## Task 2 - 221232

Determine for all 30-tuples $\left(a_{1}, a_{2}, \ldots, a_{30}\right)$ of (not necessarily distinct) positive integers $a_{i}(i=1, \ldots, 30)$, which satisfy

$$
\sum_{i=1}^{30} a_{i}=1983
$$

the greatest value that the greatest common divisor $d$ of the numbers $a_{i}$ can take.
Since all $a_{i}$ are naturally divisible by $d$, this also applies to their sum 1983 with the prime factorization $1983=3 \cdot 661$, i.e., it must be

$$
d \in\{1,3,661,1983\}
$$

Since $d=1983$ and $d=661$ can obviously be ruled out with 30 positive summands, the next to consider in terms of a maximum is $d=3$, and thus the problem here is naturally easy to solve, for example with the assignment

$$
a_{1}=a_{2}=\ldots=a_{29}=3, a_{30}=1983-29 \cdot 3=1896
$$