1970 AMC 12 Problem 33

First sum the digits from $0000$ through $9999$

In each of the four positions, every digit appears exactly $1000$ times

Write the integers from $0$ through $9999$ using four digits with leading zeros. In each of the four positions, every digit $0,$ $1,$ $\ldots,$ $9$ appears $1000$ times. Thus their total digit sum is $$\begin{gathered} 4\cdot1000(0+1+\cdots+9)\\ =4000\cdot45\\ =180000. \end{gathered}$$ Replacing $0$ by $10000$ adds $1,$ so the requested sum is $180001.$

Therefore, the correct answer is A.