Let $A = \{n \in \mathbb{N} | n^2 \le n+10000\}$, $B = \{3k+1 | k \in \mathbb{N}\}$ and $C = \{2k | k \in \mathbb{N}\}$. Then the sum of all the elements of the set $A \cap (B-C)$ is equal to ________.
Show Hint
To characterize a set like B-C, analyze the properties of its elements. Here, numbers in B are $3k+1$ and numbers in C are even. So B-C contains numbers of the form $3k+1$ that are odd, which simplifies to numbers of the form $6j+1$.
Step 1: Find set $A$.
\[
n^2 \le n + 10000
\Rightarrow n^2 - n - 10000 \le 0
\]
\[
n=\frac{1\pm\sqrt{1+40000}}{2}
=\frac{1\pm\sqrt{40001}}{2}
\]
Since $\sqrt{40001}\approx200$, the inequality holds for:
\[
1 \le n \le 100
\]
\[
A=\{1,2,3,.......,100\}
\]
Step 2: Find set $B-C$.
\[
B=\{3k+1:k\in\mathbb{N}\},\qquad
C=\{2k:k\in\mathbb{N}\}
\]
Elements of $B-C$ are those of the form $3k+1$ which are odd.
This happens when $k$ is even, say $k=2j$.
\[
B-C=\{6j+1:j\in\mathbb{N}\}
=\{7,13,19,25,.......\}
\]
Step 3: Find $A\cap(B-C)$.
\[
6j+1\le100 \Rightarrow j\le16
\]
\[
A\cap(B-C)=\{7,13,19,.......,97\}
\]
This is an arithmetic progression with:
\[
a=7,\ d=6,\ n=16
\]
Step 4: Find the sum.
\[
S=\frac{n}{2}(a+l)
=\frac{16}{2}(7+97)
=8\times104
=832
\]
\[
\boxed{832}
\]
