Question

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

$B$ and $C$ will contain three digit numbers of the form $9 k +2$ and $9 k +\ell$ 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, $B =\{101,109, \ldots, 992\}$
$\therefore S(B)=\frac{100}{2}(101+992)=54650$
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 $\ell \neq 2$
then $B \cap C=\phi$
$\therefore S ( B \cup C )= S ( B )+ S ( C )=400 \times 274$
$\Rightarrow 54650+ \displaystyle\sum_{k=11}^{110} 9 k+\ell=109600$
$\Rightarrow 9 \displaystyle\sum_{ k =11}^{110} k +\sum_{ k =11}^{110} \ell=54950$
$\Rightarrow 9\left(\frac{100}{2}(11+110)\right)+\ell(100)=54950$
$\Rightarrow 54450+100 \ell=54950$
$\Rightarrow \ell=5$