Let $$A = \{n \in N : n \text{ is a 3-digit number}\}$$, $$B = \{9k + 2 : k \in N\}$$ 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 ______

We have $$A = \{n \in \mathbb{N} : n \text{ is a 3-digit number}\} = \{100, 101, \ldots, 999\}$$, $$B = \{9k + 2 : k \in \mathbb{N}\}$$, and $$C = \{9k + l : k \in \mathbb{N}\}$$ where $$0 < l < 9$$.

The set $$B$$ consists of all natural numbers that leave remainder $$2$$ when divided by $$9$$. Similarly, $$C$$ consists of all natural numbers that leave remainder $$l$$ when divided by $$9$$.

The 3-digit numbers with remainder $$r$$ modulo $$9$$ (for $$1 \leq r \leq 8$$) form an arithmetic progression with first term $$99 + r$$, last term $$990 + r$$, and common difference $$9$$. The number of such terms is $$\frac{(990 + r) - (99 + r)}{9} + 1 = 100$$.

The sum of the 3-digit numbers with remainder $$r$$ is $$\frac{100}{2} \times (99 + r + 990 + r) = 50(1089 + 2r)$$.

For $$B$$, the remainder is $$r = 2$$, so the sum of 3-digit numbers in $$B$$ is $$50(1089 + 4) = 50 \times 1093 = 54650$$.

If $$l \neq 2$$, then $$B$$ and $$C$$ pick out different residue classes, and $$A \cap (B \cup C)$$ is the union of two disjoint sets. The total sum is $$54650 + 50(1089 + 2l)$$.

We are given that this sum equals $$274 \times 400 = 109600$$. So $$54650 + 50(1089 + 2l) = 109600$$.

Simplifying, $$54650 + 54450 + 100l = 109600$$, which gives $$109100 + 100l = 109600$$, so $$100l = 500$$ and $$l = 5$$.

Hence, the answer is $$5$$.