Question

一个等差数列有 $n \gt 3$ 项，首项为 $a$ ，公差为 $d \gt 1$ 。Carl 正确写下了这个数列中的所有项，除了有一项相差了 $1$ 。他写下的各项之和为 $222$ 。求 $a + d + n$ 。

An arithmetic sequence of positive integers has $n \gt 3$ terms, initial term $a,$ and common difference $d \gt 1.$ Carl wrote down all the terms in this sequence correctly except for one term, which was off by $1.$ The sum of the terms he wrote down was $222.$ What is $a + d + n?$

$24$

$20$

$22$

$28$

$26$

答案：B

知识点：等差数列丢番图方程质因数分解

难度评级：2380

解答：

真实和为 $S = na + \frac{n(n-1)}{2}d$ 。因为一项相差 $1$ ，写下的总和满足 $222 = S \pm 1$ ，所以 $S = 221$ 或 $223$ 。又因为 $2S = n\bigl(2a + (n-1)d\bigr)$ ，所以 $n$ 整除 $2S$ 。由 $a \ge 1$ 、 $d \ge 2$ 还可得 $2S \ge 2n^2$ ，即 $n^2 \le S$ 。当 $S = 223$ 时， $446$ 的因数中没有位于 $3$ 到 $\sqrt{223}$ 之间的数。当 $S = 221 = 13 \cdot 17$ 时， $442$ 的因数中只有 $n = 13$ 落在这个范围内。于是 $2a + 12d = 34$ ，即 $a + 6d = 17$ 。由于 $a$ 、 $d$ 是正整数且 $d \gt 1$ ，只能有 $a = 5$ 、 $d = 2$ 。因此 $a + d + n = 5 + 2 + 13 = 20$ ，正确答案是 B。

The true sum is $S = na + \frac{n(n-1)}{2}d.$ Since one term is off by $1,$ the written total satisfies $222 = S \pm 1,$ so $S = 221$ or $223.$ Also $2S = n\bigl(2a + (n-1)d\bigr),$ so $n$ divides $2S.$ Since $a \ge 1$ and $d \ge 2,$ we have $2S \ge 2n^2,$ hence $n^2 \le S.$ For $S = 223,$ no divisor of $446$ lies between $3$ and $\sqrt{223}.$ For $S = 221 = 13 \cdot 17,$ the only divisor of $442$ in this range is $n = 13.$ Thus $2a + 12d = 34,$ or $a + 6d = 17.$ Since $a$ and $d$ are positive integers with $d \gt 1,$ we get $a = 5,$ $d = 2.$ Then $a + d + n = 5 + 2 + 13 = 20.$ Thus, B is the correct answer.

2023 AMC 10B 第 23 题

答案：B

难度评级：2380

解答：

其他年份的第 23 题