9.6. Petya and Vasya came up with ten quadratic trinomials. Then Vasya sequentially named consecutive natural numbers (starting from some number), and Petya substituted each named number into one of the trinomials of his choice and wrote down the obtained values on the board from left to right. It turned out that the numbers written on the board form an arithmetic progression (in this exact order). What is the maximum number of numbers Vasya could have named?

(A. Golev)
Answer: 20 numbers.

Solution. We will show that Petya could substitute no more than two numbers into each quadratic polynomial $P(x)$. Indeed, let the $n$-th term of the resulting arithmetic progression be $a n+b$, and the $n$-th of Vasya's consecutive numbers be $k+n$. Then Petya could substitute this number into $P(x)$ if $P(k+n)=a n+b$, and this quadratic equation in $n$ has no more than two roots.

Therefore, the total number of numbers could not be more than 20. It remains to show that 20 numbers could have resulted. Suppose, for example, the polynomials $P_{k}(x)=(x-(2 k-1))(x-2 k)+x$ for $k=1,2, \ldots, 10$ were chosen, and Vasya named the numbers $1,2, \ldots, 20$. Since $P_{k}(2 k-1)=2 k-1$ and $P_{k}(2 k)=2 k$, Petya could have obtained the numbers $1,2, \ldots, 20$ in sequence.