3. Let $m$ and $n$ be positive integers, and $m>n$. If the last two digits of $9^{m}$ and $9^{n}$ are the same, then the minimum value of $m-n$ is $\qquad$
3. 10 .

From the problem, we know that $9^{m}-9^{n}=9^{n}\left(9^{m-n}-1\right)$ is a multiple of 100. Therefore, $9^{m-n}-1$ is also a multiple of 100. Hence, the last two digits of $9^{m-n}$ are 01.

Clearly, $m-n$ is an even number. Let $m-n=2 t$ (where $t$ is a positive integer). Then $9^{m-n}=9^{2 t}=81^{t}$.

It is easy to see that the last two digits of $81^{2}, 81^{3}, 81^{4}, 81^{5}$ are 61, 41, 21, 01, respectively.
Thus, the minimum value of $t$ is 5.
Therefore, the minimum value of $m-n$ is 10.