Let $A=\left\{n\in N:nisa3-digitnumber\right\}, B=\left\{9k+2;k\in N\right\}$ and $C=\left\{9k+l;k\in N\right\}$ for some l (0 < l < 9) 

          If the sum of all elements of the set $A\cap \left(B\cup C\right)$is 274 × 400, then l is equal to ____

$$\begin{gathered} 3 digit numbers of the form 9k+1 are \left\{101,109,...,992\right\} \\ Sum is equal to \frac{100}{2}\left(101+992\right)=50\times 1093 \end{gathered}$$

$3 digit numbers of the form 9k+l are \left\{99+l, 108+l, ...,990+l\right\}$    

$Sum of the above numbers is \frac{100}{2}\left(99+l+990+l\right)=50\left(1089+2l\right)$     

$Given the sum of all the elements of the set A\cap \left(B\cup C\right)=274\times 400$

$\begin{array}{rcl}50(1093+1089+2l)& =& 274\times 400\\ 50\times 2182+100l& =& 274\times 400\\ 1091+l& =& 1096\\ l& =& 5\end{array}$