Animated Solution for Mathematics - Sets and Relations: The number of 4-digit numbers which are neither multiple of 7 nor multiple of 3 is .

Imagine you are standing in front of a massive, infinite library. You are tasked with a specific mission: to count the four-digit numbers that are 'pure'—meaning they are not divisible by $3$ and not divisible by $7$.

At first glance, this feels like a daunting task. How do we count what isn't there? This is the beauty of set theory: we define our universe, identify the 'intruders,' and subtract them.

Inclusion-Exclusion

Let $U$ be the set of all $4$-digit numbers. We know that the smallest $4$-digit number is $1000$ and the largest is $9999$.

The total count, $n(U)$, is calculated as:

Now, let $A$ be the set of numbers divisible by $3$, and $B$ be the set of numbers divisible by $7$. We want to find the numbers that are in neither $A$ nor $B$. Mathematically, we are looking for $n(A^c\cap B^c)$.

By De Morgan's Law, this is equivalent to:

Arithmetic Progressions

To find $n(A\cup B)$, we use the Principle of Inclusion-Exclusion:

For set $A$, the first multiple of $3$ after $1000$ is $1002$, and the last is $9999$. Using the formula $l=a+(n-1)d$, where $d=3$:

We repeat this for set $B$ (multiples of $7$). The first multiple is $1001$ and the last is $9996$. Solving the progression:

Double Counting

If we simply add $n(A)$ and $n(B)$, we have counted the numbers divisible by both $3$ and $7$ twice. These are the multiples of their Least Common Multiple, $LCM(3,7)=21$.

We must find $n(A\cap B)$. The first multiple of $21$ is $1008$, and the last is $9996$:

Now, we calculate the union:

We take our total universe of $9000$ and subtract the $3857$ numbers that are divisible by either $3$ or $7$.

The final count is:

You have successfully navigated the logic of sets and sequences. Remember, in JEE Advanced, it is rarely about brute force; it is about the elegance of the method. The final answer is 5143.