elementary number theory - Last 3 digits of $3^{999}$

I know that it's $3^{999} \mod 1000$ and since $\varphi(1000) = 400$ and $3^{400}\equiv1 \mod1000$ it will be equivalent to $3^{199} \mod 1000$ but what should I do from then? Or am I wrong about this from the start?

Answer

Using Carmichael function will be beneficial here as

$\displaystyle\lambda(1000)=100$

$$\implies 3^{100n}\equiv1^n\pmod{1000}\equiv1$$ for any integer $n$

As $(3,1000)=1,$ this implies $$3^{100n-1}\equiv3^{-1}$$

As $\displaystyle 999\equiv-1\pmod{1000}\implies3^{-1}\equiv-333\equiv1000-333$