2025 AMC 8 第 16 题

先试着解答 2025 AMC 8 第 16 题,然后核对你的答案与视频与精心整理的解答,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 2025 AMC 8 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。

16.

从 11 到 1010 中选出五个不同整数,并从 1111 到 2020 中选出五个不同整数。没有两个选出的数相差正好 1010。这十个选出的数字之和是多少?
Five distinct integers from 11 to 1010 are chosen, and five distinct integers from 1111 to 2020 are chosen. No two numbers differ by exactly 10.10. What is the sum of the ten chosen numbers?
9595
100100
105105
110110
115115
答案:C
知识点:双射配对与分组
小提示:
把 11 到 1010 中每个数与大 1010 的数配对
Pair each number from 11 to 1010 with the number 1010 larger
大提示:
被选中的高位区间数字对应于低位区间中未被选中的数字
The chosen upper numbers correspond to the unchosen lower numbers
视频讲解:
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文字解答:
把 11 到 1010 的整数称为低位区间,把 1111 到 2020 的整数称为高位区间。
从低位区间选出的 55 个不同数字,会排除高位区间中比它们正好大 1010 的数。高位区间只有 1010 个数,所以还剩 10−5=510 - 5 = 5 个没有被排除。
我们必须从高位区间选 55 个不同数字,所以选中的高位区间数字正好就是未被排除的那些。它们各自比低位区间中一个未选数字大 1010。
因此,选中的 55 个高位区间数字之和,比低位区间中 55 个未选数字之和正好多 5×10=505 \times 10 = 50。
所以这 1010 个选中数字的总和等于 5050,再加上低位区间中选中和未选数字的总和 1+2+⋯+101 + 2 + \dots + 10。
从 11 到 1010 的和是 10(10+1)2=55\frac{10 (10+1)}{2} = 55,所以答案是 50+55=10550 + 55 = 105,选 C。
Call the integers from 11 to 1010 inclusive the lower range, and call the integers from 1111 to 2020 inclusive the higher range.
Each of the 55 distinct numbers chosen from the lower range blocks out the number in the higher range that is exactly 1010 more than itself. There are only 1010 numbers in the higher range, so there are only 10−5=510 - 5 = 5 numbers not yet blocked.
We need to choose 55 distinct numbers from the higher range, so the numbers chosen from the higher range are precisely those which are not yet blocked. They are each exactly 1010 more than a not-chosen number in the lower range.
So, the sum of the 55 distinct numbers chosen from the higher range is exactly 5×10=505 \times 10 = 50 more than the sum of the 55 not-chosen numbers in the lower range.
The sum of all 1010 chosen numbers is therefore equal to 5050 plus the sum of all chosen and not-chosen numbers in the lower range 1+2+⋯+10.1 + 2 + \dots + 10.
The sum of the numbers from 11 to 1010 is 10(10+1)2=55,\frac{10 (10+1)}{2} = 55, so the answer is 50+55=105,50 + 55 = 105, or choice C.
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