Animated Solution for Mathematics - Sets and Relations: Let $A=\{n\in [100,700]\cap \mathbb{N}:n\text{ is neither a multiple of 3 nor a multiple of 4}\}.Thenthenumberofelementsin$A$ is

Welcome, fellow traveler, to the beautiful world of combinatorics. Today, we are not just solving a problem; we are embarking on a journey to organize chaos.

We have a set of numbers, $A$, defined as all integers $n$ in the range $[100,700]$ that are neither multiples of 3 nor 4. It sounds simple, but counting what is 'not' there is often the most elegant way to see what is.

Imagine a vast, orderly library containing every integer from 100 to 700. This is our Universal Set, $U$.

To find the size of this set, we must be precise. A common pitfall is to simply calculate $700-100$, which gives 600. But remember, we are including both 100 and 700.

Think of it like a fence: if you have a fence 100 meters long with posts every meter, you have 101 posts, not 100. Thus, our total count is:

We have 601 numbers in our library.

We want to find the numbers that are neither multiples of 3 nor 4. Trying to pick these out one by one would be a nightmare.

Instead, let's use the power of complementary counting. We will find the 'bad' numbers—those that are multiples of 3 or 4—and subtract them from our total.

Mathematically, we are looking for $N(M_3^c\cap M_4^c)$, which is equivalent to:

We are essentially finding the area of the rectangle outside our two overlapping circles in a Venn diagram.

Now, let's hunt for the multiples. First, the multiples of 3 ($M_3$). The first multiple of 3 in our range is 102, and the last is 699.

This is a perfect Arithmetic Progression (A.P.) where the first term $a=102$ and the common difference $d=3$. Using the formula for the $n$-th term, $a_n=a+(n-1)d$, we set:

Solving this, we find $597=3(n-1)$, which simplifies to $199=n-1$, giving us $n=200$. There are 200 multiples of 3.

Next, the multiples of 4 ($M_4$). The first is 100, and the last is 700. Here, $a=100$ and $d=4$. Setting:

We get $600=4(n-1)$, so $150=n-1$, meaning $n=151$. There are 151 multiples of 4.

Here is where the magic happens. If we simply add 200 and 151, we have overcounted. Numbers like 12, 24, and 36 are multiples of both 3 and 4.

They are hiding in both groups! We must identify these common multiples, which are multiples of $\text{LCM}(3,4)=12$. The first multiple of 12 in our range is 108, and the last is 696.

Using our A.P. formula again:

This yields $588=12(n-1)$, so $49=n-1$, giving $n=50$. We have 50 common multiples.

Now, we apply the Principle of Inclusion-Exclusion:

We have identified 301 numbers that are multiples of 3 or 4. To find the numbers that are neither, we simply subtract this from our universal set:

And there it is! 300 numbers, standing tall, neither divisible by 3 nor 4. You have successfully navigated the logic of sets. Keep this clarity, and no counting problem will ever intimidate you again.