Let  $X=\{1,2,\mathrm{3......100}\}$  and Y be a subset of X such that the sum of no two elements in Y is divisible by 7. If the maximum possible number of elements in Y is

Let  $Y_i$  be the subset of X such that  $y_i=7m+i,m\in I$
 $$\begin{gathered} Y_0=\{7,14,\mathrm{......98}\},n\left(Y_0\right)=14 \\ {Y}_1=\{1,8,\mathrm{5......99}\},n\left(Y_1\right)=15 \\ {Y}_2=\{2,9,\mathrm{16......100}\},n\left(Y_2\right)=15 \\ {Y}_3=\{3,10,\mathrm{16......94}\},n\left(Y_3\right)=14 \\ {Y}_4=\{4,11,\mathrm{18......95}\},n\left(Y_4\right)=14 \\ {Y}_5=\{5,\mathrm{12......96}\},n\left(Y_5\right)=14 \\ {Y}_6=\{6,\mathrm{13......97}\},n\left(Y_6\right)=14 \end{gathered}$$
The largest Y will consist of (!) an element of  ${\text{Y}}_{\text{0}}\left(\text{ii}\right){\text{Y}}_{\text{1}}\left(\text{iii}\right){\text{Y}}_{\text{2}}\left(\text{iv}\right){\text{Y}}_{\text{3}}$  or  ${\text{Y}}_4$
$\Rightarrow$ The maximum possible number of elements in Y=1+15+15=45