jmc

By the Binomial Theorem, we have $$\begin{array}{rl}301^4& =(3(100)+1{)}^4\\ & =\left(\genfrac{}{}{0ex}{}{4}{0}\right)\cdot 3^4\cdot 100^4\cdot 1^0+\left(\genfrac{}{}{0ex}{}{4}{1}\right)\cdot 3^3\cdot 100^3\cdot 1^1\\ & +\left(\genfrac{}{}{0ex}{}{4}{2}\right)\cdot 3^2\cdot 100^2\cdot 1^2+\left(\genfrac{}{}{0ex}{}{4}{3}\right)\cdot 3^1\cdot 100^1\cdot 1^3\\ & +\left(\genfrac{}{}{0ex}{}{4}{4}\right)\cdot 3^0\cdot 100^0\cdot 1^4.\end{array}$$ All but the last two terms are divisible by $10000=100^2$, so we only have to consider the remainder when the last two terms are divided by 10,000. The last two terms are $$\begin{array}{rl}\left(\genfrac{}{}{0ex}{}{4}{3}\right)& \cdot 3^1\cdot 100^1\cdot 1^3+\left(\genfrac{}{}{0ex}{}{4}{4}\right)\cdot 3^0\cdot 100^0\cdot 1^4\\ & =4\cdot 3\cdot 100\cdot 1+1\cdot 1\cdot 1\cdot 1\\ & =1200+1=\boxed{1201}.\end{array}$$