1. Interpret the sets

Given $A=\{n\in \mathbb N\mid n^2\le n+10000\},$ $B=\{3k+1\mid k\in \mathbb N\},$ $C=\{2k\mid k\in \mathbb N\}.$

We need the sum of all elements of $A\cap (B-C).$ Here $B-C$ means the set difference $B\setminus C$, i.e. elements of $B$ which are not in $C$.

2. Find set $A$

We solve $n^2\le n+10000$ which gives $n^2-n-10000\le 0.$

Solve the quadratic equation: $n^2-n-10000=0.$

Its discriminant is $D=1+4\cdot 10000=40001=201^2.$

So the roots are $n=\frac{1\pm 201}{2}.$ Thus, $n=101 \quad \text{or} \quad n=-100.$

Since the quadratic opens upward, $-100\le n\le 101.$ Because $n\in \mathbb N$, $A=\{1,2,3,\dots,101\}$ (assuming $\mathbb N=\{1,2,3,\dots\}$; even if $0\in\mathbb N$, it does not affect the final sum).

3. Find $B-C$

Set $B$ consists of numbers of the form $3k+1$. We want those which are not even, since $C$ is the set of even natural numbers.

So we need numbers of the form $3k+1$ that are odd.

Check parity:

• If $k$ is even, then $3k$ is even, so $3k+1$ is odd.
• If $k$ is odd, then $3k$ is odd, so $3k+1$ is even.

Hence $B-C$ consists of all odd numbers of the form $3k+1$. These are precisely numbers congruent to $1 \pmod 6$: $B\setminus C=\{6m+1\mid m\in \mathbb N_0\}.$

So within $A$, we need numbers $\le 101$ and of the form $6m+1.$

These are $1,7,13,19,25,31,37,43,49,55,61,67,73,79,85,91,97.$

Next term would be $103>101$, so stop here.

4. Sum the elements

This is an arithmetic progression with:

• first term $a=1$
• last term $l=97$
• common difference $d=6$

Number of terms: $n=\frac{97-1}{6}+1=16+1=17.$

Sum: $S=\frac{n}{2}(a+l)=\frac{17}{2}(1+97)=\frac{17}{2}\cdot 98=17\cdot 49=833.$

Thus, $\sum \big(A\cap(B-C)\big)=833.$

5. Compare with stored answer

Stored correct answer: $832$

Our derived answer is $833$.

So the stored answer appears to be incorrect.