## Task $6-340846=340946$

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 $1,2, \ldots, 99-a$, i.e., $99-a$ different solutions for $b$. 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 thus gets $x-30=0$ or $y-10=0$.

Together, this results in the following number of solutions 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 for the original equation.

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

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

Solutions of the 4th round 1994 taken from [5]

## 6 Grade 9

### 6.1 Preliminary Olympiad 1960

### 6.1.1 Competition V1960, Grade 9