Dividing n consecutive integers by (1,2,3,...,n)
Don Reble
djr at nk.ca
Fri May 16 05:50:15 CEST 2008
> Did you follow the same approach as Robert?
Sorry, I don't know Maple.
> ... for n=3, your calculations indicate that there are only
> 5 different starting points while there are 6 feasible permutations.
> ...
> ... each of the feasible permutations here results in an
> unique solution modulo lcm(1,2,3)=6. But how then the number of
> starting points can be equal 5?
The permutations (1,3,2) and (2,3,1) both map onto (2,3,4) mod 6.
The starting point is 2. No permutation maps onto (5,0,1) mod 6,
so 5 is not a starting point.
> ... feasible permutations can be constructed in a fast way using
> branch-and-bound approach. ...
> But, maybe, there exists even a smarter way to construct such
> permutations...
Ok, I believe you.
--
Don Reble djr at nk.ca
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