2016 AMC 8

Complete problem set with solutions and individual problem pages

Problem 19 Hard

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

  • A.

    360

  • B.

    388

  • C.

    412

  • D.

    416

  • E.

    424

Answer:E

Solution 1

Let n be the 13~\text{th} 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 13~\text{th} integer, we wish to find the 25~\text{th}, which is 400+2(25-13)=\boxed{\textbf{(E)}\ 424}.

 

Solution 2

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}.

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