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2003 AMC 10B Problem 3

2003 AMC 10B Problem 3AMC 10 Step by Step

Name: Date: September 13, 2026

Sequences & Series · difficulty 1 of 5 · about 2 min · position P1-10

The sum of $5$ consecutive even integers is $4$ less than the sum of the first $8$ consecutive odd counting numbers. What is the smallest of the even integers?

A)
$6$
B)
$8$
C)
$10$
D)
$12$
E)
$14$

Answer

Key insight

The first 8 odd numbers sum to 64, so the five even integers sum to 60; their middle term is the average 12, so the smallest is 8.

Solution

The first $8$ odd counting numbers are $1, 3, 5, \dots, 15$ . Their sum is $8^2 = 64$ (or, pairing first with last, $8 \cdot \frac{1 + 15}{2} = 64$ ).

So the five consecutive even integers add up to $64 - 4 = 60$ . Five equally spaced numbers have average equal to the middle one, so the middle even integer is $\frac{60}{5} = 12$ . The five integers are $8, 10, 12, 14, 16$ , and the smallest is $8$ .

Check: $8 + 10 + 12 + 14 + 16 = 60$ .

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

Why this works

Two standard facts do all the work: $1 + 3 + \cdots + (2n-1) = n^2$ , and the sum of an odd number of consecutive terms of an arithmetic sequence is the count times the middle term. Using the middle term as the unknown avoids the algebra of $n + (n+2) + \cdots + (n+8)$ entirely.

Alternative approach

Let the smallest be $n$ . Then $n + (n+2) + (n+4) + (n+6) + (n+8) = 5n + 20 = 60$ , so $n = 8$ .

The trap

Solving for the middle term 12 (or the sum 60) and forgetting the question asks for the smallest of the five.

Common mistakes

Solving for the middle term $12$ (or the sum $60$ ) and forgetting the question asks for the smallest of the five.
Taking the first $8$ odd numbers to end at $17$ or computing their sum as $8 \cdot 8 = 64$ by luck but then adding $4$ instead of subtracting.

Techniques

Set up the equation/formula and compute; no special trick needed