5. (3 points) If the square of a two-digit number has only the tens digit as 0, then there are $\qquad$ such two-digit numbers.
【Analysis】Let the two-digit number that meets the condition be $\overline{\mathrm{ab}}$. The tens digit of the square of the two-digit number $\overline{\mathrm{ab}}$ equals the unit digit of the sum of the unit digit of $2 a b$ and the tens digit of $2 b$, which is 0. Since the square of $a b$ has only the tens digit as 0, $b \neq 0$. Then, we discuss the cases when $b$ takes values from $1 \sim 9$ to solve the problem.

【Solution】Solution: Let the two-digit number that meets the condition be $\overline{\mathrm{ab}}$.
The tens digit of the square of the two-digit number $\overline{\mathrm{ab}}$ equals the unit digit of the sum of the unit digit of $2 a b$ and the tens digit of $b^{2}$, which is 0. Since the square of $a b$ has only the tens digit as 0, $b \neq 0$.

When $b$ takes values from $1 \sim 9$,
the tens digit of $b^{2}$ are $0, 0, 0, 1, 2, 3, 4, 6, 8$.
The unit digit of $2 a b$ is as follows:
When $a$ is 1, they are $2, 4, 6, 8, 0, 2, 4, 6, 8$;
When $a$ is 2, they are $4, 8, 2, 6, 0, 4, 8, 2, 6$;
When $a$ is 3, they are $3, 6, 9, 2, 5, 8, 1, 4, 7$;
When $a$ is 4, they are $8, 6, 4, 2, 0, 8, 6, 4, 2$;
When $a$ is 5, they are $0, 0, 0, 1, 2, 3, 4, 6, 8$;
When $a$ is 6 or 7, they are the same as when $a$ is 1 or 2;
When $a$ is 8, they are $6, 2, 8, 4, 0, 6, 2, 8, 4$;
When $a$ is 9, they are the same as when $a$ is 4, which are $8, 6, 4, 2, 0, 8, 6, 4, 2$.
Therefore, such two-digit numbers are $47, 48, 49, 51, 52, 53, 97, 98, 99$, a total of 9.