Find the remainder

$\begin{aligned} 1989 \cdot 1990 \cdot 1991 + 1993^3 & \equiv 1\cdot 2\cdot 3 + 5^3 \text{ (mod 7)} \\ & \equiv 6 + 125 \text{ (mod 7)} \\ & \equiv 6 + 6 \text{ (mod 7)} \\ & \equiv \boxed{5} \text{ (mod 7)} \end{aligned}$

For $\text{1989 = 1 mod(7)}$ And for $\text{1990 = 2 mod(7)}$ and $\text{1991 = 3 mod(7)}$ , Then For $\text{1993 = 4 mod 7}$ or $1993^3= \text{64 mod(7)}$ = $\text{1 mod(7)}$ . Now Multiplying First 3 and adding with the fourth we get $1989·1990·1991+1993^3$ = $\text{7 mod (7)}$ = $\text{0 mod 7}$