Let $A=\{n\in N:n$ is a $3 –$ digit number]   

$B=\{9k+2:k\in N\}$ 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 $3\mathcal{l}$ is equal to_______.

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

Also $B=\{101,109,\mathrm{.....992}\}$

$\therefore S\left(B\right)=\frac{100}{2}(101+992)=54650$

Case – I: If $\mathcal{l}=2$ then $B\cap C=B$

$\therefore S(B\cup C)=S\left(B\right)$

Which is not possible as given sum is $274\times 400=109600$

Case – II: If $\mathcal{l}\ne 2$ then $B\cap C=\varphi$

$\therefore S(B\cup C)=S\left(B\right)+S\left(C\right)=400\times 274$

$\Rightarrow 54650+\sum _{k=11}^{110}9k+\mathcal{l}=109600\Rightarrow 9\sum _{k=11}^{110}k+\sum _{k=11}^{110}\mathcal{l}=54950$

$\Rightarrow 9\left(\frac{100}{2}(11+110)\right)+\mathcal{l}\left(100\right)=54950\Rightarrow 54450+100\mathcal{l}=54950\Rightarrow \mathcal{l}=5$