1. Find all polynomials $P(x)$ such that $(x+100) P(x)-x P(x+1)=1$ for every real number $x$.
1. The substitution $P(x)=Q(x)+\frac{1}{100}$ gives $(x+100) Q(x)=x Q(x+1)$. By substituting $x=0$ and $x=-100$, we get $Q(0)=Q(-99)=0$. Further, for $x=-1$ and $x=-99$, we get $99 Q(-1)=-Q(0)=0$ and $0=Q(-99)=-99 Q(-98)$, so $Q(-1)=Q(-98)=0$. Similarly, for $x=-2$ and $x=-98$, we get $Q(-2)=Q(-97)=0$, and so on. After 50 iterations, we will have $Q(0)=Q(-1)=Q(-2)=\cdots=Q(-99)=0$, so $Q(x)=x(x+1)(x+2) \cdots(x+99) R(x)$.

Substituting into the original equation for $Q$ gives $R(x)=R(x+1)$, from which $R(x)=c$ is a constant. Therefore, $P(x)=c x(x+1)(x+2) \cdots(x+99)+\frac{1}{100}$.