A polynomial P(x) of degree 3n has the value 2 at x= 0, 3, 6,..., 3n, and:
the value 1 at x= 1, 4, 7, ... , 3n-2, and:
the value 0 at x= 2, 5, 8, ... , 3n-1.

Given that P(3n+1) = 730, find n.

Well, an unknown polynomial of degree 3n has 3n+1 unknown coefficients.  At first blush, therefore, it seems like any positive n can be made to work.

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For instance, let n = 1.  Then P(0) = 2, P(1) = 1, P(2) = 0.

P(x) = (x-2)(ax(x-1)-1) for any a

Then P(4) = 2*(12a-1) = 730
so a = 30.5

P(x) = (x-2)(30.5x(x-1)-1) satisfies the problem requirements

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But I expect that any higher n can be made to work also.

Posted by Steve Herman on 2016-05-07 15:30:40