# 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 largest number. Then, $x+(x-2)+(x-4)+\cdots +(x-48)=10000$. Factoring this gives $2\left(\frac{x}{2} + \left(\frac{x}{2} - 1\right) + \left(\frac{x}{2} - 2\right) +\cdots + \left(\frac{x}{2} - 24\right)\right)=10,000$. Grouping like terms gives $25\left(\frac{x}{2}\right) - 300=5000$, and continuing down the line, we find $x=\boxed{\textbf{(E)}\ 424}$.

~MrThinker

## Solution 3

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 $376+48 = \boxed{\textbf{(E)}\ 424}$.

Note: After we combine like terms, you would have an arithmetic sequence from $2$ to $48$ (because $24 \cdot 2 = 48$ to get last term), which would look like $2 + 4 + 6...+46 + 48$ To calculate the sum of the numbers we can use the formula $S_n = n(\frac{a_1 + a_n}{2})$. This simplifies to $24 \cdot 25$, giving us $600$, which is what AfterglowBlaziken did.

~AfterglowBlaziken ~ Note by probab2023

## Solution 4

Dividing the series by $2$, we get that the sum of $25$ consecutive integers is $5000$. Let the middle number be $k$ we know that the sum is $25k$, so $25k=5000$. Solving, $k=200$. $2k=400$ is the middle term of the original sequence, so the original last term is $400+\frac{25-1}{2}\cdot 2=424$. So the answer is $\boxed{\textbf{(E)}\ 424}$.

~vadava_lx

## Video Solution (CREATIVE THINKING + ANALYSIS!!!)

~Education, the Study of Everything

~savannahsolver

## See Also

 2016 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 18 Followed byProblem 20 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 AJHSME/AMC 8 Problems and Solutions

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