The number of elements in the set $\{n \in \mathbb{N} : 10 \leq n \leq 100\}$ and $3^n - 3$ is a multiple of…

Given that $n\in \left[10,100\right]$ and $n\in N$.

Let us take the values of $n=1,2,3,...$ and check whether $3^n-3$ is divisible by $7$ or not.

$\Rightarrow 3^1-3=0$ is divisible by $7$

$\Rightarrow 3^2-3=3$, $3$ is not divisible by $7$

$\Rightarrow 3^6=7k+1$ is not divisible by $7$

$\Rightarrow 3^7=7\alpha +3$, 

$\Rightarrow 3^7-3=7\alpha$ is divisible by $7$

Similarly $\Rightarrow 3^{13}=\left(7k+1\right)\left(7\alpha +3\right)$

$=7\beta +3$

$\Rightarrow 3^{13}-3=7\beta$ is divisible by $7$

If we observe the pattern for $n=1,7,13$, $3^n-3$ is divisible by $7$.

That means the series forms an AP with $d=6$.

Also $n\in \left[10,100\right]$

The required progression is $13,19,......a_n$

$\Rightarrow a_n=13+\left(n-1\right)6$

But $13+\left(n-1\right)6<100$

$\Rightarrow n<15.5$

$\Rightarrow n=15\in N$

Therefore, the required answer is $15$.