Let $S= \left\{ 1,2,3,...,100 \right\}$ be a set of positive integers from $1$ to $100$. Let $P$ be a subset of $S$ such that any arithmetic progression of length 10 consisting of numbers in $S$ will contain at least a number in $P$. What is the smallest possible number of elements in $P$ ?

Denote $|P|$ as the number of elements in $P$. We shall find the smallest possible value of $|P|$.

For $|P|=16$, we have the answer by @RobertIsrael below.

However, for $|P|<16$, I can neither find such set $P$ nor prove that $|P|$ cannot be less than $16$. So my question is:

Is it true that $|P| \geq 16$? How can I prove it? If not, what is the minimum amount of elements in $P$ ?

Also, I am wondering that:

If we replace 10 with an even number $n$,and $100$ with $n^2$, can we find the minimum of $|P|$ ?

Any answers or comments will be appreciated. If this question should be closed, please let me know. If this forum cannot answer my question, I will delete this question immediately.