Key Concepts and Formulas

1. Set Operations: Understanding of set builder notation, set difference ($A-B$), and set intersection ($A \cap B$).
2. Quadratic Inequalities: Solving inequalities of the form $ax^2 + bx + c \le 0$.
3. Modular Arithmetic: Characterizing numbers based on their remainders when divided by a certain integer.
4. Arithmetic Progression (AP): Formulas for the $n$-th term ($a_n = a_1 + (n-1)d$) and the sum of the first $n$ terms ($S_n = \frac{n}{2}(a_1 + a_n)$).

Step-by-Step Solution

Step 1: Determine the elements of Set A

The set $A$ is defined as $A = \{n \in \mathbb{N} \mid n^2 \le n + 10,000\}$. We need to find all natural numbers $n$ satisfying this inequality.

1. Rearrange the inequality: $n^2 - n - 10,000 \le 0$
2. Find the roots of the quadratic equation $n^2 - n - 10,000 = 0$. Using the quadratic formula: $n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-10,000)}}{2(1)} = \frac{1 \pm \sqrt{1 + 40,000}}{2} = \frac{1 \pm \sqrt{40,001}}{2}$
3. Approximate the roots: $\sqrt{40,001} \approx 200.0025$. The roots are approximately: $n_1 \approx \frac{1 - 200.0025}{2} \approx -99.501$ $n_2 \approx \frac{1 + 200.0025}{2} \approx 100.501$
4. Determine the inequality's solution: Since the coefficient of $n^2$ is positive, the parabola opens upwards, so the inequality $n^2 - n - 10,000 \le 0$ holds for $n$ between the roots: $-99.501 \le n \le 100.501$
5. Apply the constraint $n \in \mathbb{N}$: Natural numbers are positive integers. Therefore, the natural numbers satisfying the inequality are $n = 1, 2, 3, \ldots, 100$. So, $A = \{1, 2, 3, \ldots, 100\}$.

Step 2: Determine the elements of Set B and Set C

1. Set B: $B = \{3k + 1 \mid k \in \mathbb{N}\}$. Since $k \in \mathbb{N}$, $k$ starts from $1$. $B = \{3(1)+1, 3(2)+1, 3(3)+1, \ldots\} = \{4, 7, 10, 13, \ldots\}$. These are numbers that leave a remainder of 1 when divided by 3, i.e., $n \equiv 1 \pmod 3$.

2. Set C: $C = \{2k \mid k \in \mathbb{N}\}$. Since $k \in \mathbb{N}$, $k$ starts from $1$. $C = \{2(1), 2(2), 2(3), \ldots\} = \{2, 4, 6, 8, \ldots\}$. These are all positive even natural numbers.

Step 3: Determine the elements of Set B - C

The set $B - C$ contains elements that are in $B$ but not in $C$. An element $n$ is in $B - C$ if:

• $n \in B$, meaning $n$ is of the form $3k+1$ for some $k \in \mathbb{N}$. This implies $n \equiv 1 \pmod 3$.
• $n \notin C$, meaning $n$ is not an even number. This implies $n$ is odd, so $n \equiv 1 \pmod 2$.

We need to find numbers $n$ satisfying both congruences: $n \equiv 1 \pmod 3$ $n \equiv 1 \pmod 2$ Since the moduli (3 and 2) are coprime and the remainders are the same, the solution is $n \equiv 1 \pmod{\text{lcm}(3,2)}$, which is $n \equiv 1 \pmod 6$. So, elements of $B - C$ are of the form $6j + 1$ for some integer $j$.

Now we need to find the range of $j$. Since $n \in B$, the smallest element of $B$ is 4. So, $6j+1 \ge 4 \implies 6j \ge 3 \implies j \ge 0.5$. Since $j$ must be an integer to generate the sequence, the smallest integer value for $j$ is 1. Thus, $B - C = \{6j + 1 \mid j \in \mathbb{N}\} = \{7, 13, 19, 25, \ldots\}$.

Step 4: Determine the elements of $A \cap (B - C)$

We are looking for elements that are common to set $A$ and set $B - C$. $A = \{1, 2, 3, \ldots, 100\}$ $B - C = \{6j + 1 \mid j \in \mathbb{N}\}$

We need elements $n$ such that $n \in A$ and $n \in B-C$. So, $n$ must be of the form $6j+1$ where $j \in \mathbb{N}$, and $1 \le n \le 100$. Substituting $n = 6j+1$ into the inequality for $A$: $1 \le 6j+1 \le 100$ Subtract 1 from all parts: $0 \le 6j \le 99$ Divide by 6: $0 \le j \le 16.5$ Since $j \in \mathbb{N}$, the possible integer values for $j$ are $1, 2, 3, \ldots, 16$. The elements of $A \cap (B - C)$ are obtained by substituting these values of $j$ into $6j+1$:

• For $j=1$: $n = 6(1)+1 = 7$
• For $j=2$: $n = 6(2)+1 = 13$
• ...
• For $j=16$: $n = 6(16)+1 = 97$

So, $A \cap (B - C) = \{7, 13, 19, \ldots, 97\}$.

Step 5: Calculate the sum of all elements of $A \cap (B - C)$

The set $A \cap (B - C) = \{7, 13, 19, \ldots, 97\}$ is an arithmetic progression.

• First term ($a_1$) = 7
• Common difference ($d$) = $13 - 7 = 6$
• Number of terms ($N$) = 16 (since $j$ ranged from 1 to 16)
• Last term ($a_N$) = 97

The sum of an arithmetic progression is given by $S_N = \frac{N}{2}(a_1 + a_N)$. $S_{16} = \frac{16}{2}(7 + 97)$ $S_{16} = 8(104)$ $S_{16} = 832$

Summary

The problem involved finding the intersection of three sets defined by different properties. Set A was defined by a quadratic inequality, and its elements were natural numbers up to 100. Set B contained numbers of the form $3k+1$ ($k \in \mathbb{N}$), and Set C contained even natural numbers. The set difference $B-C$ was found to contain numbers of the form $6j+1$ ($j \in \mathbb{N}$). The intersection $A \cap (B-C)$ yielded an arithmetic progression $\{7, 13, \ldots, 97\}$. The sum of this arithmetic progression was calculated using the standard formula.

The final answer is $\boxed{832}$.