AMC8 2023

AMC8 2023 · Q25

AMC8 2023 · Q25. It mainly tests Linear inequalities, Sequences & recursion (algebra).

Fifteen integers $a_1, a_2, a_3, \dots, a_{15}$ are arranged in order on a number line. The integers are equally spaced and have the property that \[1 \le a_1 \le 10, 13 \le a_2 \le 20, \text{ and } 241 \le a_{15}\le 250.\] What is the sum of digits of $a_{14}?$

十五个整数 $a_1, a_2, a_3, \dots, a_{15}$ 按顺序排列在数线上。这些整数等间距排列，且满足 \[1 \le a_1 \le 10, 13 \le a_2 \le 20, \text{ and } 241 \le a_{15}\le 250.\] $a_{14}$ 的数字和是多少？

(A) \ 8 \ 8

(B) \ 9 \ 9

(D) \ 11 \ 11

(E) \ 12 \ 12

Answer

Correct choice: (A)

正确答案：（A）

Solution

We can find the possible values of the common difference by finding the numbers which satisfy the conditions. To do this, we find the minimum of the last two: $241-20=221$, and the maximum: $250-13=237$. There is a difference of $13$ between them, so only $17$ and $18$ work, as $17\cdot13=221$, so $17$ satisfies $221\leq 13x\leq237$. The number $18$ is similarly found. $19$, however, is too much. Now, we check with the first and last equations using the same method. We know $241-10\leq 14x\leq250-1$. Therefore, $231\leq 14x\leq249$. We test both values we just got, and we can realize that $18$ is too large to satisfy this inequality. On the other hand, we can now find that the difference will be $17$, which satisfies this inequality. The last step is to find the first term. We know that the first term can only be from $1$ to $3$ since any larger value would render the second inequality invalid. Testing these three, we find that only $a_1=3$ will satisfy all the inequalities. Therefore, $a_{14}=13\cdot17+3=224$. The sum of the digits is therefore $\boxed{\textbf{(A)}\ 8}$.

通过寻找满足条件的公差来确定可能的公差值。针对后两个项，最小值为 $241-20=221$，最大值为 $250-13=237$。差值为 $13$，因此只有 $17$ 和 $18$ 可行，因为 $17\cdot13=221$，满足 $221\leq 13x\leq237$。类似地 $18$ 也满足，但 $19$ 太大。现在用前后两项检查。已知 $241-10\leq 14x\leq250-1$，即 $231\leq 14x\leq249$。测试后发现 $18$ 太大不满足，而 $17$ 满足。最后确定首项。由于首项只能为 $1$ 到 $3$（否则第二个不等式不成立），测试发现只有 $a_1=3$ 满足所有条件。因此 $a_{14}=13\cdot17+3=224$。数字和为 $2+2+4=8$，即 $\boxed{\textbf{(A)}\ 8}$。

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