Remainder-4

We first deduce out the cyclicity of remainders for $3^n \pmod {11}$ , where $n \in N$ .

$3^{1}\equiv3\pmod{11}$

$3^{2}\equiv9\pmod{11}$

$3^{3}\equiv5\pmod{11}$

$3^{4}\equiv4\pmod{11}$

$3^{5}\equiv1\pmod{11}$

So , $3^{5}\equiv 1\pmod{11}$ .

$3^{1994}\equiv3^{5\times398 + 4}\equiv(3^{5})^{398}\times3^{4}\equiv1^{398}\times81\equiv81\pmod{11}$

Hence, $3^{1994}\equiv81\pmod{11}$

$\Rightarrow (3^{1994} + 2\equiv81 + 2\equiv83\equiv6\pmod{11}$