Question

Let $A=\{n\in N:n$ is a 3 -digit number $\}$
$B=\{9k+2:k\in N\}$
and $C=\{9k+l:k\in N\}$ 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

Answer: 5

$B$ and $C$ will contain three digit numbers of the form $9k+2$ and $9k+\ell$ respectively. We need to find sum of all elements in the set $B\cup C$ 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, $B=\{101,109,\dots ,992\}$
$\therefore S(B)=\frac{100}{2}(101+992)=54650$
Case-I : If $\ell =2$
then $B\cap C=B$
$\therefore S(B\cup C)=S(B)$
which is not possible as given sum is
$274\times 400=109600$
Case-II: If $\ell \ne 2$
then $B\cap C=\varphi$
$\therefore S(B\cup C)=S(B)+S(C)=400\times 274$
$\Rightarrow 54650+\sum _{k=11}^{110}9k+\ell =109600$
$\Rightarrow 9\sum _{k=11}^{110}k+\sum _{k=11}^{110}\ell =54950$
$\Rightarrow 9\left(\frac{100}{2}(11+110)\right)+\ell (100)=54950$
$\Rightarrow 54450+100\ell =54950$
$\Rightarrow \ell =5$