Question

Let $A=\left\{n\in N:nisa3-digitnumber\right\}$,$B=\left\{9k+2:k\in N\right\}$ and $C=\left\{9k+l:k\in N\right\}$ for some $l(0<l<9)$. If the sum of all the elements of the set $A\cap (B\cup C)$ is $274\times 400$, then $l$ is equal to

Answer 1

The value of $I$ is equal to $5$

Explanation for the correct answer:

Finding the sum of elements of set $B$

The given set

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

Considering $S\left(k\right)$denotes sum of elements of set $k$.

$\because S(B\cup C)=S\left(B\right)+S\left(C\right)-S(B\cap C)...\left(i\right)$

Put $k=100$ in set $B$

$B=\left\{101,109,...,992\right\}$

So number of elements of set $B$is $100$

Now sum of elements of set $B$

$\Rightarrow S\left(B\right)=\frac{100}{2}(101+992)=54650$

Considering Case First: If $l=2$

$\because$ $C=\left\{9k+l:k\in N\right\}$

Then $B\cap C=B$

$\therefore S(B\cup C)=S\left(B\right)$

Which is not possible because $274\times 400=109600$ is given and $S\left(B\right)=54650$

Considering Case Second: If $l\ne 2$

$$\begin{gathered} \Rightarrow B\cap C=\varphi \\ \therefore S(B\cup C)=S\left(B\right)+S\left(C\right) \\ =400\times 274 \\ \Rightarrow 54650+\sum _{k=11}^{110}\left(9k+l\right)=109600 \\ \Rightarrow 9\sum _{k=11}^{110}k+\sum _{k=11}^{110}l=54950 \\ \Rightarrow 9\left(\frac{100}{2}(11+110)\right)+l\left(100\right)=54950 \\ \Rightarrow 54450+100l=54950 \\ \Rightarrow l=5 \end{gathered}$$

Hence, the required value of $l=5$