Power Up

Observe that the last digit of the power of $3$ repeat in the cycle $3,9,7,1\cdots$ . So, the last digit of $3^{2016}$ will be the $2016^{th}$ digit in the given sequence. As $2016$ when divided by $4$ leaves ni remainder, $1$ .

When $9018^{th}$ is divided by $4$ it leaves $2$ as remainder, so, $b=9$ .

So, $a+b=1+9=10$ .

We know $3^2\equiv -1\pmod{10}$ , thus $3^{2016}\equiv (3^2)^{1008}\\\equiv 1$

Similarly, $3^{9018}\equiv (3^2)^{4009}\\\equiv -1\equiv 9$

Thus, $a=1$ and $b=9$ , hence $a+b=1+9=\boxed{10}$ .

$\phi(10) = 4$ where $\phi(n)$ is the Euler-Phi function.
According to Euler's theorem,
$a^{\phi(n)} \equiv \pmod{n} , gcd(a,n) = 1$
$\therefore 3^{4} \equiv 1 \pmod{10}$
$3^{2016} \equiv (3^{4})^{504} \equiv 1 \pmod{10}$
$\therefore a = 1$

$3^{9018} \equiv 3^{2} \cdot 3^{9016} \equiv 9\cdot (3^{4})^{2254} \equiv 9 \cdot 1 \equiv 9 \pmod{10}$
$\therefore b = 9$
$\therefore a + b = 10$ .

Note : What I actually did was calculated the remainder when the number was divided by 10, which is actually the last digit.

$3 ^ { 2016 } \rightarrow 2016 \equiv 0 ( \mod 4 )$

$3 ^ { 9018 } \Rightarrow 9018 \equiv 2 ( \mod 4 )$

So, $3 ^ { n }$ is repeating by the pattern of last digits, 3, 9, 7, and 1, so the answer will be $a = 1, b = 9, \therefore a + b = 1 + 9 = \boxed { 10 }$