Question 16.

There are multiple ways of solving these sorts of questions. One method is to look for powers of the term in the numerator that leave a remainder of 1 or -1 when divided by the denominator. 

Noting down the powers of 3, 3, 9, 27, 81, 243

243 is one such number, 242 is multiple of 11 (11 times 22), hence 243 will leave a remainder of 1 when divided by 11. 

243 is 3 raised to power 5; we can rewrite the given term as $\dfrac{3^{330}\times\ 3^3}{11}$

The overall remainder will be $\left[\dfrac{3^{330}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$

$\left[\dfrac{3^{5\times\ 66}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$

$\left[\dfrac{243^{66}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$

$1^{66}\times\ \left[\dfrac{27}{11}\right]_R$

$1\times\ 5$

$5$

Therefore, Option A is the correct answer.