\section*{Problem 6 - \(340946=340846\)}

How many pairs \((x, y)\) of integers \(x, y\) that satisfy the inequality \(|x-30|+|y-10|<100\) are there in total?
}

We first consider the inequality \(\left(^{*}\right) a+b<100\) with \(a>0\) and \(b>0\). For a fixed \(a\), there are exactly the solutions \(1,2, \ldots, 99-a\) for \(b\), i.e., \(99-a\) different solutions. In total, there are thus

\[
\sum_{a=1}^{99}(99-a)=99 \cdot 99-\sum_{a=1}^{99} a=99 \cdot 99-\frac{99 \cdot 100}{2}=99 \cdot(99-50)=99 \cdot 49
\]

solutions in this case.

Now back to the inequality from the problem statement:

For each solution \((a, b)\) of the inequality \(\left({ }^{*}\right)\) with \(a, b \neq 0\), one obtains four solutions of the inequality from the problem statement, since one can choose \(x-30= \pm a\) and independently \(y-10= \pm b\). If one or both values \(a, b\) are zero, however, one can only choose one sign and gets \(x-30=0\) or \(y-10=0\).

Thus, the following numbers of solutions result for the original equation:

If \(a=b=0\), one gets exactly \(1 \cdot 1=1\) solution for the original equation.

If \(a=0\) and \(b \neq 0\), or vice versa, one gets exactly \(99 \cdot 2\) solutions in each case, so together \(99 \cdot 4\) solutions of the original equation.

And if both \(a\) and \(b\) are different from 0, one gets from \((99 \cdot 49) \cdot 4=99 \cdot 196\) solutions of the original equation.

In total, we thus obtain that the inequality from the problem statement has exactly \(99 \cdot 196+99 \cdot 4+1=\) 19801 integer solutions.