TITLE: Smallest set such that all arithmetic progression will always contain at least a number in a set
QUESTION [12 upvotes]: 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.

REPLY [11 votes]: Using a tabu search procedure, I have found a solution for $|P|=17$, namely ${1, 11, 18, 25, 31, 32, 33, 36, 44, 51, 58, 65, 69, 70, 77, 84, 91}$.  I don't know if this is optimal. 
EDIT: Found a solution for $|P|=16$, namely 
$$10, 15, 22, 29, 36, 43, 53, 55, 56, 57, 58, 68, 73, 74, 84, 91$$