AMC10 2016 B

AMC10 2016 B · Q18

AMC10 2016 B · Q18. It mainly tests Arithmetic misc, 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构成的递增序列之和？

(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 $345=3\cdot5\cdot23$. 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 $5\cdot23=115$, $5$ integers with median $3\cdot23=69$, $3\cdot5=15$ integers with median $23$, and $23$ integers with median $3\cdot5=15$. 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串整数的个数乘以它们的中位数。注意 $345=3\cdot5\cdot23$。如果和中包含奇数个整数，那么中位数与整数个数必须是 $345$ 的互补因子。唯一的可能是：$3$ 个整数且中位数为 $5\cdot23=115$；$5$ 个整数且中位数为 $3\cdot23=69$；$3\cdot5=15$ 个整数且中位数为 $23$；以及 $23$ 个整数且中位数为 $3\cdot5=15$。若包含更多整数，则会迫使其中一些整数为负数。若和中包含偶数个整数，设为 $2k$，则中位数为 $\frac{j}{2}$，其中 $k$ 与 $j$ 是 $345$ 的互补因子。可能性为：$2$ 个整数且中位数为 $\frac{345}{2}$；$6$ 个整数且中位数为 $\frac{115}{2}$；$10$ 个整数且中位数为 $\frac{69}{2}$。同样，若包含更多整数，则会迫使其中一些整数为负数。因此共有 $7$ 种解。

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