A number theory problem by Snehashis Mukherjee

$1000 = 2^{3} \cdot 5^{3}$
$\phi(1000) = 400$ where $\phi$ is the Euler Phi function.
$\therefore 7^{400} \equiv 1 \pmod{1000}$
$\therefore 7^{10000} \equiv 1 \pmod{1000}$
$\therefore 7 \cdot 7^{9999} \equiv 1 \pmod{1000}$
Let $7^{9999} = x$
$7x \equiv 1 \pmod{1000}$
Multiplying by $143$
$1001x \equiv 143 \pmod{1000}$
$\therefore x \equiv 143 \pmod{1000}$
Thus, the last 3 digits are $143$