Let A = { n∈N: 'n' is a 3-digit number}B={9k+2:k∈N}C={9k+l;k∈N}for some l(0<l<9)If the sum of elements of the set A∩(B∪C) is 61×1800 ; then 'l ' is equal to_______

A∩(B∪C) = There digit numbers which are of the form 9k + 2 or 9k + l.Sum of elements of A∩(B∪C) =∑k=11110(9k+2)+∑k=11110(9k+l)(ifl>2) =(1091+l)100 ⇒l=7

Let A = { n∈N: 'n' is a 3-digit number}B={9k+2:k∈N}C={9k+l;k∈N}for some l(0<l<9)If the sum of elements of the set A∩(B∪C) is 61×1800 ; then 'l ' is equal to_______

A∩(B∪C) = There digit numbers which are of the form 9k + 2 or 9k + l.Sum of elements of A∩(B∪C) =∑k=11110(9k+2)+∑k=11110(9k+l)(ifl>2) =(1091+l)100 ⇒l=7

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Let A = { $n \in N :^{'} n^{'}$ is a 3-digit number}
$B = {9 k + 2 : k \in N}$
$C = {9 k + l; k \in N}$ for some $l (0 < l < 9)$
If the sum of elements of the set $A \cap (B \cup C)$ is $61 \times 1800$ ; then ' $l$ ' is equal to_______

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Detailed Solution

$A \cap (B \cup C)$ = There digit numbers which are of the form 9k + 2 or 9k + $l$ .
Sum of elements of $A \cap (B \cup C)$
$= \sum_{k = 11}^{110} (9 k + 2) + \sum_{k = 11}^{110} (9 k + l) (i f l > 2)$
$= (1091 + l) 100$
$\Rightarrow l = 7$