HMMT February

Algebra Difficulty 5.0 AIME, harder Find the answer

Let $P(x)$ be a polynomial with degree 2008 and leading coefficient 1 such that $P(0)=2007, P(1)=2006, P(2)=2005, \ldots, P(2007)=0$ . Determine the value of $P(2008)$ . You may use factorials in your answer.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Consider the polynomial $Q(x)=P(x)+x-2007$ . The given conditions tell us that $Q(x)=0$ for $x=0,1,2, \ldots, 2007$ , so these are the roots of $Q(x)$ . On the other hand, we know that $Q(x)$ is also a polynomial with degree 2008 and leading coefficient 1 . It follows that $Q(x)=x(x-1)(x-2)(x-3) \cdots(x-2007)$ . Thus $P(x)=x(x-1)(x-2)(x-3) \cdots(x-2007)-x+2007$ Setting $x=2008$ gives the answer.

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