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 ________.

In this problem, we're dealing with 3-digit numbers in set $A$, and subsets $B$ and $C$ which represent numbers of specific forms. <br/><br/>1. First, we consider the numbers of the form $9k + 2$ (Set $B$) within the 3-digit range, which starts at 101 and ends at 992. <br/><br/>2. We calculate the sum of these numbers, denoted as $s_1$. To calculate $s_1$, you use the formula for the sum of an arithmetic series : <br/><br/> $(n/2) \times (\text{{first term}} + \text{{last term}})$ <br/><br/> Here, $n$ is the total count of such numbers. These are 3-digit numbers of the form $9k + 2$, and we can find the total count by subtracting the smallest such number (101) from the largest (992), dividing the result by 9 (because we're considering numbers with a difference of 9), and then adding 1. <br/><br/> The sum $s_1$ is calculated as follows : <br/><br/> $(100/2) \times (101 + 992) = 54650$ <br/><br/>3. According to the problem, the sum of all elements of the set $A \cap (B \cup C)$ is $274 \times 400 = 109600$. <br/><br/> Since the set $A \cap (B \cup C)$ is the union of two disjoint sets (the set of all three-digit numbers of form $9k + 2$ and the set of all three-digit numbers of form $9k + l$), we can write this sum as : <br/><br/> $s_1$(for numbers of the form 9k + 2) + $s_2$ (for numbers of the form 9k + l) = 109600 <br/><br/>4. Solving this equation for $s_2$ (the sum of numbers of the form $9k + l$), we get : <br/><br/> $s_2 = 109600 - s_1 = 109600 - 54650 = 54950$ <br/><br/>5. The sum $s_2$ can be expressed as $(n/2) \times (\text{{first term}} + \text{{last term}})$, where $n$ is the count of numbers of the form $9k + l$. The first term here is the smallest 3-digit number of this form, which is $99 + l$, and the last term is the largest such number, which is $990 + l$. <br/><br/> We equate this to $s_2$ to solve for $l$ : <br/><br/> $54950 = (100/2)[(99 + l) + (990 + l)]$ <br/><br/>6. Simplifying this equation, we get : <br/><br/> $2l + 1089 = 1099$ <br/><br/> Solving for $l$, we find : <br/><br/> $l = 5$ <br/><br/>So, the correct answer is 5.