2003 AMC 10A Problem 1

Adding up $2003$ numbers in each list would be slow, so compare the two lists side by side instead.

1. Write out the start of each list:

1. Look at each column. The even number is always exactly $1$ more than the odd number under it: $2 - 1 = 1$, $4 - 3 = 1$, $6 - 5 = 1$, and so on up to $4006 - 4005 = 1$.
2. So the even list is "the odd list, with $1$ added to every number." Every one of the $2003$ columns adds an extra $1$.
3. The total difference is $2003 \times 1 = 2003$.

The answer is $\boxed{\textbf{(D)}\ 2003}$.

When you subtract two sums that have the same number of terms, you can subtract them column by column instead of finding each total. Here every column gives the same small difference, so the answer is just (difference per column) times (number of columns). Look for this whenever two lists march along side by side.

Use known totals. The sum of the first $n$ odd numbers is $n^2$ (for example $1 + 3 + 5 = 9 = 3^2$). The sum of the first $n$ even numbers is twice $1 + 2 + \cdots + n$, which is $n(n+1)$ (for example $2 + 4 + 6 = 12 = 3 \cdot 4$). With $n = 2003$ the difference is
$$2003 \cdot 2004 - 2003 \cdot 2003 = 2003 \cdot (2004 - 2003) = 2003.$$

The $k$-th even counting number is $2k$ and the $k$-th odd counting number is $2k - 1$. Line the two lists up term by term:
$$(2 - 1) + (4 - 3) + (6 - 5) + \cdots + (4006 - 4005).$$
Each of the $2003$ parentheses equals $1$, so the difference of the two sums is $2003 \cdot 1 = 2003$.

The answer is $\boxed{\textbf{(D)}\ 2003}$.