1. (10 points) Petya added several consecutive odd numbers. It turned out that the resulting sum is 20 times greater than the largest addend and 60 times greater than the smallest. Find which numbers Petya added.
Answer: $29,31,33, \ldots, 87$.

Solution. Let the first number be $n$, and the last number be $n+2k$. There are a total of $k+1$ numbers. The numbers form an arithmetic progression, the sum of which is $(n+k)(k+1)$. We obtain the system $\left\{\begin{array}{c}(n+k)(k+1)=20(n+2k), \\ (n+k)(k+1)=60n .\end{array}\right.$ Subtracting the equations, we get $k=n$. Then the last equation becomes $2k(k+1)=60k$, from which $k=29$. Therefore, the required numbers are $29,31, \ldots, 87$.

Grading Criteria. Correct system obtained - 6 points, correct and justified solution - 10 points; there are arithmetic errors - minus 2 points.