Let $$A = \{n \in N \mid n^2 \leq n + 10000\}$$, $$B = \{3k + 1 \mid k \in N\}$$ and $$C = \{2k \mid k \in N\}$$, then the sum of all the elements of the set $$A \cap (B - C)$$ is equal to _________.

First, recall that in JEE problems the symbol $$N$$ denotes the set of positive integers $$\{1,2,3,\ldots\}$$; the number $$0$$ is not included.

We begin with the description of the set $$A$$. By definition

$$A=\{\,n\in N\mid n^{2}\le n+10000\,\}.$$

To find all such $$n$$ we rearrange the inequality:

$$n^{2}\le n+10000 \;\Longrightarrow\; n^{2}-n-10000\le 0.$$

We apply the quadratic-formula roots of $$n^{2}-n-10000=0$$. The formula is stated first:

If $$ax^{2}+bx+c=0,\text{ then }x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$$

For the present quadratic $$a=1,\;b=-1,\;c=-10000$$, so

$$n=\dfrac{-(-1)\pm\sqrt{(-1)^{2}-4(1)(-10000)}}{2(1)} =\dfrac{1\pm\sqrt{1+40000}}{2} =\dfrac{1\pm\sqrt{40001}}{2}.$$

The discriminant $$40001$$ lies between $$200^{2}=40000$$ and $$201^{2}=40401$$, therefore

$$200<\sqrt{40001}<201.$$

Consequently

$$\dfrac{1-\sqrt{40001}}{2}\approx -99.5 \quad\text{and}\quad \dfrac{1+\sqrt{40001}}{2}\approx 100.5.$$

Because $$n$$ must be a positive integer, the inequality $$n^{2}-n-10000\le 0$$ restricts $$n$$ to

$$1\le n\le 100.$$

Hence

$$A=\{1,2,3,\ldots,100\}.$$

Next we examine the set $$B$$:

$$B=\{\,3k+1\mid k\in N\}=\{\,4,7,10,13,\ldots\}.$$

The set $$C$$ is

$$C=\{\,2k\mid k\in N\}=\{\,2,4,6,8,\ldots\},$$

the collection of all positive even integers.

The expression $$B-C$$ means “all elements of $$B$$ that are not in $$C$$.” Because every element of $$C$$ is even, an element $$3k+1$$ is removed whenever it is even. Let us determine the parity of $$3k+1$$:

Parity $$(3k) = \text{Parity}(k)\;(\text{since }3\text{ is odd}).$$

Therefore

$$3k+1$$ is odd $$\Longleftrightarrow k$$ is even $$.$$

Only those $$k$$ that are even survive the subtraction $$B-C$$. Write $$k=2m$$ with $$m\in N$$; then

$$3k+1=3(2m)+1=6m+1.$$

Thus

$$B-C=\{\,6m+1\mid m\in N\,\}=\{\,7,13,19,25,\ldots\}.$$

Now we intersect with $$A$$. We need all numbers of the form $$6m+1$$ that do not exceed $$100$$:

$$6m+1\le100 \;\Longrightarrow\; 6m\le99 \;\Longrightarrow\; m\le16.5.$$

Because $$m$$ is a positive integer, $$m$$ ranges from $$1$$ to $$16$$ inclusive. For each $$m$$ we obtain

$$\begin{aligned} m=1&\;:\;6(1)+1=7,\\ m=2&\;:\;6(2)+1=13,\\ m=3&\;:\;6(3)+1=19,\\ &\;\;\vdots\\ m=16&:;;6(16)+1=97. \end{aligned}$$

Collecting every value we have the set

$$A\cap(B-C)=\{7,13,19,25,31,37,43,49,55,61,67,73,79,85,91,97\}.$$

This is an arithmetic progression whose first term is $$a_{1}=7$$, common difference is $$d=6$$, and the number of terms is $$n=16$$.

The sum $$S_{n}$$ of the first $$n$$ terms of an arithmetic progression is given by the formula

$$S_{n}=\dfrac{n}{2}\bigl(2a_{1}+(n-1)d\bigr).$$

Substituting $$n=16,\;a_{1}=7,\;d=6$$ we obtain

$$S_{16}=\dfrac{16}{2}\bigl(2\cdot7+(16-1)\cdot6\bigr) =8\bigl(14+15\cdot6\bigr) =8\bigl(14+90\bigr) =8\times104 =832.$$

Hence, the correct answer is Option 832.