7. Given the polynomial $P(x)=x^{5}+a_{4} x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}$, it is known that $P(2014)=1, P(2015)=2, P(2016)=3, P(2017)=4$, $P(2018)=5$. Find $P(2013)$.
Answer: $-5!=-120$. Solution. Consider the polynomial $P(x)-x+2013$, the numbers $2014, \ldots 2018$ are its roots. Therefore (since the leading coefficient is 1) it can be represented as $P(x)-x+2013=$ $(x-2014)(x-2015) \cdot \ldots \cdot(x-2018)$. Substituting $x=2013$, we get $P(2013)-2013+2013=-120$.