$A=\{n\in N:nisa3-digitnumber\}$ $B=\{9k+2;k\in N\}$  and $C=\{9k+\mathcal{l}:k\in N\}$  for some $\mathcal{l}(0<\mathcal{l}<9)$  if the sum of all the elements of the set $A\cap (B\cup C)$  is $274\times 400$ then $\mathcal{l}$  is equal to ___

B and C will contain three digit number of the form $9k+2$ and  $9k+\mathcal{l}$ respectively.
We need to find the sum of all elements in the set $B\cup C$ respectively. $S(B\cup C)=S\left(B\right)+S\left(C\right)-S(B\cap C)$

Where $S\left(K\right)$  denotes sum of elements of set k. $B=\{101,109,\mathrm{...992}\};S\left(B\right)=\frac{100}{2}[101+992]=54650$

Case (i); If  $\mathcal{l}=2$  $B\cap C=B$  $S(B\cup C)=S\left(B\right)=54650\ne 274\times 400$

 which is not possible
Case (ii): If $\mathcal{l}\ne 2$then  $\mathcal{l}\ne 2$
$S(B\cup C)=S\left(B\right)+S\left(C\right)=400\times 274$ $54650+\sum _{k=11}^{110}(9k+\mathcal{l})=109600$ $9\sum _{k=11}^{110}k+\sum _{k=11}^{110}\mathcal{l}=54950$$9\left[\frac{100}{2}(11+110)\right]+\mathcal{l}\left[100\right]=54950$ $54450+100\mathcal{l}=54950$$\mathcal{l}=5$