Note that for any $x$ we have $5^{4x} \equiv 625 \pmod {10000}$. Hence $5^{1996} \equiv 625 \pmod {10000}$, meaning that $5^{1998} \equiv 625 \cdot 25 \equiv 5625 \pmod {10000}$. Multiplying and taking the last four digits gives $\boxed{3125}$.

jasperE3 wrote: Determine the last four decimal digits of the number $1997\cdot5^{1998}$. bruh this is 8th grade I think I'm gonna throw up

Instead of multiplying directly, do $1997 \cdot 5625 = 2000(225 \cdot 25) - 3(5625)$ and note that the first term is $0$ mod $10^4$, so our answer is $-3 \cdot 5625 \equiv 3125 \pmod{10^4}$