Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.9 AIME Prove it United States

Problem:
Let $P$ be a polynomial with $P(1)=P(2)=\cdots=P(2007)=0$ and $P(0)=2009!$ . $P(x)$ has leading coefficient $1$ and degree $2008$ . Find the largest root of $P(x)$ .

Solution

Solution:
$P(0)$ is the constant term of $P(x)$ , which is the product of all the roots of the polynomial, because its degree is even. So the product of all $2008$ roots is $2009!$ and the product of the first $2007$ is $2007!$ , which means the last root is $\frac{2009!}{2007!}=2009 \cdot 2008=4034072$ .

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