AMC8 2016 8, Problem 19

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

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 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 $400+2(25-13)=\fbox{424}$.

Answer: 424