2. A table of numbers with 20 rows and 15 columns, $A_{1}, \ldots, A_{20}$ are the sums of the numbers in the rows, $B_{1}, \ldots, B_{15}$ are the sums of the numbers in the columns.

a) Is it possible that $A_{1}=\cdots=A_{20}=B_{1}=\cdots=B_{15}$?

b) If the equalities in part a) are satisfied, what is the sum $A_{1}+\cdots+A_{20}+$ $B_{1}+\cdots+B_{15}?$
Let $A_{i}=B_{j}=X$ for $i=1, \ldots 20$ and $j=1, \ldots, 15$. Consider the sum $S$ of all elements in the table. We have $S=20 X=15 X, X=0$ and $A_{1}+\cdots+A_{20}+B_{1}+\cdots+B_{15}=0$. An example of such a table is, for instance, a table consisting entirely of zeros. There is no need to consider other cases.

Answer: a) - yes, b) - 0.