Question 216, Given $2^{2013}<5^{867}<2^{2014}$, how many integer pairs $(\mathrm{m}, \mathrm{n})$ satisfy: $5^{\mathrm{n}}<2^{\mathrm{m}}<$ $2^{\mathrm{m}+2}<5^{\mathrm{n}+1}$ ~ where $1 \leq \mathrm{m} \leq 2012$.
Question 216, Solution: First, we prove that for any non-negative integer $n$, the interval $\left(5^{n}, 5^{n+1}\right)$ contains either two or three powers of 2.

If there exists a non-negative integer $n$ such that the interval $\left(5^{n}, 5^{n+1}\right)$ contains only one power of 2, let's say $2^{m}$, then $2^{m-1} \leq 5^{n}\frac{5^{n+1}}{5^{n}}=5$, which is impossible; if there exists a non-negative integer $n$ such that the interval $\left(5^{n}, 5^{n+1}\right)$ contains at least four powers of 2, let's say $5^{n}\frac{2^{m+3}}{2^{m}}=8$, which is also impossible. Therefore, the above statement is proven.

Returning to the original problem, according to the conditions, we know that $2^{2012}<2^{2013}<5^{867}<2^{2014}$. We need to find how many of the 867 intervals $\left(5^{0}, 5^{1}\right)$, $\left(5^{1}, 5^{2}\right), \cdots,\left(5^{866}, 5^{867}\right)$ contain exactly three powers of 2. Let's say there are $\mathrm{m}$ intervals that contain exactly three powers of 2, and $\mathrm{n}$ intervals that contain exactly two powers of 2. Then $\left\{\begin{array}{c}m+n=867 \\ 3 m+2 n=2013\end{array}\right.$, solving this gives $m=279$.