AMC10 2003 B

AMC10 2003 B · Q3

AMC10 2003 B · Q3. It mainly tests Arithmetic sequences basics, Arithmetic misc.

The sum of 5 consecutive even integers is 4 less than the sum of the first 8 consecutive odd counting numbers. What is the smallest of the even integers?

5 个连续偶整数的和比前 8 个连续奇数的和少 4。求这些偶整数中最小的那个。

(A) 6 6

(B) 8 8

(D) 12 12

(E) 14 14

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let $n$ be the smallest of the even integers. Since $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64$, we have $60 = n + (n + 2) + (n + 4) + (n + 6) + (n + 8) = 5n + 20$, so $n = 8$.

设 $n$ 是最小的偶整数。因为 $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64$，所以 $60 = n + (n + 2) + (n + 4) + (n + 6) + (n + 8) = 5n + 20$，因此 $n = 8$。

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