2016 AMC 10B 第 18 题

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18.

345345 可以用多少种方式写成两个或更多个连续正整数的递增序列之和?

In how many ways can 345345 be written as the sum of an increasing sequence of two or more consecutive positive integers?

 1\ 1

 3\ 3

 5\ 5

 6\ 6

 7\ 7

答案:E
知识点:等差数列因数个数
难度评级:1660
解答:

设序列长度为 ss,首项为 xx345=s(x+s12),345=s\left(x+\frac{s-1}{2}\right), s(2x+s1)=690.s(2x+s-1)=690.

因此 ss 必须是 690690 的因子,而且 x=690/ss+12x=\frac{690/s-s+1}{2} 必须是正整数。检查可能的因子长度,得到 只有这些长度能使 xx 为正整数。 s=2,3,5,6,10,15,23.s=2,3,5,6,10,15,23.

因此共有 77 种表示。

所以正确答案是 E

Suppose the sequence has length ss and first term xx. Then 345=s(x+s12),345=s\left(x+\frac{s-1}{2}\right), or s(2x+s1)=690.s(2x+s-1)=690.

Thus ss must be a divisor of 690690, with x=690/ss+12x=\frac{690/s-s+1}{2} a positive integer. Checking the possible divisor lengths gives s=2,3,5,6,10,15,23.s=2,3,5,6,10,15,23. These are the only lengths that keep xx positive and integral.

Therefore there are 77 representations.

Thus, the correct answer is E.

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