(9) If $a_{i} \equiv b_{i}(\bmod m), i=0,1,2, \cdots, n$, then $a_{n} x^{n}+a_{n \cdots 1} x^{n-1}+\cdots+a_{1} x+a_{0} \equiv b_{n} x^{n}+$ $b_{n-1} x^{n-1}+\cdots+b_{1} x+b_{0}(\bmod m)$. In particular, let $f(x)=c_{n} x^{n}+c_{n-1} x^{n-1}+\cdots+c_{1} x+c_{0}\left(c_{i} \in\right.$ $\mathbf{Z})$, if $a \equiv b(\bmod m)$, then $f(a) \equiv f(b)(\bmod m)$.