Question

求有序整数对 $(x, y)$ 的个数，其中 $x$ 和 $y$ 都在 $-100$ 到 $100$ 之间（包含端点），并满足 $12x^2 - xy - 6y^2 = 0$ 。

Find the number of ordered pairs $(x, y),$ where both $x$ and $y$ are integers between $-100$ and $100,$ inclusive, such that $12x^2 - xy - 6y^2 = 0.$

答案

答案：117

知识点：丢番图方程因式分解区间内整数计数

难度评级：2110

解答：

方程可分解为所以每个解都满足 $4x = 3y$ 或 $3x = -2y$ 。 $\begin{gathered} 12x^2 - xy - 6y^2 \\ = (3x + 2y)(4x - 3y) \\ = 0, \end{gathered}$

$4x = 3y$ 的整数解为 $(x, y) = (3t, 4t)$ ；限制 $|4t| \le 100$ 给出 $-25 \le t \le 25$ ，即 $51$ 对。 $3x = -2y$ 的整数解为 $(x, y) = (2t, -3t)$ ；限制 $|3t| \le 100$ 给出 $-33 \le t \le 33$ ，即 $67$ 对。这两个族只在 $(0, 0)$ ，重合，所以总数为 $51 + 67 - 1 = 117$ 。

The equation factors as $\begin{gathered} 12x^2 - xy - 6y^2 \\ = (3x + 2y)(4x - 3y) \\ = 0, \end{gathered}$ so every solution has $4x = 3y$ or $3x = -2y.$

Integer solutions of $4x = 3y$ are $(x, y) = (3t, 4t);$ the constraint $|4t| \le 100$ gives $-25 \le t \le 25,$ or $51$ pairs. Integer solutions of $3x = -2y$ are $(x, y) = (2t, -3t);$ the constraint $|3t| \le 100$ gives $-33 \le t \le 33,$ or $67$ pairs. The families overlap only at $(0, 0),$ so the count is $51 + 67 - 1 = 117.$

2025 AIME I 第 4 题

答案：117

难度评级：2110

解答：

其他年份的第 4 题