3. Let $d$ be the greatest common divisor of ten natural numbers whose sum is 1001. What is the largest value that $d$ can take?
Answer: 91.

Solution. If each of the numbers is divisible by $d$, then their sum is also divisible by $d$. Therefore, $d$ is a divisor of the number 1001. Let's factorize the latter into prime factors: $1001=7 \cdot 11 \cdot 13$ and list all its divisors:

$$
1,7,11,13,77,91,143,1001
$$

Each of the ten numbers (from the problem statement) is not less than $d$. Therefore, their sum, which is 1001, is not less than $10d$. From the inequality $1001 \geqslant 10d$ we get $d \leqslant 101$. From the list $(*)$ the largest possible value is 91.

It is not difficult to come up with a corresponding example (it is unique): nine numbers are equal to 91, and one more is 182.

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