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2003 AMC 10A Problem 1

2003 AMC 10A Problem 1AMC 10 Step by Step

Name: Date: September 13, 2026

Sequences & Series · difficulty 1 of 5 · about 1 min · position P1-10

What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers?

A)
$0$
B)
$1$
C)
$2$
D)
$2003$
E)
$4006$

Answer

Key insight

Pair the k-th even number 2k with the k-th odd number 2k-1; every pair differs by exactly 1, and there are 2003 pairs.

Solution

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 $\boxed{\textbf{(D)}\ 2003}$ .

Why this works

Subtracting two sums with the same number of terms can be done term by term, and here every term-wise difference is constant. Whenever two sequences are "shifted copies" of each other, compare corresponding terms instead of evaluating each sum separately.

Alternative approach

Arithmetic series: the evens sum to $2(1 + 2 + \cdots + 2003) = 2003 \cdot 2004$ , and the odds sum to $2003^2$ . The difference is $2003(2004 - 2003) = 2003$ .

The trap

Computing each sum with the arithmetic-series formula and slipping on the arithmetic, or answering 1 by looking at a single pair.

Common mistakes

Computing each sum with the arithmetic-series formula and slipping on the arithmetic, or answering 1 by looking at a single pair.
Starting the even numbers at $0$ ; the counting numbers begin at $1$ , so the first even counting number is $2$ (though the term-by-term difference would still be $1$ per pair).

Techniques

Set up the equation/formula and compute; no special trick needed · Exploit symmetry to reduce work or pair up objects