An algebra puzzle

Algebra Level 3

We know that

\((1-\frac{1}{2^2})*(1-\frac{1}{3^2})*(1-\frac{1}{4^2})*\dots*(1-\frac{1}{n^2})=k\)

where \(n\) is a positive integer.

If \(k\) can't be \(\frac{1}{2}\) and can't be \(\frac{1008}{2017}\), then choose 0.

If \(k\) can't be \(\frac{1}{2}\), but it can be \(\frac{1008}{2017}\), then choose 1.

If \(k\) can't be \(\frac{1008}{2017}\), but it can be \(\frac{1}{2}\), then choose 2.

If \(k\) can be \(\frac{1}{2}\) and \(\frac{1008}{2017}\) too, then choose 3.

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