16. Given $a, b \in \mathbf{N}^{*}$, and in decimal notation, the number $a^{a} \cdot b^{b}$ ends with exactly 98 zeros. Find all pairs $(a, b)$ that minimize $ab$.

 untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 

Note: The last sentence is a note and not part of the problem statement, so it is not translated.
16. Let the powers of 2 and 5 in the prime factorizations of $a$ and $b$ be $\alpha_{1}, \beta_{1}$ and $\alpha_{2}, \beta_{2}$, respectively. Then, by the given condition, one of the equations $a \cdot \alpha_{1} + b \cdot \alpha_{2} = 98$ or $a \cdot \beta_{1} + b \cdot \beta_{2} = 98$ must hold. If $a \cdot \beta_{1} + b \cdot \beta_{2} = 98$, then when $\beta_{1}, \beta_{2}$ are both positive integers, the left side of the equation is a multiple of 5, while the right side is not; when exactly one of $\beta_{1}, \beta_{2}$ is zero, taking the equation modulo 5 still leads to a contradiction. Therefore, it must be that $a \cdot \alpha_{1} + b \cdot \alpha_{2} = 98$. In this case, if one of $\alpha_{1}, \alpha_{2}$ is zero, without loss of generality, let $\alpha_{2} = 0$, then $\alpha_{1} > 0$. If $\alpha_{1} \geqslant 2$, then $4 \mid a \cdot \alpha_{1}$. Since $4 \times 98$, this leads to a contradiction. Hence, $\alpha_{1} = 1$, which implies $a = 98$. By the problem's condition, in this case, $a \cdot \beta_{1} + b \cdot \beta_{2} \geqslant 98$, and since $a = 98$, it follows that $\beta_{1} = 0$. Thus, $b \cdot \beta_{2} \geqslant 98$, and the smallest possible value for $b$ is $3 \times 5^{2} = 75$.

In the equation $a \cdot \alpha_{1} + b \cdot \alpha_{2} = 98$, if $\alpha_{1}, \alpha_{2}$ are both positive integers, without loss of generality, let $\alpha_{1} \geqslant \alpha_{2} \geqslant 1$. If $\alpha_{2} \geqslant 2$, then $4 \mid (a \cdot \alpha_{1} + b \cdot \alpha_{2})$, leading to a contradiction. Therefore, $\alpha_{2} = 1$. From this, we can also deduce that $\alpha_{1} > 1$. We set $a = 2^{a_{1}} \cdot 5^{\beta_{1}} \cdot x$ and $b = 2 \cdot 5^{\beta_{2}} \cdot y$, then we should have
$$2^{a_{1}-1} \cdot 5^{\beta_{1}} \cdot x \cdot \alpha_{1} + 5^{\beta_{2}} \cdot y = 49$$

It follows that one of $\beta_{1}, \beta_{2}$ must be zero. Combining this with $a \cdot \beta_{1} + b \cdot \beta_{2} > 98$, if $\beta_{2} = 0$, then the number of trailing zeros in $a^{a}$ is greater than 98, leading to a contradiction. If $\beta_{1} = 0$, then the number of trailing zeros in $b^{b}$ is greater than 98, also leading to a contradiction.

Therefore, the pairs $(a, b)$ that satisfy the conditions and minimize $a b$ are $(98, 75)$ or $(75, 98)$.