Dividing n consecutive integers by (1,2,3,...,n)

[Don Reble]

> 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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