Existence of two natural number satisfying a given condition in a given set Suppose $A=\{1,2,\dots,112\}$, $B \subset A$ and the number of elements in $B$ is greater or equal to 37. Then, is it true that there always exist two elements $x,y \in B$ such that $x-y\in \{9,10,19\}$? Thanks in advance.
 A: Let's prove the following ...
Claim: Let $S$ be a subset of $X=\{1,2,\dots , 28\}$ for which $a-b\not\in\{9,10,19\}$ for all $a,b\in S$.  Then $|S|\le 9$.
By way of contradiction, assume that $|S|\ge 10$.  Now arrange the elements of $X$ as follows.
\begin{array}{rrrrrrrrrr}
1&2&3&4&5&6&7&8&9&\\
10&11&12&13&14&15&16&17&18&19\\
20&21&22&23&24&25&26&27&28&
\end{array} 
Note that $S$ can contain at most one element from each column.  However, as $|S|\ge 10$, $S$ must contain at least one element from every column.  In particular, $19\in S$ is forced.
Now consider the following arrangement of the elements of $X$.
\begin{array}{rrrrrrrrrr}
1&2&3&4&5&6&7&8&9&10\\
11&12&13&14&15&16&17&18&19\\
20&21&22&23&24&25&26&27&28&
\end{array} 
By using the exact same reasoning as above, we see that $10\in S$ is forced.  But this is a contradiction because $19-10=9$.  Thus $|S|\le 9$ as claimed.
We may repeat this argument on the three remaining subintervals $\{29,\dots,56\}$, $\{57,\dots,84\}$, and $\{85,\dots,112\}$.  This shows there are at most $9+9+9+9=36$ elements of $\{1,2,\dots,112\}$ with the desired avoidance property.  Hence $|A|\ge 37$ will force the difference of some pair of elements of $A$ to be $9$, $10$ or $19$.  
