10.6. Petya and Vasya came up with ten polynomials of the fifth degree. Then Vasya, in turn, called out consecutive natural numbers (starting from some number), and Petya substituted each called number into one of the polynomials 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 formed an arithmetic progression (in this exact order). What is the maximum number of numbers Vasya could have called out?
Answer: 50 numbers.

Solution. We will show that Petya could substitute no more than five numbers into each 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 is a fifth-degree equation in $n$, so it has no more than five roots.

Thus, the total number of numbers could not be more than 50. It remains to show that 50 numbers could have resulted. For example, let the polynomials

$$
P_{k}(x)=x+(x-(5 k-4)) \cdot(x-(5 k-3)) \cdot \ldots \cdot(x-5 k)
$$

for $k=1,2, \ldots, 10$, and Vasya named the numbers $1,2, \ldots, 50$. Since $P_{k}(5 k-i)=5 k-i$ for $i=0,1, \ldots, 4$, Petya could have obtained the numbers $1,2, \ldots, 50$ in sequence.