Animated Solution for Mathematics - Sets and Relations: Let $A=\{n\in \mathbb{N}:\text{H.C.F.}(n,45)=1\}$ and Let $B=\{2k:k\in \{1,2,\dots ,100\}\}$. Then the sum of all the elements of $A\cap B$ is ________.

We define Set $A$ as $A=\{n\in \mathbb{N}:\text{H.C.F.}(n,45)=1\}$. The condition that the Highest Common Factor is $1$ implies that $n$ and $45$ are coprime.

Since the prime factorization of $45$ is $45=3^2\times 5$, for $n$ to be coprime to $45$, it must not share any prime factors with $45$. Therefore, $n$ cannot be divisible by $3$ and $n$ cannot be divisible by $5$.

Set $B$ is defined as $\{2k:k\in \{1,2,\dots ,100\}\}$. This represents the set of the first $100$ even numbers: $\{2,4,6,\dots ,200\}$.

This set acts as our restricted universe. We are only concerned with elements within this specific range.

The Filtered Reality

We seek the sum of elements in $A\cap B$. This requires finding all even numbers up to $200$ that are not divisible by $3$ and not divisible by $5$.

Since all elements in $B$ are already even (multiples of $2$), the condition "not divisible by $3$" effectively means "not divisible by $6$" (since $2\times 3=6$). Similarly, "not divisible by $5$" becomes "not divisible by $10$" (since $2\times 5=10$).

To find the sum, we take the total sum of Set $B$ and subtract the sums of multiples of $6$ and $10$. However, multiples of $30$ (which are multiples of both $6$ and $10$) are subtracted twice.

To correct this, we must add the sum of multiples of $30$ back into the total. The formula is:

First, we calculate the total sum of Set $B$. This is an arithmetic progression with $100$ terms:

Next, we calculate the sum of multiples of $6$ ($6,12,\dots ,198$), which contains $33$ terms:

Then, we calculate the sum of multiples of $10$ ($10,20,\dots ,200$), which contains $20$ terms:

Finally, we calculate the sum of multiples of $30$ ($30,60,\dots ,180$), which contains $6$ terms:

Substituting these values into our inclusion-exclusion formula:

The final sum of the numbers satisfying both conditions is 5264.