3. The sum of 20 natural numbers is 2002. Find the greatest value that their GCD can take.
Solution. Let $a_{1}, a_{2}, \ldots, a_{20}$ be natural numbers for which

$$
a_{1}+a_{2}+\ldots+a_{20}=2002
$$

If $d=\operatorname{GCD}\left(a_{1}, a_{2}, \ldots, a_{20}\right)$, then $a_{i}=d k_{i}, i=1,2, \ldots, 20$ where $k_{i} \in \mathbb{N}, i=1,2, \ldots, 20$

Thus

$$
d\left(k_{1}+k_{2}+\ldots+k_{20}\right)=2002=2 \cdot 7 \cdot 11 \cdot 13
$$

Since $s=k_{1}+k_{2}+\ldots+k_{20}$ (sum of twenty natural numbers) and $d$ are natural numbers for which $s \cdot d=2 \cdot 7 \cdot 11 \cdot 13$ and $s \geq 20$, the smallest value that the number $s=k_{1}+k_{2}+\ldots+k_{20}$ can take is $2 \cdot 11=22$. According to this, the largest value of $d$ is 91

The sequence of numbers

$$
a_{1}=91, a_{2}=91, a_{3}=91, \ldots, a_{19}=91, a_{20}=273
$$

has the required properties.