52 The number of integer solutions to the equation $x^{1988}+y^{1988}+z^{1988}=7^{1990}$ is
A. 0
B. 1
C. $2^{3} \cdot P_{3}^{3}$
D. Not less than 49
52 A. Suppose the original equation has an integer solution $(x, y, z)=(m, n, r)$, then
$$
m^{1988}+n^{1988}+r^{1988}=7^{1990} .
$$

On one hand, since $7^{1990}=7^{1 \times 497+2}$, the last digit of $7^{1990}$ is 9; on the other hand, for any $t \in \mathbf{Z}$, the last digit of $t^{4}$ can only be one of $0,1,5,6$. Therefore, the last digit of $t^{1988}=\left(t^{4}\right)^{497}$ can only be one of $0,1,5,6$. Since $m, n, r \in \mathbf{Z}$, the last digits of $m^{1988}, n^{1988}, r^{1988}$ can only be one of $0,1,5,6$. However, the sum of any three (with repetition allowed) of these four digits does not have a last digit of 9. Therefore, the original equation has no integer solutions.