Find the last three digits of $383^{101}$ We have to find out $383^{101} \equiv ? \pmod {1000}$. 
I know that $383^2 ≡ 689 \pmod {1000}$
$383^5≡143 \pmod {1000}$
I know that $ϕ(1000)=400 >101 $  from Euler.
It definitely can't help me.
I don't know how to continue.
I can't use the Chinese Remainder Theorem.
 A: Answer: $\boxed{383}$. Let's solve it:
$\phi(125)=100$ and $\phi(8)=4$. Least common multiple of $100$ and $4$ is $100$. 
$$383^{100} \equiv 1 \pmod{125}$$ and $$383^{4} \equiv 1 \pmod{8}$$ Therefore $383^{100} \equiv 1 \pmod{1000}$. Hence we yields 
$$383^{101} \equiv 383 \pmod{1000} $$
A: I don't know what the expected approach to this problem is, but using $101 = 50 + 51$, $50 = 25 + 25$, $51 = 50 + 1$, $25 = 24 + 1$, $24 = 12 + 12$, $12 = 6 + 6$, $6 = 3 + 3$, $3 = 2 + 1$, you should be able to solve it by hand in a few minutes.
A: The units digit of $383^1$ is $3$.
The units digit of $383^2$ is $9$ ($3 \times 3 = 9$).
The units digit of $383^3$ is $7$ ($3 \times 9 = 27$).
The units digit of $383^4$ is $1$ ($3 \times 7 = 21$).
Calculating,
$\; 383^4 \equiv 721 \pmod{1000}$
Calculating,
$\; 383^{101} \equiv 383 \cdot 721^{25} \equiv 383 \cdot (1 + 720)^{25}  \equiv 383 \cdot (1 + 2^4 \cdot 3^2 \cdot 5^1)^{25} \equiv$
$\quad\quad\quad\quad  \; 383 \; \bigr(1 + \binom{25}{1}\,2^4 \cdot 3^2 \cdot 5^1  + \binom{25}{2}\,2^8 \cdot 3^4 \cdot 5^2  + 0\bigr) \equiv$
$\quad\quad\quad\quad  \; 383 \; \bigr(1 + 0 + 0 + 0 \bigr) \equiv 383 \pmod{1000}$
