AMC8 2016

AMC8 2016 · Q19

AMC8 2016 · Q19. It mainly tests Sequences & recursion (algebra), Fractions.

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

25 个连续偶数的和是 10,000。这些 25 个连续偶数中最大的是多少？

(A) 360 360

(B) 388 388

(D) 416 416

(E) 424 424

Answer

Correct choice: (E)

正确答案：（E）

Solution

The average of the 25 even integers is $10000/25=400$. So 12 consecutive even integers will be larger than 400 and 12 consecutive even integers will be smaller than 400. The sum $376+378+\cdots+398+400+402+\cdots+422+424=10000$. The largest of these numbers is $424$.

这25个偶整数的平均数是 $10000/25=400$。因此有12个连续偶数大于400，12个连续偶数小于400。于是 $376+378+\cdots+398+400+402+\cdots+422+424=10000$。其中最大的数是 $424$。

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