JEE Main 2021 — Mathematics Algebra

$B$B and $C$C will contain three digit numbers of the form $9k+2$9k+2 and $9k+l$9k+l respectively. We need to find sum of all elements in the set $B\cup C$B∪C effectively.

Now, $S(B\cup C)=S(B)+S(C)-S(B\cap C)$S(B∪C)=S(B)+S(C)−S(B∩C) where $S(k)$S(k) denotes sum of elements of set $k.$k.

Also, $B={101,110,\ldots \ldots ,992}$B=101,110,……,992

$$\therefore S(B)=\frac{100}{2}(101+992)=54650$$∴S(B)=2100​(101+992)=54650

Case-I: If $l=2$l=2

then $B\cap C=B$B∩C=B

$$\therefore S(B\cup C)=S(B)$$∴S(B∪C)=S(B)

which is not possible as given sum is

$$274\times 400=109600.$$274×400=109600.

Case-II If $l\neq 2$l=2

then $B\cap C=\phi$B∩C=ϕ

$$\therefore S(B\cup C)=S(B)+S(C)=400\times 274$$∴S(B∪C)=S(B)+S(C)=400×274

$$\Rightarrow 54650+\sum _{k=1}^{110}9k+l=109600$$⇒54650+k=1∑110​9k+l=109600

$$\Rightarrow 9\sum _{k=11}^{110}k+\sum _{k=11}^{110}l=54950$$⇒9k=11∑110​k+k=11∑110​l=54950

$$\Rightarrow 9(\frac{100}{2}(11+110))+l(100)=54950$$⇒9(2100​(11+110))+l(100)=54950

$$\Rightarrow 54450+100l=54950$$⇒54450+100l=54950

$$\Rightarrow l=5$$⇒l=5