 larger ones.
4. Doh!

5. Both eqaul 855, but I don't know about a general method. Just played around with a few numbers and got this to work.
6. Subscribe.
7. Presumibly there's somewhat of a method for products (prime decomposition?)

As for sums, I don't know. I may not know enough tools to be able to find a method of doing this. Maybe modular arithmetic :\?
8. (Original post by USB)
It is the sum. Not the products :/

Try going for 3 smaller ones equal to 2 larger ones.
By the looks of it they've written 'x' but meant '+', because they're the correct numbers.
9. (Original post by nuodai)
By the looks of it they've written 'x' but meant '+', because they're the correct numbers.
Hmm that would mean there is more than one solution.
10. (Original post by USB)
Hmm that would mean there is more than one solution.
There's no reason why there shouldn't be!
11. There are two solutions, up to ordering. Easiest method is to write a program.
12. (Original post by ghostwalker)
There are two solutions, up to ordering. Easiest method is to write a program.
Is there any method a human could do?
13. (Original post by Tallon)
Is there any method a human could do?
Not that I'm aware of; though that doesn't mean there isn't one.

Having said that, there are only 2520 combinations to check.
14. (Original post by Tallon)
Is there any method a human could do?
You could draw the 5-dimensional hypersurface described by the set of coordinates satisfying the equation , overlay an integer coordinate lattice and observe, by inspection, the points of intersection.

EDIT: Oh wait, a human!

There probably is a way of doing this without a brute force method. But in this case, cheating by writing a computer program to investigate the 2520 possible cases is probably the easiest thing to do.
15. (Original post by nuodai)
By the looks of it they've written 'x' but meant '+', because they're the correct numbers.
Not unless my calculator's wrong...

By my count:

Anyway, knocking up a quick maple program, the answers I got were:

Spoiler:
Show

And additionally, if you ignore the fact they have to be "distinct" numbers (thanks ghostwalker),

This isn't really answering the question IMHO though, I'd still like to know why these work.

EDIT: Hey, that's cool, I never noticed that

and

however
16. (Original post by StephenNeill)
Spoiler:
Show

Check the question; third option's not valid.
17. (Original post by ghostwalker)
Check the question; third option's not valid.
Whoops!

Thanks.
18. I used the internet quickly to solve it.

There are only 45 ways to sum two distinct cubes from 1 to 10 and they are readily accessible from the sum of two distinct cubes page here:

http://oeis.org/A024670/b024670.txt

I then cross-checked them against the sum of three distinct cubes page here:

http://oeis.org/A024975/b024975.txt

It only took a couple of minutes to figure out which cubes were needed to produce the matching answers.
19. Looking at this as a systematic search problem, I wrote 1000, 720, ... 8, 1 in a vertical table and set about considering all pairs, starting with 1000+729. A simple visual check will then find whether there could be three to match. I did wonder about mod 2 and mod 3 which are good for cubes, but the numbers are so small it hardly helps.

But my first approach was to play around with factorising x^3 - y^3 and stuff - I got nowhere.

Then I started to wonder what would happen if the upper bound were not 10, but large. For example, any multiple of 6 can be written as the sum of four cubes - usually two positive and two negative - but not distinct.

And then Wolfram showed me that there are similat sequences with distinct cubes.

So it's an interesting problem. Better mathmos may look for a general method and fail, whilst a good systematic search is quite quick.

Where did the problem originate?
20. (Original post by ian.slater)
Where did the problem originate?
One reference, not necessarily the origin, is the National Mathematics Magazine, 1938, Problem Department, possibly Walter B. Clarke.

Difficult to tell as it's on Jstor, and I don't have full access.

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