Little math challenge (no complete automatism allowed)

[Valentin Albillo]

(04-27-2018 07:27 AM)EdS2 Wrote:  Why cyclic - is that obvious??

Well, I have lots of experience solving 'combinatorial' math puzzles (like the many such in Project Euler) and I've developed the heuristic of trying cyclic permutations first, whether I solve them using a computer or not.

That said, and as I've already solved the challenge as per Pier's requirements, I've done a little computer-assisted research to see whether the 1-27 sequence of integers can be rearranged in 3 groups of unequal length that nevertheless still add up to 2310 each.

To help me in that regard, I've coded a 5-line generalized subprogram for the HP-71B which accepts a list of integers, the value they should add up to (no repeated values allowed), and the maximum element allowable, and outputs each solution sequence in order and the number of solutions found. For instance:

- decompositions of 200 as a sum of squares from 1\(^2\) up to 10\(^2\):

>RUN
? 200,10

1 : 1 2 3 4 5 8 9 (i.e.: 1\(^2\)+2\(^2\)+3\(^2\)+4\(^2\)+5\(^2\)+8\(^2\)+9\(^2\) = 200)
2 : 1 3 4 5 6 7 8
3 : 1 3 4 5 7 10
4 : 3 5 6 7 9
5 : 6 8 10

- same but with elements up to 14\(^2\):

>RUN
? 200,14

1 : 1 2 3 4 5 8 9
2 : 1 2 3 4 7 11
3 : 1 2 5 7 11
4 : 1 3 4 5 6 7 8
5 : 1 3 4 5 7 10
6 : 2 4 6 12
7 : 2 14
8 : 3 5 6 7 9
9 : 6 8 10

Applying it to Pier's problem, i.e.: solutions which add up to 2310 using squares up to 27\(^2\), we get:

>RUN
? 2310,27

1 : 1 2 3 4 5 6 7 8 9 10 11 12 13 17 19 20 21
2 : 1 2 3 4 5 6 7 8 9 10 11 12 14 16 18 20 22
3 : 1 2 3 4 5 6 7 8 9 10 11 12 15 16 17 19 23
4 : 1 2 3 4 5 6 7 8 9 10 11 13 14 15 17 21 22
5 : 1 2 3 4 5 6 7 8 9 10 11 13 14 15 18 19 23
6 : 1 2 3 4 5 6 7 8 9 10 11 13 15 16 23 25
7 : 1 2 3 4 5 6 7 8 9 10 11 13 16 17 19 27
8 : 1 2 3 4 5 6 7 8 9 10 11 15 17 19 20 23
9 : 1 2 3 4 5 6 7 8 9 10 11 17 19 23 25
10 : 1 2 3 4 5 6 7 8 9 10 12 13 14 16 22 26
... ... ... ... ...
29256 : 13 14 16 17 18 20 26
29257 : 13 14 16 22 23 26
29258 : 14 15 22 26 27
29259 : 14 16 17 18 19 20 22
29260 : 14 17 18 21 22 24
29261 : 15 16 18 20 23 24
29262 : 15 20 22 24 25
29263 : 16 18 23 24 25
29264 : 16 19 21 24 26
29265 : 16 20 21 22 27

So we get 29265 solutions in all of various lengths, all of which add up to 2310 and use no squares greater than 27\(^2\). Now it's just a matter of writing a simple routine to scan them and check whether there are three (or more) which share no common elements and there you are. I won't spoil the fun by providing the solution, of course.

By the way, I said "generalized subprogram" because it works for any arbitrary sequence of integers (such as cubes, Fibonacci numbers, primes, ...), not just squares.

Last, some trivia:

There are 154 solutions to express 2310 as a sum of 4 squares or less but all of them require either elements greater than 27\(^2\) or else repeated elements. If no repetitions are allowed, there are just 142 solutions, the last being:

2310 = 19\(^2\)+21\(^2\)+22\(^2\)+32\(^2\)

Finally, using my subprogram to find decompositions of 2018 using squares up to 18\(^2\), finds the following three solutions:

>RUN
? 2018,18

1 : 2 3 6 8 9 10 11 12 13 14 15 16 17 18
2 : 2 4 5 6 7 8 10 11 12 13 14 15 16 17 18
3 : 7 8 9 10 11 12 13 14 15 16 17 18

and the last is particularly noteworthy as it's the sum of 12 consecutive squares beginning with 7\(^2\) and ending with 18\(^2\):

2018 = 7\(^2\)+8\(^2\)+9\(^2\)+10\(^2\)+11\(^2\)+12\(^2\)+13\(^2\)+14\(^2\)+15\(^2\)+16​\(^2\)+17\(^2\)+18\(^2\)

Regards.
V.
.