1. Given $A=\{1,2,3\}, B=\{4,5,6,7\}$ mapping $f: A \rightarrow B$, if when $x \in A$, $x + f(x) + x \cdot f(x)$ is an odd number, the number of such mappings $f$ is ( ).
A. 12
B. 20
C. 24
D. 32
1. D

Given $x+f(x)+x f(x)=(x+1)(f(x)+1)-1=$ odd number, hence $(x+1)(f(x)+1)$ is even. Therefore, the only possible images of 2 under $f$ are 5 or 7. If the image of 2 is 5, then there are 16 such $f$ functions. Similarly, if the image is 7, there are also 16 such $f$ functions, making a total of 32.