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Step-by-Step Solution
Step 1: Understand the Sets A, B, and C
• The set A is given by A = \{n \in \mathbb{N} \mid n^2 \leq n + 10000\} .
• The set B is B = \{3k + 1 \mid k \in \mathbb{N}\} .
• The set C is C = \{2k \mid k \in \mathbb{N}\} .
Step 2: Describe the Set B − C
Elements of B are of the form 3k + 1 , while elements of C are even numbers 2k .
• B - C means we take elements that are in B but not even. So we are looking for elements of the form 3k + 1 that are odd.
• For small values of k , these elements are 4, 7, 10, 13, 16, 19, etc. Among these, only the odd ones remain in B - C . Those are 7, 13, 19, 25, and so on.
Step 3: Determine the Set A Explicitly
We need all n (natural numbers) such that n^2 \leq n + 10000 . Rewrite it as:
n^2 - n \leq 10000 \quad \Longrightarrow \quad n(n - 1) \leq 10000 .
Observe that 100 \times 99 = 9900 \leq 10000 and (101 \times 100) = 10100 > 10000 .
Hence, n can go up to 100 . Thus:
A = \{1, 2, 3, \ldots, 100\} .
Step 4: Find A ∩ (B − C)
• From the sets above, B - C contains numbers like 7, 13, 19, \dots (all of the form 3k + 1 that are odd).
• Within A , we only consider numbers from 1 to 100 . So in B - C , the numbers between 1 and 100 are:
\{7, 13, 19, 25, 31, 37, 43, 49, 55, 61, 67, 73, 79, 85, 91, 97\} .
Thus,
A \cap (B - C) = \{7, 13, 19, 25, 31, 37, 43, 49, 55, 61, 67, 73, 79, 85, 91, 97\} .
Step 5: Calculate the Sum of Elements in A ∩ (B − C)
The terms listed form an arithmetic progression with first term a_1 = 7 , common difference d = 6 , and last term a_{16} = 97 . The number of terms is 16 .
Using the formula for the sum of an arithmetic progression:
S = \frac{n}{2} \bigl(a_1 + a_n\bigr) , where n is the number of terms.
So,
S = \frac{16}{2} (7 + 97) = 8 \times 104 = 832 .
Final Answer
The sum of all elements in A \cap (B - C) is \boxed{832} .
