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, $x+(x-2)+(x-4)+\cdots +(x-50)=10000$ . Factoring this gives $2\left(\frac{x}{2} + \left(\frac{x}{2} - 1\right) + \left(\frac{x}{2} - 2\right) +\cdots + \left(\frac{x}{2} - 25\right)\right)=10000$ . 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

Video Solution

https://youtu.be/NHdtjvRcDD0

~savannahsolver

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2016 AMC 8 Problems/Problem 19

Contents

Problem

Solution 1

Solution 2

Video Solution

See Also