AMC12 2019 A

AMC12 2019 A · Q4

AMC12 2019 A · Q4. It mainly tests Sequences & recursion (algebra), GCD & LCM.

What is the greatest number of consecutive integers whose sum is 45?

和为 45 的连续整数最多有几个？

(A) 9 9

(B) 25 25

(D) 90 90

(E) 120 120

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): The sum $(-44)+(-43)+\cdots+43+44+45=45$. This sum has 90 consecutive integers. There is no longer list because for the sum of consecutive integers to be positive, there must be more positive integers than negative integers. Further, if there are more than 90 consecutive integers as part of a list that sums to a positive number, then there must be a positive integer greater than 45 that is not cancelled out by its additive inverse.

答案（D）：和为$(-44)+(-43)+\cdots+43+44+45=45$。这个和包含90个连续整数。不存在更长的列表，因为要使连续整数之和为正，正整数的个数必须多于负整数的个数。进一步，如果某个连续整数列表包含超过90个整数且其和为正，那么必然存在一个大于45的正整数没有被它的加法逆元抵消。

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