Question

有多少种方式可将 $345$ 写成两个或更多连续正整数的递增序列之和？

In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?

$1$

$3$

$5$

$6$

$7$

答案：E

知识点：等差数列因数个数分类讨论

难度评级：1800

小提示：

一串 $k$ 个连续整数的和等于 $k$ 乘以中位数，且 $345=3\cdot5\cdot23$ 。

A run of $k$ consecutive integers equals $k$ times its median, and $345=3\cdot5\cdot23$

大提示：

当 $k$ 为奇数时，中位数为整数因数；当 $k$ 为偶数时，中位数为半整数。两类都要计数，并保证所有项为正。

For odd $k,$ the median is an integer factor; for even $k,$ the median is a half-integer. Count both while keeping all terms positive

解答：

连续整数之和等于项数乘以中位数。若项数为奇数，中位数是 $345$ 的整数因数，得到项数为 $3$ （中位数 $115$ ）、 $5$ （中位数 $69$ ）、 $15$ （中位数 $23$ ）和 $23$ （中位数 $15$ ）的序列。若项数为偶数，中位数为半整数；符合要求的项数为 $2,6,10$ ，对应的中位数依次为 $172.5,57.5,34.5$ 。更长的序列会含有非正项。因此共有 $4+3=7$ 种。

所以正确答案是 E。

A sum of consecutive integers equals the count times the median. For an odd number of terms, the median is an integer divisor of $345,$ giving runs of $3$ (median $115$ ), $5$ (median $69$ ), $15$ (median $23$ ), and $23$ (median $15$ ) terms. For an even number of terms the median is a half-integer. The positive possibilities have lengths $2,6,10,$ with respective medians $172.5,57.5,34.5.$ Longer divisor-based runs would force a nonpositive first term. This gives $4+3=7$ ways.

Thus, the correct answer is E.

2016 AMC 12B 第 16 题

答案：E

难度评级：1800

小提示：

大提示：

解答：

其他年份的第 16 题