AMC12 2016 B

AMC12 2016 B · Q16

AMC12 2016 B · Q16. It mainly tests Arithmetic sequences basics, Primes & prime factorization.

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

345 可以用多少种方式表示为两个或更多个连续正整数的递增序列之和？

(A) 1 1

(B) 3 3

(D) 6 6

(E) 7 7

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): A sum of consecutive integers is equal to the number of integers in the sum multiplied by their median. Note that $345 = 3\cdot 5\cdot 23$. If there are an odd number of integers in the sum, then the median and the number of integers must be complementary factors of $345$. The only possibilities are $3$ integers with median $5\cdot 23=115$, $5$ integers with median $3\cdot 23=69$, $3\cdot 5=15$ integers with median $23$, and $23$ integers with median $3\cdot 5=15$. Having more integers in the sum would force some of the integers to be negative. If there are an even number of integers in the sum, say $2k$, then the median will be $\frac{j}{2}$, where $k$ and $j$ are complementary factors of $345$. The possibilities are $2$ integers with median $\frac{345}{2}$, $6$ integers with median $\frac{115}{2}$, and $10$ integers with median $\frac{69}{2}$. Again, having more integers in the sum would force some of the integers to be negative. This gives a total of $7$ solutions.

答案（E）：一串连续整数的和等于该和中整数的个数乘以它们的中位数。注意 $345=3\cdot 5\cdot 23$。如果和中整数个数为奇数，那么中位数与整数个数必须是 $345$ 的一对互补因子。唯一的可能性是：$3$ 个整数，中位数为 $5\cdot 23=115$；$5$ 个整数，中位数为 $3\cdot 23=69$；$3\cdot 5=15$ 个整数，中位数为 $23$；以及 $23$ 个整数，中位数为 $3\cdot 5=15$。若包含更多整数，就会迫使其中一些整数为负数。若和中整数个数为偶数，设为 $2k$，则中位数为 $\frac{j}{2}$，其中 $k$ 与 $j$ 是 $345$ 的一对互补因子。可能性为：$2$ 个整数，中位数为 $\frac{345}{2}$；$6$ 个整数，中位数为 $\frac{115}{2}$；$10$ 个整数，中位数为 $\frac{69}{2}$。同样，若包含更多整数，就会迫使其中一些整数为负数。因此共有 $7$ 种解法。

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