2016 AMC 8 Problems/Problem 19

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

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

Solution

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 simply by $25n$ 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 $400+2 \cdot (15-13)=424$ .

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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2016 AMC 8 Problems/Problem 19

Solution