might be best to add a test case to guard against test suite hacking
Far too few test cases. In fact, a serious issue with this problem is that there are too few parasitic numbers that will fit into even 64 bits for most values of n.
I've added some test cases, that will make it a bit more difficult to game this problem.
Cases 5 and 7 are identical, and I believe they are both wrong. 5 x 142857 = 714285. Isn't this a parasitic number? What am I missing?
Can the problem creator disqualify "solutions" that cheat? There should be a point penalty (-200) associated with cheating too.
There is 4, not 5
I mis-typed. 4 x 142857 = 571428, which is a shift right of 2. Incidentally, both n = 4 and 5 both make parasitic pairs with 142857.
Thus it is not a 4-parasitic number. It doesn't meet the definition. Nevertheless is a great example of a cyclic number. You can get different cyclic permutations when multiply 142857 by 1, 2, 3, 4, 5 and 6. For 5 you got shift by one digit, therefore it's 5-parasitic.
Ahhh, I see! Thanks, Jan - I get it now!
Not actually a general solution. Takes advantage of limited test cases.
Back to basics 8 - Matrix Diagonals
Back to basics 17 - white space
Matrix with different incremental runs
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The Hitchhiker's Guide to MATLAB
Factorize THIS, buddy
Tune your guitar
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