# Difference between revisions of "2013 AMC 8 Problems/Problem 17"

## Problem

The sum of six consecutive positive integers is 2013. What is the largest of these six integers?

$\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350$

## Solution 1

The mean of these numbers is $\frac{\frac{2013}{3}}{2}=\frac{671}{2}=335.5$. Therefore the numbers are $333, 334, 335, 336, 337, 338$, so the answer is $\boxed{\textbf{(B)}\ 338}$

## Solution 2

Let the $4^{\text{th}}$ number be $x$. Then our desired number is $x+2$.

Our integers are $x-3,x-2,x-1,x,x+1,x+2$, so we have that $6x-3 = 2013 \implies x = \frac{2016}{6} = 336 \implies x+2 = \boxed{\textbf{(B)}\ 338}$.

## Solution 3

Let the first term be $x$. Our integers are $x,x+1,x+2,x+3,x+4,x+5$. We have, $6x+15=2013\implies x=333\implies x+5=\boxed{\textbf{(B)}\ 338}$