Question:

Step 1: To find the number of elements in set $A$, we first need to understand what the set $A$ represents. Set $A$ consists of all natural numbers $n$ between 100 and 700 (inclusive) that are neither multiples of 3 nor multiples of 4. Step 2: The total number of natural numbers between 100 and 700 (inclusive) can be calculated by subtracting the lower limit from the upper limit and then adding 1, which gives us $700 - 100 + 1 = 601$. Step 3: Next, we need to find the number of multiples of 3 and 4 within this range. To do this, we can find the first and last multiples of 3 and 4 within the range [100, 700]. The first multiple of 3 in this range is 102, and the last multiple is 699. The first multiple of 4 is 100, and the last multiple is 700. Step 4: However, instead of directly counting the multiples of 3 and 4, the given solution takes a different approach. It calculates the total count of numbers that are either multiples of 3 or 4 (or both) by considering the count of numbers from 1 to 700 and then subtracting the count of numbers from 1 to 100 (excluding the end points that are multiples of 3 or 4), but adjusted for the multiples of 3 and 4. Step 5: The solution then directly provides the count of numbers that are not multiples of 3 or 4 by subtracting the count of numbers that are multiples of 3 or 4 from the total count of numbers in the range [100, 700]. This is given as $|A| = 601 - 301 = 300$. Step 6: The calculation $601 - 301 = 300$ implies that there are 301 numbers between 100 and 700 (inclusive) that are either multiples of 3 or 4. By subtracting this from the total count of numbers in the range (601), we get the count of numbers that are neither multiples of 3 nor 4, which is 300. Therefore, the number of elements in set $A$ is 300, which corresponds to Option 3. The final answer is $\boxed{300}$, and this matches Option 3.