Question

从 $1$ 到 $10$ 中选出五个不同整数，并从 $11$ 到 $20$ 中选出五个不同整数。没有两个选出的数相差正好 $10$ 。这十个选出的数字之和是多少？

Five distinct integers from $1$ to $10$ are chosen, and five distinct integers from $11$ to $20$ are chosen. No two numbers differ by exactly $10.$ What is the sum of the ten chosen numbers?

$95$

$100$

$105$

$110$

$115$

答案：C

知识点：双射配对与分组

难度评级：1650

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文字解答：

把 $1$ 到 $10$ 的整数称为低位区间，把 $11$ 到 $20$ 的整数称为高位区间。

从低位区间选出的 $5$ 个不同数字，会排除高位区间中比它们正好大 $10$ 的数。高位区间只有 $10$ 个数，所以还剩 $10 - 5 = 5$ 个没有被排除。

我们必须从高位区间选 $5$ 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 $10$ 。

因此，选中的 $5$ 个高位区间数字之和，比低位区间中 $5$ 个未选数字之和正好多 $5 \times 10 = 50$ 。

所以这 $10$ 个选中数字的总和等于 $50$ ，再加上低位区间中选中和未选数字的总和 $1 + 2 + \dots + 10$ 。

从 $1$ 到 $10$ 的和是 $\frac{10 (10+1)}{2} = 55$ ，所以答案是 $50 + 55 = 105$ ，选 C。

Call the integers from $1$ to $10$ inclusive the lower range, and call the integers from $11$ to $20$ inclusive the higher range.

Each of the $5$ distinct numbers chosen from the lower range blocks out the number in the higher range that is exactly $10$ more than itself. There are only $10$ numbers in the higher range, so there are only $10 - 5 = 5$ numbers not yet blocked.

We need to choose $5$ 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 $10$ more than a not-chosen number in the lower range.

So, the sum of the $5$ distinct numbers chosen from the higher range is exactly $5 \times 10 = 50$ more than the sum of the $5$ not-chosen numbers in the lower range.

The sum of all $10$ chosen numbers is therefore equal to $50$ plus the sum of all chosen and not-chosen numbers in the lower range $1 + 2 + \dots + 10.$

The sum of the numbers from $1$ to $10$ is $\frac{10 (10+1)}{2} = 55,$ so the answer is $50 + 55 = 105,$ or choice C.

2025 AMC 8 第 16 题

答案：C

难度评级：1650

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