Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.7 AMC 10/12 Find the answer Italy

The polynomial $p(x)$ has the following property: for every triple of integers $a, b, c$ such that $a+b+c=2022$ we have $p(a)+p(b)+p(c)=p(674)$ . It is also known that $p(0)=-2696$ . What is $p(2022)$ ?

Pick one

Solution

Solution:

The answer is $\mathbf{(C)}$ . Substituting $a=b=c=674$ (integers that indeed satisfy $a+b+c=2022$ ) we obtain $3p(674)=p(674)$ , that is $p(674)=0$ . Substituting then $a=b=0$ and $c=2022$ we obtain
$2p(0)+p(2022)=p(674)=0 \Rightarrow p(2022)=-2p(0)=5392.$

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.