2003 AMC 10B Problem 3

Question

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?

6

8

10

12

14

The first

8

odd counting numbers sum to

8^2

Write the even integers as

n,

n+2,

n+4,

n+6,

n+8

and set their sum equal to

64-4

The first

8

odd counting numbers sum to

1+3+\cdots+15=64.

Letting

n

be the smallest even integer,

\begin{gathered} n+(n+2)+(n+4) \\ {}+(n+6)+(n+8) \\ = 5n+20 = 60, \end{gathered}

so

n=8.

Thus, the correct answer is B.

Problem 3 in other years