2002 AMC 10P Problems/Problem 2

Problem 2

The sum of eleven consecutive integers is $2002.$ What is the smallest of these integers?

$\text{(A) }175 \qquad \text{(B) }177 \qquad \text{(C) }179 \qquad \text{(D) }180 \qquad \text{(E) }181$

Solution 1

We can use the sum of an arithmetic series to solve this problem.

Let the first integer equal $a.$ The last integer in this string will be $a+10.$ Plugging in $n=11, a_1=a,$ and $a_n=a+10$ into $\frac{n(a_1 + a_n)}{2}=2002,$ we get:

\begin{align*} \frac{11(a + a+10)}{2}&=2002 \\ 11(2a+10)&=4004 \\ 2a+10&=364 \\ 2a&=354 \\ a&=177\\ \end{align*}

Thus, our answer is $\boxed{\textbf{(B) }177}$

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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.

Page

Toolbox

Search

2002 AMC 10P Problems/Problem 2

Problem 2

Solution 1

See Also