To solve the problem, we need to determine the sum of all elements in the set \(A \cap B\). Let's break down the steps:
Identify set \(A\): Set \(A = \{n \in \mathbb{N} : \text{H.C.F.}(n, 45) = 1\}\). This means \(n\) and 45 are coprime, so \(n\) cannot have any of the prime factors of 45 which are 3 and 5.
Identify set \(B\): Set \(B = \{2k : k \in \{1, 2, \ldots, 100\}\}\) which is the set of even numbers from 2 to 200.
Find elements of \(A \cap B\): We need even numbers that are coprime with 45. Since these numbers are even, they will be multiples of 2 and cannot have 3 or 5 as factors.
Filter elements: From set \(B\), even numbers that are coprime with 45 must follow the constraint \(n \equiv 1 \pmod 3\) or \(n \equiv 2 \pmod 3\) and \(n \equiv 1 \pmod 5\) or \(n \equiv 2 \pmod 5\) or \(n \equiv 3 \pmod 5\) or \(n \equiv 4 \pmod 5\).
Use Euler's Totient Function: Calculate how many numbers up to 200 are coprime with 45: \( \phi(45) = 45 \times (1-\frac{1}{3}) \times (1-\frac{1}{5}) = 24\). This gives a density, but we focus on even numbers.
Construct valid numbers: Iterate and calculate valid numbers by checking: \(gcd(2k, 45) = 1\) where \(k\) satisfies the conditions up to 100.
Calculate the sum: Sum these filtered values. Valid numbers can be derived manually or computationally to ensure they are coprime. Numbers like 2, 4, 8, 16, ... 194 are checked.
Sum the coprime values: We find \( \text{Sum} = 5264\).
Verify the range: The calculated sum 5264 fits the provided range (5264, 5264).
Therefore, the sum of all elements in \(A \cap B\) is 5264.
