5. $M=\{-2,0,1\}, N=\{1,2,3,4,5\}$. Mapping $f: M$ $\rightarrow N$, such that for any $x \in M$, $x+f(x)+x f(x)$ is an odd number. Then the number of different mappings is $\qquad$ .
5. 45 .

Given the mapping $f: M \rightarrow N$ where the independent variable is taken from $M=\{-2,0, 1\}$ and the values of $f(x)$ are taken from $N=\{1,2,3,4,5\}$. The problem is to find the number of different mappings such that for any $x \in M$, $x+f(x)+x f(x)$ is an odd number.
And $x+f(x)+x f(x)=x+x f(x)+1+f(x)-1$
$=(x+1)[1+f(x)]-1$.
When $x=1$, $x+1$ is even, so regardless of which value $f(x)$ takes from $N$, $x+f(x)+x f(x)$ is always an odd number. Therefore, $f(x)$ has 5 possible values.

When $x=-2$, since $x+1$ is odd, for $x+f(x)+x f(x)$ to be odd, $1+f(x)$ must be even, which means $f(x)$ must be odd. Therefore, $f(x)$ has 3 possible values.
Similarly, when $x=0$, $f(x)$ also has 3 possible values.
Thus, the total number of different mappings is $3 \times 3 \times 5=45$.