Let $$A = \sum_{i=1}^{10}\sum_{j=1}^{10} \min\{i, j\}$$ and $$B = \sum_{i=1}^{10}\sum_{j=1}^{10} \max\{i, j\}$$. Then $$A + B$$ is equal to ______

We need $$A + B$$ where $$A = \sum_{i=1}^{10}\sum_{j=1}^{10} \min\{i,j\}$$ and $$B = \sum_{i=1}^{10}\sum_{j=1}^{10} \max\{i,j\}$$.

Since for any $$i, j$$, $$\min\{i,j\} + \max\{i,j\} = i + j$$, substituting into the double sum gives $$A + B = \sum_{i=1}^{10}\sum_{j=1}^{10} (\min\{i,j\} + \max\{i,j\}) = \sum_{i=1}^{10}\sum_{j=1}^{10} (i + j)\,. $$ This can be split as $$\sum_{i=1}^{10}\sum_{j=1}^{10} i + \sum_{i=1}^{10}\sum_{j=1}^{10} j = 10\sum_{i=1}^{10} i + 10\sum_{j=1}^{10} j\,. $$ Since $$\sum_{k=1}^{10} k = 55$$, this becomes $$10 \times 55 + 10 \times 55 = 550 + 550 = 1100\,. $$ Therefore the answer is $$\boxed{1100}$$.