Question

序列 $1, x, y, z$ 是等差数列，序列 $1, p, q, z$ 是等比数列。两个序列都严格递增且只含整数，并且 $z$ 尽可能小。 $x + y + z + p + q$ 的值是多少？

The sequence $1, x, y, z$ is arithmetic. The sequence $1, p, q, z$ is geometric. Both sequences are strictly increasing and contain only integers, and $z$ is as small as possible. What is the value of $x + y + z + p + q?$

$66$

$91$

$103$

$132$

$149$

答案：E

知识点：等差数列等比数列模运算

难度评级：1500

解答：

从等差数列可知 $z = 1 + 3d$ ，所以 $z \equiv 1 \pmod 3$ 。从等比数列可知 $z = p^3$ ，其中整数公比 $p \ge 2$ 。为了让 $z$ 最小，测试 $p = 2, 3, 4$ 。只有 $p = 4$ 可行，因为 $p^3 = 64 \equiv 1 \pmod 3$ 。于是 $d = 21$ ，两个序列为 $1, 22, 43, 64$ 和 $1, 4, 16, 64$ 。因此 $x + y + z + p + q$ $= 22 + 43 + 64 + 4 + 16$ $= 149$ 。所以正确答案是 E。

From the arithmetic sequence, $z = 1 + 3d,$ so $z \equiv 1 \pmod 3.$ From the geometric one, $z = p^3$ for some integer ratio $p \ge 2.$ We want the smallest such $z,$ so test $p = 2, 3, 4.$ Only $p = 4$ works, since $p^3 = 64 \equiv 1 \pmod 3.$ That forces $d = 21,$ and the sequences are $1, 22, 43, 64$ and $1, 4, 16, 64.$ So $x + y + z + p + q$ $= 22 + 43 + 64 + 4 + 16$ $= 149.$ Thus, E is the correct answer.

2025 AMC 10A 第 11 题

答案：E

难度评级：1500

解答：

其他年份的第 11 题