8. You have a $10 \times 10$ grid of squares. You write a number in each square as follows: you write $1,2,3, \ldots, 10$ from left to right across the top row, then $11,12, \ldots, 20$ across the second row, and so on, ending with a 100 in the bottom right square. You then write a second number in each square, writing $1,2, \ldots, 10$ in the first column (from top to bottom), then $11,12, \ldots, 20$ in the second column, and so forth.
When this process is finished, how many squares will have the property that their two numbers sum to $101 ?$
Solution: 10
The number in the $i$ th row, $j$ th column will receive the numbers $10(i-1)+j$ and $10(j-1)+i$, so the question is how many pairs $(i, j)(1 \leq i, j \leq 10)$ will have
$$
101=[10(i-1)+j]+[10(j-1)+i] \quad \Leftrightarrow \quad 121=11 i+11 j=11(i+j) .
$$

Now it is clear that this is achieved by the ten pairs $(1,10),(2,9),(3,8), \ldots,(10,1)$ and no others.