JEE Main 2023 — Sets And Relations Question with Solution

To determine the number of elements in the given set, we need to find how many natural numbers $n$ between $10$ and $100$ (inclusive) satisfy the condition that $3^n-3$ is a multiple of $7$.

Recall that for any integers $a$ and $b$, $a$ is a multiple of $b$ if there exists an integer $k$ such that $a=bk$. So in our case, we need to find how many $n$ satisfy the equation $3^n-3=7k$ for some integer $k$.

Notice that $3^n-3=3(3^{n-1}-1)$. We want this expression to be a multiple of 7. Let's explore a few powers of 3 modulo 7:

$3^1\equiv 3(\mathrm{mod}7)$

$3^2\equiv 9\equiv 2(\mathrm{mod}7)$

$3^3\equiv 27\equiv 6(\mathrm{mod}7)$

$3^4\equiv 81\equiv 4(\mathrm{mod}7)$

$3^5\equiv 243\equiv 5(\mathrm{mod}7)$

$3^6\equiv 729\equiv 1(\mathrm{mod}7)$

We observe that $3^n(\mathrm{mod}7)$ follows a cycle of length 6. So, $3^{n-1}(\mathrm{mod}7)$ also follows the same cycle, but shifted:

$3^0\equiv 1(\mathrm{mod}7)$

$3^1\equiv 3(\mathrm{mod}7)$

$3^2\equiv 2(\mathrm{mod}7)$

$3^3\equiv 6(\mathrm{mod}7)$

$3^4\equiv 4(\mathrm{mod}7)$

$3^5\equiv 5(\mathrm{mod}7)$

We want $3(3^{n-1}-1)\equiv 0(\mathrm{mod}7)$, which means that $3^{n-1}-1\equiv 0(\mathrm{mod}7)$. From the cycle above, we see that this is true when $n-1$ is a multiple of 6, or equivalently, when $n$ is one more than a multiple of 6.

Now let's find the multiples of 6 between 10 and 100:

$12,18,24,30,36,42,48,54,60,66,72,78,84,90,96$

Adding 1 to each of these values, we get the set of natural numbers $n$ that satisfy the given condition:

$13,19,25,31,37,43,49,55,61,67,73,79,85,91,97$

There are 15 elements in this set. Therefore, the number of elements in the given set is $\boxed{15}$.