2001 AMC 10 Problem 13

1. Count the letters: $A$ through $J$ is ten letters, and they are all different digits. There are only ten digits, $0$ to $9$, so every digit is used exactly once. So $A$, $B$, $C$ are whatever three digits are left after choosing $DEF$ and $GHIJ$.
2. Start with the part that has the fewest choices. $GHIJ$ is four odd digits in a row, going down. The odd digits are $9, 7, 5, 3, 1$, so $GHIJ$ is $9753$ or $7531$.
3. $DEF$ is three even digits in a row, going down. The even digits are $8, 6, 4, 2, 0$, so $DEF$ is $864$, $642$ or $420$.
4. That is only $2 \times 3 = 6$ cases. For each, list the three digits left over and add them. We need a sum of $9$.

1. Only one row works: the left-over digits are $0, 1, 8$. In decreasing order, $ABC = 810$.
2. The full number is $810\text{-}642\text{-}9753$, and $A = 8$.

The answer is $\boxed{\textbf{(E)}\ 8}$.

The key fact is hidden in "each letter represents a different digit": with ten letters, all ten digits appear. That turns the problem around. Instead of searching for $A$, $B$, $C$ directly, you pick the parts with very few options (the runs of odd and even digits), and $A$, $B$, $C$ are forced. When a problem uses every item exactly once, think about what is left over.