2. In a row, 10 natural numbers are written in ascending order such that each number, except the first, is divisible by one of the preceding numbers. The first number is not equal to 1, and the sum of all 10 numbers is 275. Restore these numbers.

Let the first number be $x$, then each of the remaining 9 numbers must be divisible by $x$, and thus the sum of all numbers is also divisible by $x$. Therefore, $x$ must be a divisor of the number $275=5^{2} \cdot 11$.

If $x$ is greater than 5, the smallest possible sum would be $11+$ $11 \cdot 2+11 \cdot 3+\cdots+11 \cdot 9+11 \cdot 10=605$, which is already greater than 275. Therefore, only $x=5$ will work. In this case, the only possible sum of the numbers is $5+5 \cdot 2+5 \cdot 3+\cdots+5 \cdot 9+5 \cdot 10=275$.
Answer: $5,10,15,20,25,30,35,40,45,50$.

Criteria: correct answer without explanation - 2 points.