Prove the following
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 $\frac{100}{2}(1093)=\mathrm{s}_{1}=54650$

$274 \times 400=s_{1}+s_{2}$

$274 \times 400=\frac{100}{2}[101+992]+\mathrm{s}_{2}$

$274 \times 400=50 \times 1093+s_{2}$

$s_{2}=109600-54650$

$s_{2}=54950$

$s_{2}=54950=\frac{100}{2}[(99+\ell)+(990+\ell)]$

$1099=2 \ell+1089$

$\ell=5$

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