# Solve this following

Question:

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 Solution:$\mathrm{B}$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\$