1997 AMC 12 第 20 题

先试着解答 1997 AMC 12 第 20 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1997 AMC 12 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

20.

下列哪个整数可以表示为 100100 个连续正整数之和?
Which one of the following integers can be expressed as the sum of 100100 consecutive positive integers?
1,627,384,9501{,}627{,}384{,}950
2,345,678,9102{,}345{,}678{,}910
3,579,111,3003{,}579{,}111{,}300
4,692,581,4704{,}692{,}581{,}470
5,815,937,2605{,}815{,}937{,}260
答案:A
知识点:arithmetic sequences整除性
小提示:
将这些数写成 a+1,a+2,…,a+100a+1,a+2,\ldots,a+100
Write the terms as a+1,a+2,…,a+100a+1,a+2,\ldots,a+100
大提示:
它们的和模 100100 必须与 5050 同余
Their sum must be congruent to 5050 modulo 100100
解答:
它们的和为 100a+(1+⋯+100)100a+(1+\cdots+100) =100a+5050=100a+5050,所以末两位是 5050。只有 1,627,384,9501,627,384,950 具有这一性质,并且对应的 a=1,627,384,950−5050100a=\frac{1,627,384,950-5050}{100} 是正整数。因此正确答案是 A。
The sum is 100a+(1+⋯+100)100a+(1+\cdots+100) =100a+5050,=100a+5050, so it ends in 50.50. Only 1,627,384,9501,627,384,950 has that property, and it gives the positive integer a=1,627,384,950−5050100.a=\frac{1,627,384,950-5050}{100}. Thus A is correct.
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