Steps

1) Find all the factors possible for the number

2) Take cuberoot of the given number

3) If it’s not an integer then ceil it to next value.

4) Find all the numbers starting from that number which can perfectly divide that number, if you get a prime number then that’s your first number else the minimum number from all the possible divisors is the first number.

5) Next for remaining 2 numbers divide and find the remaining amount for which we need to solve.

6) Then take the square root, and if it’s not an integer then ceil it to next value.

7) Repeat step 4 and you will get your second number

8) Now you can easily find the third number.

George, we’re using different definitions of “easy” and “hard”. I’m happy with an algorithm that’s polynomial in V, the volume of the box, and brute force does that well enough: just try all the triples of numbers less than V (there are only V^3 of them) and see which is best. Cryptographers want (or, perhaps, don’t want) an algorithm that’s polynomial in the *number of digits* of V. I think that’s the source of confusion here.

I was looking at the problem and, as many already have, realise it’s based on integer factorisation. In effect what you are doing is finding all integer factorizations of the number and finding the one with the most similar numbers (most cube), as in closest to cbrt(k^3) sides. You want to find an algorithm which can do this quickly (scaling nicely) and works with all integers with no error. In affect what you were trying to find is a Polynomial [P] algorithm for integer factorization.

If the 1000 teachers had been able to solve this problem outright with a Polynomial [P] algorithmn (not brute force – NP) that works for all numbers then they might have earned $1 million dollars, saved millions of lives and changed our society irrevocably. They would have definitely in the process have made most of the encryption methods used on the internet irrelevant. It is an open problem in the field of computer science by itself (Wikipedia [1] – “Can integer factorization be done in polynomial time on a classical computer?”.) Granted with this problem it would not have to find all factors of the number just 3 that work but let’s say we had a semiprime number (made by multiplying two primes — Wikipedia [2] — as is used in Public Key Cryptography – Wikipedia [3]) the algorithm;
Let’s say 3 * 5 as both primes produce 15 a semiprime.
If we use this supposed algorithm on it would produce 3 * 5 * 1 as that is the only way you can factor 15.
We can generalise this to any semiprime.
Let a and b be the number multiplied to get the semiprime
When passed through the algorithm we would expect some permutation of (a, b, 1)

In effect we can find the prime factorization of any semiprime. If this algorithm worked fast, it means that public key cryptography which relies on one-way feature of multiplication + factorisation of semiprimes would be broken. This would contribute to another open problem in computer science which is whether one way functions exist: Wikipedia [4] — One-way function.

The reason this might earn $1-million-dollars and save millions of lives is because: depending on the solution some believe it would also solve P vs. NP which a $1-million-dollar solution prize is awarded if they find a solution by the clay maths institute as part of their 7 millennium maths problems (Clay-Math [5].) If it did prove P = NP it would mean things like protein folding would be possible instantly (or at least very fast) undoubtedly saving millions of lives. It would mean that hypothetically a computer would be able create a masterpiece in any art form (as currently creating a masterpiece is in effect a brute force for computers which is an NP problem.)

Looking at the responses to this problem currently it seems most of the responses algorithms work with numbers up to a certain value then start faltering. If they want to be able to do this with all numbers with no faults and with an improved brute force – the best algorithm that we have for integer factorization at the moment is the general number field sieve (Wikipedia [6] – GNFS.) If you wanted to solve the problem for this, then you would find all factors for this number with this algorithm and then compare them and choose ones which are closest to together to create the number (closest to cbrt(k^3) sides.)

In effect what I am saying is there is currently no way to create a non-brute force solution that works with all sizes of cuboid without solving the semiprime factorization problem in computer science, something that has been baffling computer scientists and mathematicians alike for centuries.
So in short if the collection of 1000 teachers / various people viewing this on the web had found a fast algorithm to do this then they would have:
– Broken most cryptography used on the web (PKC)
– Contributed to solving whether one-way functions really exist
– Save millions of lives
– Contributed to solving P vs. NP and possibly depending on the algorithm found may have won $1 million dollars from solving it
– In the process saving society irrevocably — all artistic pieces would be able to be remade using a computer hypothetically
– Made a huge discovery in the field of computer science, computational complexity and mathematics as a whole!
So, if anyone actually finds a fast algorithm for this then seriously tell someone… now!

For those unfamiliar with P vs. NP I recommend this video:
https://youtu.be/YX40hbAHx3s

If you are interested in looking at current challenges related to semiprime factorization:
https://en.wikipedia.org/wiki/RSA_Factoring_Challenge

Please do check this for yourself I am only a secondary school student so may be wrong!

[1]: https://en.wikipedia.org/wiki/List_of_unsolved_problems_in_computer_science
[2]: https://en.wikipedia.org/wiki/Semiprime
[3]: https://en.wikipedia.org/wiki/Public-key_cryptography
[4]: https://en.wikipedia.org/wiki/One-way_function
[5]: http://www.claymath.org/millennium-problems/p-vs-np-problem
[6]: https://en.wikipedia.org/wiki/General_number_field_sieve

-Bob

– Obtain the prime factorization and put in descending order.

– Begin filling three bins, A, B, and C by placing the largest prime factor in A, the next largest in B, and third largest in C.

– The next prime factor in descending order will be placed in the bin whose product of prime factors already contained in that bin is the smallest.

– Continue in descending order until all prime factors are binned.

(To handle the cases where there are fewer than three prime factors, start with a 1 in each bin.)

Example: 1260 = 2 x 2 x 3 x 3 x 5 x 7

Bin A: 7 x 2 = 14
Bin B: 5 x 2 = 10
Bin C: 3 x 3 = 9

7, 5 and 3 would go in first. Then the second 3 would go into Bin C with the first 3, then a 2 would go in Bin B with the 5 and the last 2 would go in Bin A with the 7.

If there are two bins with the same product of primes, you could select the “lower” letter (A over B, etc.).

-Bob

-Example run-

p(594720);

144 divisors: [84, 80, 90, 96, 72, 70, 105, 63, 60, 59, 112, 56, 118, 120, 48, 45, 126, 42, 40, 36, 35, 32, 30, 140, 28, 144, 24, 21, 20, 18, 16, 15, 14, 12, 10, 9, 160, 8, 7, 6, 5, 4, 3, 2, 1, 168, 177, 180, 210, 224, 236, 240, 252, 280, 288, 295, 315, 336, 354, 360, 413, 420, 472, 480, 504, 531, 560, 590, 630, 672, 708, 720, 826, 840, 885, 944, 1008, 1062, 1120, 1180, 1239, 1260, 1416, 1440, 1652, 1680, 1770, 1888, 2016, 2065, 2124, 2360, 2478, 2520, 2655, 2832, 3304, 3360, 3540, 3717, 4130, 4248, 4720, 4956, 5040, 5310, 5664, 6195, 6608, 7080, 7434, 8260, 8496, 9440, 9912, 10080, 10620, 12390, 13216, 14160, 14868, 16520, 16992, 18585, 19824, 21240, 24780, 28320, 29736, 33040, 37170, 39648, 42480, 49560, 59472, 66080, 74340, 84960, 99120, 118944, 148680, 198240, 297360, 594720]

cube root: 84.09513032

z: 84
value: 22032
removed: [60, 84, 118]

z: 80
value: 21914
removed: [63, 80, 118]

z: 90
removed: [59, 90, 112]

z: 96
removed: [59, 96, 105]

z: 72
value: 21796
removed: [70, 72, 118]

z: 56
removed all z less than or equal to 56, bounded by: 22161.97904

z: 120
removed all z greater than or equal to 120, bounded by: 21851.72727

divisors examined: 7
[70, 72, 118]

z: 84
value: 22032
removed: [60, 84, 118]

z: 80
value: 21914
removed: [63, 80, 118]

z: 90
removed: [59, 90, 112]

z: 96
removed: [59, 96, 105]

z: 72
value: 21796
removed: [70, 72, 118]

z: 56
removed all z = 120, bounded by: 21851.72727

divisors examined: 7
[70, 72, 118]