# Neither quadratic, nor cubic, it's 2011-ic!

**Algebra**Level 5

\[p(x) = x^{2011} + \displaystyle \sum_{n=1}^{2010} {a_{n} x^{2011-n}} + 1\]

Let \(x_{1}, x_{2}, x_{3}, x_{4}, \cdots , x_{2011}\) be the roots of the monic polynomial above and let \( y_{1}, y_{2}, y_{3}, y_{4}, \cdots , y_{2011}\) be defined as:

\[\begin{align} y_{1} & = x_{1}x_{2} x_{3} ... x_{10} \\ y_{2} & = x_{2} x_{3} x_{4} ... x_{11} \\ y_{3} & = x_{3} x_{4} x_{5} ... x_{12} \\ \vdots & = \ \vdots \\ y_{2010} & = x_{2010} x_{2011} x_{1} ... x_{8} \\ y_{2011} & = x_{2011} x_{1} x_{2} ... x_{9} \end{align}\]

If the value of \({\left( y_{1} y_{2} y_{3} ... y_{2011}\right)}^{2011^{2010}}\) can be expressed as \(\log_{10}{a}\). Find \(a\).

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