# 2016 AMC 8 Problems/Problem 19

## Problem

The sum of $25$ consecutive even integers is $10,000$. What is the largest of these $25$ consecutive integers?

$\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424$

## Solution 1

Let $n$ be the 13th consecutive even integer that's being added up. By doing this, we can see that the sum of all 25 even numbers will simplify to $25n$ since $(n-2k)+\dots+(n-4)+(n-2)+(n)+(n+2)+(n+4)+ \dots +(n+2k)=25n$. Now, $25n=10000 \rightarrow n=400$. Remembering that this is the 13th integer, we wish to find the 25th, which is $400+2(25-13)=\boxed{\textbf{(E)}\ 424}$.

## Solution 2

Let $x$ be the smallest number. The equation will become, $x+(x+2)+(x+4)+\cdots +(x+48)=10,000$. After you combine like terms, you get $25x+(50*12)=10,000$ which turns into $10,000-600=25x$. $25x=9400$, so $x=376$. Then, you add $376+48 = \boxed{\textbf{(E)}\ 424}$.

~AfterglowBlaziken

~savannahsolver