Let $$A = \{n \in [100, 700] \cap \mathbb{N} : n$$ is neither a multiple of 3 nor a multiple of 4 $$\}$$. Then the number of elements in $$A$$ is

We need to find the number of integers in $$[100, 700]$$ that are neither multiples of 3 nor multiples of 4. From 100 to 700 inclusive there are $$700 - 100 + 1 = 601$$ integers. To determine how many are multiples of 3, note that the number of multiples of 3 up to n is $$\lfloor n/3 \rfloor$$, so in this range one finds $$\lfloor 700/3 \rfloor - \lfloor 99/3 \rfloor = 233 - 33 = 200$$ multiples of 3. Similarly, the count of multiples of 4 is given by $$\lfloor 700/4 \rfloor - \lfloor 99/4 \rfloor = 175 - 24 = 151$$. Since LCM(3, 4) = 12, the integers divisible by both 3 and 4 correspond to multiples of 12, namely $$\lfloor 700/12 \rfloor - \lfloor 99/12 \rfloor = 58 - 8 = 50$$.

Applying the inclusion-exclusion principle, the total number of integers that are multiples of 3 or 4 is $$|A \cup B| = |A| + |B| - |A \cap B| = 200 + 151 - 50 = 301$$. Therefore, the complementary set of integers that are neither multiples of 3 nor multiples of 4 has size $$601 - 301 = 300$$.

The correct answer is Option 3: 300.