jmc

We can write $9^{105}=(10-1{)}^{105}.$ Then by the Binomial Theorem, $$(10-1{)}^{105}=10^{105}-(\genfrac{}{}{0ex}{}{105}{1})10^{104}+(\genfrac{}{}{0ex}{}{105}{2})10^{103}-\cdots +(\genfrac{}{}{0ex}{}{105}{102})10^3-(\genfrac{}{}{0ex}{}{105}{103})10^2+(\genfrac{}{}{0ex}{}{105}{104})10-1.$$All the terms up to $\left(\genfrac{}{}{0ex}{}{105}{102}\right)10^3$ are divisible by $10^3,$ so for the purpose of finding the last three digits, we can ignore them. We are left with $$\begin{array}{rl}-\left(\genfrac{}{}{0ex}{}{105}{103}\right)10^2+\left(\genfrac{}{}{0ex}{}{105}{104}\right)10-1& =-\left(\genfrac{}{}{0ex}{}{105}{2}\right)10^2+\left(\genfrac{}{}{0ex}{}{105}{1}\right)10-1\\ & =-\frac{105\cdot 104}{2}\cdot 10^2+105\cdot 10-1\\ & =-546000+1050-1\\ & =-546000+1049.\end{array}$$Hence, the last three digits are $\boxed{049}.$