AMC12 2006 A

AMC12 2006 A · Q8

AMC12 2006 A · Q8. It mainly tests Arithmetic sequences basics, Basic counting (rules of product/sum).

How many sets of two or more consecutive positive integers have a sum of $15$?

有多少组由两个或更多连续正整数组成的数列，其和为 $15$？

(A) 1 1

(B) 2 2

(D) 4 4

(E) 5 5

Answer

Correct choice: (C)

正确答案：（C）

Solution

Notice that if the consecutive positive integers have a sum of $15$, then their average (which could be a fraction) must be a divisor of $15$. If the number of integers in the list is odd, then the average must be either $1, 3,$ or $5$, and $1$ is clearly not possible. The other two possibilities both work: - $1 + 2 + 3 + 4 + 5 = 15$ - $4 + 5 + 6 = 15$ If the number of integers in the list is even, then the average will have a $\frac{1}{2}$. The only possibility is $\frac{15}{2}$, from which we get: - $15 = 7 + 8$ Thus, the correct answer is $\boxed{\textbf{(C) }3}.$

注意到若若干个连续正整数的和为 $15$，则它们的平均数（可能是分数）必须是 $15$ 的一个因数。若数列中整数个数为奇数，则平均数必须是 $1, 3,$ 或 $5$，而 $1$ 显然不可能。其余两种情况都可行： - $1 + 2 + 3 + 4 + 5 = 15$ - $4 + 5 + 6 = 15$ 若数列中整数个数为偶数，则平均数将带有 $\frac{1}{2}$。唯一可能是 $\frac{15}{2}$，从而得到： - $15 = 7 + 8$ 因此正确答案是 $\boxed{\textbf{(C) }3}.$

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