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 $\mathrm{B} \cap \mathrm{C}=\phi$
$\therefore \quad S(B \cup C)=S(B)+S(C)=400 \times 274$
$\Rightarrow 54650+\sum_{\mathrm{k}=11}^{110} 9 \mathrm{k}+\ell=109600$
$\Rightarrow 9 \sum_{\mathrm{k}=11}^{110} \mathrm{k}+\sum_{\mathrm{k}=11}^{110} \ell=54950$
$\Rightarrow 9\left(\frac{100}{2}(11+110)\right)+\ell(100)=54950$
$\Rightarrow 54450+100 \ell=54950$
$\Rightarrow \ell=5$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.