Little math problem(s) February 2019

[pier4r]

(02-23-2019 04:06 PM)Thomas Klemm Wrote:  This online LaTeX Equation Editor may help.

Cheers
Thomas

Thanks. I know it, I didn't have much time and I decided it to post it and beautify it later. Editing something is always easier than editing an empty post.

Including your formatting in the post.

[pier4r]

(02-23-2019 04:23 PM)Albert Chan Wrote:  If ... meant any positive integers, with no relation to each other, S1 and S2 is not enough.

If ... meant I = {i1, i1 + 1, ... i1 + N-1}, I may be recovered from S1, S2.

No the elements are not related to each other. But please expose every other variant you have got. This is the intended effect, one propose a problem and then it starts explorations in every direction with results reported.

Also while deducing from S1 and S2 the original I is nice, it is a bonus. The question is whether, given a S1 and a S2, we can generate then with only one I, although we don't know which or there are multiple I that generate them.

This arise from me doing stats on the monthly bills.
From a month to another I wondered: "could it be that I sum two different sets of numbers and I get the same S1 and S2?". Doing something eventually produces questions, that produce actions, that produce questions, etc...

[Albert Chan]

(02-23-2019 08:15 PM)pier4r Wrote:  From a month to another I wondered: "could it be that I sum two different sets of numbers and I get the same S1 and S2?"

This generated 2 sets of 26 numbers that get the same S1, S2, using Pythorean triplets:
13² + 84² = 36² + 77² = 85²
16² + 63² = 33² + 56² = 65²

>>> lst = [13,84] * 5 + [33,56] * 8
>>> print sum(lst), sum([i*i for i in lst])
1197 69925

>>> lst = [36,77] * 5 + [16,63] * 8
>>> print sum(lst), sum([i*i for i in lst])
1197 69925

[pier4r]

nice counterexample and approach. The idea of the pythagorean triples was effective.