7 、Given the function $f:\{0,1,2, \ldots, 2017\} \rightarrow N$, if for all possible integers $x$ we have: $f(4 x+2)=f(4 x+1) ; f(5 x+3)=f(5 x+2) ; f(7 x+5)=f(7 x+4)$, then $f(x)$ can take at most $\qquad$ different values.
7.1138

Analysis: For the function $f$, construct a function $g$, defined as: $g(0)=1, g(i+1)=\left\{\begin{array}{l}1, f(i+1) \neq f(i) \\ 0, f(i+1)=f(i)\end{array}\right.$, where $i=0,1, \ldots, 2016$. Then $g$ is a mapping from $\{0,1, \ldots, 2017\}$ to $\{0,1\}$, and for all possible $x$, we have $g(4 x+2)=g(5 x+3)=g(7 x+5)=0$. Therefore, $f(x)$ can take at most $\sum_{i=0}^{2017} g(i)$ different values.
Let $g(x)=h(x+2)$, then $h$ is a mapping from $\{2,3, \ldots, 2019\} \rightarrow\{0,1\}$, and for all possible $x$, we have
$$
\begin{array}{l}
h(4 x)=h(5 x)=h(7 x)=0 \text {. Also, } \\
\left(\sum_{i=2}^{2019} h(i)\right)_{\max }=2018-\left[\frac{2019}{4}\right]-\left[\frac{2019}{5}\right]-\left[\frac{2019}{7}\right]+\left[\frac{2019}{4 \times 5}\right]+\left[\frac{2019}{4 \times 7}\right]+\left[\frac{2019}{5 \times 7}\right] \\
-\left[\frac{2019}{4 \times 5 \times 7}\right]=1138
\end{array}
$$