Difference between revisions of "2003 AMC 10B Problems/Problem 3"

Revision as of 12:09, 4 July 2013

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 $\boxed{\textbf{(B) } 8}$ .

2003 AMC 10B (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 23: / Line 23: @@
 ==See Also==
 {{AMC10 box|year=2003|ab=B|num-b=2|num-a=4}}
+{{MAA Notice}}

Page

Toolbox

Search

Difference between revisions of "2003 AMC 10B Problems/Problem 3"

Revision as of 12:09, 4 July 2013

Problem

Solution

See Also