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

~AfterglowBlaziken

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https://youtu.be/qEq4JNouMNY

~Education, the Study of Everything

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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All AJHSME/AMC 8 Problems and Solutions

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

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