Question:
Let
$A=\{n \in N: n$ is a 3 -digit number $\}$
$B=\{9 k+2: k \in N\}$ and
$\mathrm{C}:\{9 \mathrm{k}+\ell: \mathrm{k} \in \mathrm{N}\}$ for some $\ell(0<\ell<9)$
If the sum of all the elements of the set $A \cap(B \cup C)$ is $274 \times 400$, then $\ell$ is equal to
Solution:
3 digit number of the form $9 \mathrm{~K}+2$ are $\{101,109, \ldots \ldots \ldots . .992\}$
$\Rightarrow$ Sum equal to $\frac{100}{2}(1093)=\mathrm{s}_{1}=54650$
$274 \times 400=s_{1}+s_{2}$
$274 \times 400=\frac{100}{2}[101+992]+\mathrm{s}_{2}$
$274 \times 400=50 \times 1093+s_{2}$
$s_{2}=109600-54650$
$s_{2}=54950$
$s_{2}=54950=\frac{100}{2}[(99+\ell)+(990+\ell)]$
$1099=2 \ell+1089$
$\ell=5$