3. Let $d$ - the greatest common divisor of eight natural numbers, the sum of which is equal to 595. What is the largest value that $d$ can take?
Answer: 35.

Solution. If each of the numbers is divisible by $d$, then their sum is also a multiple of $d$. Therefore, $d$ is a divisor of the number 595. Let's factorize the latter into prime factors: $595=5 \cdot 7 \cdot 17$ and list all its divisors:

$$
1,5,7,17,35,85,119,595
$$

Each of the eight numbers (from the problem statement) is not less than $d$. Therefore, their sum, which is 595, is not less than $8 d$. From the inequality $595 \geqslant 8 d$ we get $d \leqslant 74$. From the list $(*)$ the largest possible value is 35.

It is not difficult to come up with a corresponding example (which is far from unique): let seven numbers be 70, and one more be 105.

Evaluation. 13 points for a complete solution. 7 points for an estimate without an example. 6 points for an example without an estimate.