Let $p_1<p_2<..<p_{10}$ the ten numbers. Consider $S=\left \{p_i-p_j,1\leq i<j\leq 10\right \}$.We have that $|S|=45$ but $p_i-p_j\in \left \{1,2,..,36\right \}$ . Suppoce that there is no two $p_i,p_j,p_k,p_l$ different in pairs such that $p_i-p_j=p_k-p_l$ (*) If there is 3 equal differences $p_i-p_j$,suppoce that $p_i-p_j=p_k-p_l=p_m-p_n$ We must have $p_i=p_l$ and $p_j=p_m$ so $p_k-p_i=p_j-p_n$ contradiction. So there is nine pairs of equals differences $a_1-b_1=c_1-d_1,...,a_9-b_9=c_9-d_9$ ,WLOG $a_1\leq a_2..\leq a_9$ We must have $a_i=d_i$ so $2a_i=b_i+c_i$.Suppoce that $a_i=a_j$ then $c_i+b_i=c_j+b_j,c_i\neq c_j$ so $c_i=b_j$ but $a_i<c_i=b_j<a_j$,contradiction. So $a_i\neq a_j$ .This means that $p_10\not \in \left \{a_1,a_2,..,a_9\right \}$ because $p_10>b_i,c_i,i=1,..,9$.Also if $p_1\in \left \{a_1,a_2,..,a_9\right \}$ then $p_1>p_j$ for some $p_j\in \left \{p_1,p_2,..,p_9 \right \}$ contradiction. So we end up in a contradiction if (*) is true. The proof is complete.