Question

十五个整数 $a_1, a_2, a_3, \cdots, a_{15}$ 按顺序排列在数轴上。 $1 \leq a_1 \leq 10,$ $13 \leq a_2 \leq 20,$ $241 \leq a_{15} \leq 250.$

这些整数等距排列。 $a_{14}$ 的数字和是多少？

Fifteen integers $a_1, a_2, a_3, \cdots, a_{15}$ are arranged in order on a number line. The integers are equally spaced and have the property that $1 \leq a_1 \leq 10,$ $13 \leq a_2 \leq 20,$ and $241 \leq a_{15} \leq 250.$

What is the sum of the digits of $a_{14}?$

$8$

$9$

$10$

$11$

$12$

答案：A

知识点：等差数列极限情形界定数字

难度评级：1950

视频讲解：

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文字解答：

设公差为 $d$ 。若取 $a_1 = 10$ 、 $a_{15} = 241$ ，则因此整数公差 $d$ 至少为 $17$ 。 $d \geq \dfrac{241 - 10}{14} = 16.5.$

另一方面，若取 $a_1 = 1$ 、 $a_{15} = 250$ ，则因为公差是整数，所以 $d$ 至多为 $17$ 。因此 $d$ 是 $17$ 。 $d \leq \dfrac{250 - 1}{14} \approx 17.8.$

注意 $17 \cdot 14 = 238$ 。为了让 $a_{15}$ 落在给定范围内， $a_1$ 至少为 $3$ 。

如果 $a_1$ 大于 $3$ ，则 $a_2$ 会大于 $20$ ，不符合条件。

所以 $a_1 = 3$ ，且 $d = 17$ 。于是 $a_{14} = 3 + 13 \cdot 17 = 224.$

数字和为 $2 + 2 + 4 = 8.$

所以正确答案是 A。

Let $d$ be the common difference. If we let $a_1 = 10$ and $a_{15} = 241,$ we see that $d \geq \dfrac{241 - 10}{14} = 16.5.$ Since all the numbers are integers, $d$ must be at least $17.$

Also, if $a_1 = 1$ and $a_{15} = 250,$ we get that $d \leq \dfrac{250 - 1}{14} \approx 17.8.$ Once again since all the numbers are integers, $d$ is at most $17.$ This tells us that $d$ is $17.$

Note that $17 \cdot 14 = 238.$ This means that $a_1$ must be at least $3$ for $a_{15}$ to be within the desired range.

If $a_1$ is greater than $3,$ however, $a_2$ becomes greater than $20,$ which is not allowed.

Now we know that $a_1 = 3$ and $d = 17.$ This tells us that $a_{14} = 3 + 13 \cdot 17 = 224.$

Therefore, sum of the digits is $2 + 2 + 4 = 8.$

Thus, A is the correct answer.

2023 AMC 8 第 25 题

答案：A

难度评级：1950

视频讲解：

文字解答：

其他年份的第 25 题