Simple modular arithmetic

From Euler's theorem, we get, $3^{\phi(100)}\equiv 3^{40} \equiv 1 \pmod {100}$ ........ $(1)$

Let, $3^{999} \equiv x \pmod {100}$

$\Rightarrow 3^{1000} \equiv 3x \pmod {100}$

$\Rightarrow (3^{40})^{25} \equiv 3x \pmod {100}$

$\Rightarrow (1)^{25} \equiv 3x \pmod {100}$ $[ From (1)]$

$\Rightarrow 1 \equiv 3x \pmod {100}$

$\Rightarrow -99 \equiv 3x \pmod {100}$

$\Rightarrow -33 \equiv x \pmod {100}$

$\Rightarrow 67 \equiv x \pmod {100}$

So, $x=\boxed{67}$

3^999 = (3^40)^24 * 3^39 , so 3^39 ≡ x mod (100)

Then, 3^40 ≡ 3x ≡ 1 ; 3x ≡ 1, trying with 201 we get x = 67

$3^{20}≡1(mod 100)$ & $3^{19}≡67(mod 100)$ So, $3^{999}=(3^{20})^{49}*3^{19}≡1^{49}*67(mod 100)$ ≡ $67(mod 100)$

$\begin{aligned} 3^{999} & \equiv 3\times 3^{2\times 499} \text{ (mod 100)} \\ & \equiv 3\times (10-1)^{499} \text{ (mod 100)} \\ & \equiv 3\times (\cdots + 499\times 10 -1) \text{ (mod 100)} \\ & \equiv 3\times 89 \equiv \boxed{67} \text{ (mod 100)} \end{aligned}$