The number of elements in the set n ∈ ℕ: 10 ≤ n ≤ 100 and 3 ^ n - 3...

Given that $n \in$\left[10, 100\right]$and$n \in N$.</p><p>Let us take the values of$n = 1, 2, 3, . . .$and check whether${3}^{n} - 3$is divisible by$7$or not.</p><p>$⇒ {3}^{1} - 3 = 0$is divisible by$7$</p><p>$⇒ {3}^{2} - 3 = 3$,$3$is not divisible by$7$</p><p>$⇒ {3}^{6} = 7k + 1$is not divisible by$7$</p><p>$⇒ {3}^{7} = 7\alpha + 3$, </p><p>$⇒ {3}^{7} - 3 = 7\alpha$is divisible by$7$</p><p>Similarly$⇒ {3}^{13} = \left(7k + 1\right)\left(7\alpha + 3\right)$</p><p>$= 7\beta + 3$</p><p>$⇒ {3}^{13} - 3 = 7\beta$is divisible by$7$</p><p>If we observe the pattern for$n = 1, 7, 13$,${3}^{n} - 3$is divisible by$7$.</p><p>That means the series forms an AP with$d = 6$.</p><p>Also$n \in \left[10, 100\right]$</p><p>The required progression is$13, 19, . . . . . . {a}_{n}$</p><p>$⇒ {a}_{n} = 13 + \left(n - 1\right)6$</p><p>But$13 + \left(n - 1\right)6 < 100$</p><p>$⇒ n < 15 . 5$</p><p>$⇒ n = 15 \in N$</p><p>Therefore, the required answer is$15.