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No 42 as a sum of three cubes - problem solved using the power of grid computing by hotbit

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· @hotbit ·
$2.04
No 42 as a sum of three cubes - problem solved using the power of grid computing
<center>
![42.png](https://cdn.steemitimages.com/DQmSghAgVrB3zXRSgYi6HNP8uUfVR3ahazL8SAgn8hg56V4/42.png)
</center>

The seemingly simple problem is: what are three numbers s, y, z, such that cubed and summed give n = 42?

![](https://cdn.steemitimages.com/DQmTiYeb2oRJXARccZc1qweXxNkbpzBwtpEAKcE4CAGocAC/image.png)

Above problem is an example of the family of mathematical puzzles called Diophantine equation.
Until recently 42 was the only number lower than 100 for which answer was not known. Andrew Booker, of the University of Bristol in England, and Andrew Sutherland of Massachusetts Institute of Technology teamed up and using Charity Engine computer grid found, that:

![](https://cdn.steemitimages.com/DQmY3WnU8HzmieQXNj4XqnxbessUu7cySwj9Hjghv5izic9/image.png)

Researchers developed a custom algorithm to accelerate computations. I have asked Prof. Andrew Booker about Charity Engine role in the project. A custom BOINC server run by the company was feeding work orders to thousands of computers within the grid. Thus work orders were only available through the Charity Engine app.

The previous puzzle, solution for number 33 was found using supercomputer resources. It shows that some scientific problems may be solved using the power of distributed computing network run by volunteer researchers using spare power of their gaming pcs or laptops at a fraction of cost or no cost at all.

If you would like to get your teeth on the problem, don't be disappointed! The next 42 is:

<center>
![114.png](https://cdn.steemitimages.com/DQmZbiwsvHG1eMEwQi6dnE6xBRoUi6P3ihbegBWR6pdp7tX/114.png)
</center>
Good luck!

Links:
https://www.youtube.com/watch?v=zyG8Vlw5aAw&t=1s
[Sums of three cubes](https://en.wikipedia.org/wiki/Sums_of_three_cubes)
[42](https://en.wikipedia.org/wiki/42_(number))
πŸ‘  , , , , , , , , , , , , , , , , , , , , , ,
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@partiko ·
Thank you so much for participating in the Partiko Delegation Plan Round 1! We really appreciate your support! As part of the delegation benefits, we just gave you a 3.00% upvote! Together, let’s change the world!
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@steem-ua ·
#### Hi @hotbit!

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Your **UA** account score is currently 2.149 which ranks you at **#23975** across all Steem accounts.
Your rank has not changed in the last three days.

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@jamescowens ·
$0.62
He he. The reason it is not easy is that the exponents are odd. This allows the sign to carry through and essentially allows the domain of solutions to be infinite. Conversely, even powered versions, such as

x^(2k) + y^(2k) + z^(2k) = N, k is a positive integer, are much easier, because

x^(2k) = N - y^(2k) - z^(2k). Since y^(2k) and z^(2k) must be >= 0, then the maximum value that any one term can have is where the other two are zero, therefore a loose bound to the number space is

x^(2k) = N ....  x = floor(N^(1/2k)). This is symmetric. (Ignoring the depression that occurs when other terms are non-zero. (The extreme case is when they are all equal... 3*x^(2k) = N .... x = floor((N/3)^(1/2k)).)

So for k = 1 and N = 100 (i.e.) x^2+y^2+z^2 = 100 would result in a search space loosely bounded by

0 <= x <= 100^(1/2) = 10
0 <= y <= 100^(1/2) = 10
0 <= z <= 100^(1/2) = 10

a cube of 0 though 10 on each side or 11^3 = 1331 as a LOOSE bound. Not very hard. Higher powers result in even a smaller search space for a given N.
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