jmc

Note that $400=4\cdot 10^2=2^2\cdot 10^2=2^4\cdot 5^2$. Therefore, $\frac{141}{400}=\frac{141}{2^4\cdot 5^2}$. If we multiply this fraction by $10^4$, we shift all digits $4$ places to the left, so $\frac{141}{2^4\cdot 5^2}\cdot 10^4=141\cdot 5^2=3525$. The last nonzero digit is therefore $\boxed{5}$.