Question

正整数数列 $1, a_2, a_3, \ldots$ 与 $1, b_2, b_3, \ldots$ 分别是递增等差数列与递增等比数列。令 $c_n = a_n + b_n$ 。存在一个整数 $k$ ，使得 $c_{k-1} = 100$ 且 $c_{k+1} = 1000$ 。求 $c_k$ 。

The sequences of positive integers $1, a_2, a_3, \ldots$ and $1, b_2, b_3, \ldots$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n = a_n + b_n.$ There is an integer $k$ such that $c_{k-1} = 100$ and $c_{k+1} = 1000.$ Find $c_k.$

答案

答案：262

知识点：等差数列等比数列模运算分类讨论

难度评级：2920

解答：

写 $a_n = 1 + (n-1)d$ 与 $b_n = r^{n-1}$ ，其中整数 $d \ge 1$ 、 $r \ge 2$ 。因为 $c_1 = 2 \lt 100$ ，所以 $k \ge 3$ ，两个条件变为 $(k-2)d + r^{k-2} = 99,$ $kd + r^k = 999.$

相减得 $2d + r^{k-3}(r-1)r(r+1) = 900$ 。三个连续整数的乘积能被 $3$ 整除，所以 $3 \mid 2d$ ，于是 $3 \mid d$ 。再由 $(k-2)d + r^{k-2} = 99$ 可知 $3 \mid r^{k-2}$ ，所以 $3 \mid r$ 。界限 $r^{k-2} \le 98$ 与 $r^k \le 998$ 只留下 $(r, k) = (3, 3)$ 、 $(3, 4)$ 、 $(3, 5)$ 、 $(3, 6)$ 、 $(6, 3)$ 、 $(9, 3)$ 。

逐一代入 $(k-2)d = 99 - r^{k-2}$ 与 $kd = 999 - r^k$ ，只有 $(r, k) = (9, 3)$ 给出一致的值 $d = 90$ 。于是 $c_3 = 1 + 2 \cdot 90 + 9^2 = 262$ 。

Write $a_n = 1 + (n-1)d$ and $b_n = r^{n-1}$ with integers $d \ge 1$ and $r \ge 2.$ Since $c_1 = 2 \lt 100,$ we have $k \ge 3,$ and the two conditions read $(k-2)d + r^{k-2} = 99,$ $kd + r^k = 999.$

Subtracting, $2d + r^{k-3}(r-1)r(r+1) = 900.$ The product of three consecutive integers is divisible by $3,$ so $3 \mid 2d,$ hence $3 \mid d.$ Then $(k-2)d + r^{k-2} = 99$ forces $3 \mid r^{k-2},$ so $3 \mid r.$ The bounds $r^{k-2} \le 98$ and $r^k \le 998$ leave only $(r, k) = (3, 3),$ $(3, 4),$ $(3, 5),$ $(3, 6),$ $(6, 3),$ $(9, 3).$

Testing each against $(k-2)d = 99 - r^{k-2}$ and $kd = 999 - r^k,$ only $(r, k) = (9, 3)$ gives a consistent value, $d = 90.$ Then $c_3 = 1 + 2 \cdot 90 + 9^2 = 262.$

2016 AIME II 第 9 题

答案：262

难度评级：2920

解答：

其他年份的第 9 题