# 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 arithmetic 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, $6x+15=2013\implies x=333\implies x+5=\boxed{\textbf{(B)}\ 338}$

## Solution 4

Since there are $6$ numbers, we divide $2013$ by $6$ to find the mean of the numbers. $\frac{2013}{6} = 335 \frac{1}{2}$. Then, $335 \frac{1}{2} + \frac{1}{2} = 336$ (the fourth number). Fifth: $337$; Sixth: $\boxed {338}$.

## Solution 5

Let the $6th$ number be $x$. Then our list is: $x-6+x-5+x-4+x-3+x-x-1=2013$. Simplifying this gets you $6x-21=2013\implies 6x=2034$, which means that $x = \boxed{\textbf{(B)}338}$

## Video Solution

https://youtu.be/E7DZ1oILqE0 ~savannahsolver

## Video Solution 2

https://youtu.be/AvOXHvvTlio Soo, DRMS, NM

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