Solve this following


Let $A=\{n \in N: n$ is a 3 -digit number $\}$

$\mathrm{B}=\{9 \mathrm{k}+2: \mathrm{k} \in \mathrm{N}\}$

and $\mathrm{C}=\{9 \mathrm{k}+l: \mathrm{k} \in \mathrm{N}\}$ for some $l(0

If the sum of all the elements of the set $\mathrm{A} \cap(\mathrm{B} \cup \mathrm{C})$ is $274 \times 400$, then $l$ is equal to



$\mathrm{B}$ and $\mathrm{C}$ will contain three digit numbers of the form $9 \mathrm{k}+2$ and $9 \mathrm{k}+\ell$ respectively. We need to find sum of all elements in the set $\mathrm{B} \cup \mathrm{C}$ effectively.

Now, $S(B \cup C)=S(B)+S(C)-S(B \cap C)$

where $S(k)$ denotes sum of elements of set $k$.

Also, B $=\{101,109, \ldots \ldots, 992\}$

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

Case-I : If $\ell=2$

then $B \cap C=B$

$\therefore \quad \mathrm{S}(\mathrm{B} \cup \mathrm{C})=\mathrm{S}(\mathrm{B})$

which is not possible as given sum is

$274 \times 400=109600$

Case-II : If $\ell \neq 2$

then $\mathrm{B} \cap \mathrm{C}=\phi$

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

$\Rightarrow 54650+\sum_{\mathrm{k}=11}^{110} 9 \mathrm{k}+\ell=109600$

$\Rightarrow 9 \sum_{\mathrm{k}=11}^{110} \mathrm{k}+\sum_{\mathrm{k}=11}^{110} \ell=54950$

$\Rightarrow 9\left(\frac{100}{2}(11+110)\right)+\ell(100)=54950$

$\Rightarrow 54450+100 \ell=54950$

$\Rightarrow \ell=5$


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