1. The sum of 5 natural numbers is 1982. What is the greatest value that their greatest common divisor can take?
Solution. Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ be those numbers and let $d$ be their greatest common divisor. Then $a_{i}=k_{i} d, i=1,2,3,4,5$ and, therefore,

$$
d\left(k_{1}+k_{2}+k_{3}+k_{4}+k_{5}\right)=1982
$$

i.e.

$$
d s=2 \cdot 991
$$

where $s=k_{1}+k_{2}+k_{3}+k_{4}+k_{5}$.

Since $s \geq 5$, we have: $s=991, d=2$ or $s=1982, d=1$. Thus, the greatest possible value that $d$ can take is 2.