\section*{Problem 2 - 320932}

How many pairs \((x, y)\) of natural numbers for which \(10 x+y<1993\) exist in total?
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For this task, it is assumed that 0 is a natural number.

Then for each natural \(x \leq 199\), there are exactly \(1993-10 x\) different possibilities for \(y\) to choose (namely 0 to \(1993-10 x-1\)), such that the inequality is satisfied. Summing this over all \(x\), we thus obtain

\(\sum_{x=0}^{199}(1993-10 x)=200 \cdot 1993-10 \cdot \sum_{x=0}^{199} x=200 \cdot 1993-10 \cdot \frac{199 \cdot 200}{2}=200 \cdot 1993-1990 \cdot 100=199600\)

pairs of natural numbers that satisfy the inequality from the problem statement.