2003 AMC 10B Problems/Problem 3

Problem

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?

$\textbf{(A) } 6 \qquad\textbf{(B) } 8 \qquad\textbf{(C) } 10 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 14$

Solution

It is a well-known fact that the sum of the first $n$ odd numbers is $n^2$. This makes the sum of the first $8$ odd numbers equal to $64$.

Let $n$ be equal to the smallest of the $5$ even integers. Then $(n+2)$ is the next highest, $(n+4)$ even higher, and so on.

This sets up the equation $n+(n+2)+(n+4)+(n+6)+(n+8)+4=64$

Now we solve:

\[5n+24=64\] \[5n=40\] \[n=8\]

Thus, the smallest integer is $\boxed{\textbf{(B) } 8}$.

See Also

2003 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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