AMC 8 2023 Problem 25

AMC8 2023 年第 25 题

数列与规律★★★★☆
Fifteen integers a1,a2,…,a15a_1, a_2, \dots, a_{15} are arranged in order on a number line. The integers are equally spaced and have the property that
1≤a1≤10,13≤a2≤20,and241≤a15≤250.1 \le a_1 \le 10,\quad 13 \le a_2 \le 20,\quad \text{and}\quad 241 \le a_{15} \le 250.
What is the sum of the digits of a14a_{14}?
十五个整数 a1,a2,…,a15a_1,a_2,\dots,a_{15}依次排列在数轴上．这些整数呈等距排列，并且具有以下特性
1≤a1≤10,13≤a2≤20,241≤a15≤250.1 \le a_1 \le 10,\quad 13 \le a_2 \le 20,\quad 241 \le a_{15} \le 250.
问 a14a_{14} 的各位数字之和是多少？
  1. A.8
  2. B.9
  3. C.10
  4. D.11
  5. E.12

正确答案 · Correct Answer

A

解析 · Solution

解法1

常规解法 · 双向夹逼公差 设公差 dd。a15−a2=13d∈[241−20,250−13]=[221,237]⇒d∈{17,18}a_{15}-a_2=13d\in[241-20,250-13]=[221,237]\Rightarrow d\in\{17,18\}。a15−a1=14d∈[241−10,250−1]=[231,249]⇒d≤17.8⇒d=17a_{15}-a_1=14d\in[241-10,250-1]=[231,249]\Rightarrow d\le17.8\Rightarrow d=17。再定 a1a_1：a2=a1+17∈[13,20]⇒a1∈{1,2,3}a_2=a_1+17\in[13,20]\Rightarrow a_1\in\{1,2,3\}；a15=a1+238∈[241,250]⇒a1=3a_{15}=a_1+238\in[241,250]\Rightarrow a_1=3。故 a14=3+13⋅17=224a_{14}=3+13\cdot17=224，数字和 2+2+4=8。

Solution 1

Elementary · squeeze the common difference Let dd be the common difference. From a15−a2=13d∈[241−20,250−13]=[221,237]a_{15}-a_2=13d\in[241-20,250-13]=[221,237] we get d∈{17,18}d\in\{17,18\}; from a15−a1=14d∈[231,249]a_{15}-a_1=14d\in[231,249] we get d≤17.8d\le17.8, so d=17. Then a2=a1+17∈[13,20]a_2=a_1+17\in[13,20] gives a1∈{1,2,3}a_1\in\{1,2,3\}, and a15=a1+238∈[241,250]a_{15}=a_1+238\in[241,250] forces a1=3a_1=3. So a14=3+13⋅17=224a_{14}=3+13\cdot17=224, with digit sum 2+2+4=8.

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