Question

Harold、Tanya 和 Ulysses 给一排很长的尖桩篱笆刷漆。

• Harold 从第一根尖桩开始，每逢第 $h$ 根刷一根；

• Tanya 从第二根尖桩开始，每逢第 $t$ 根刷一根；

• Ulysses 从第三根尖桩开始，每逢第 $u$ 根刷一根。

当正整数三元组 $(h, t, u)$ 能使每根尖桩恰好被刷一次时，称正整数 $100h + 10t + u$ 为 可刷数。求所有可刷数之和。

Harold, Tanya, and Ulysses paint a very long picket fence.

• Harold starts with the first picket and paints every $h$ th picket;

• Tanya starts with the second picket and paints every $t$ th picket; and

• Ulysses starts with the third picket and paints every $u$ th picket.

Call the positive integer $100h + 10t + u$ paintable when the triple $(h, t, u)$ of positive integers results in every picket being painted exactly once. Find the sum of all the paintable integers.

答案

答案：757

知识点：等差数列模运算分类讨论

难度评级：2840

解答：

三个等差数列 $\{1, 1 + h, \ldots\}$ 、 $\{2, 2 + t, \ldots\}$ 、 $\{3, 3 + u, \ldots\}$ 必须划分正整数。若 $h = 2$ ，Harold 会刷第 $3$ ，根，Ulysses 也会刷它，所以 $h \ge 3$ 。若 $h \ge 5$ ，考虑第 $4$ ：根尖桩：Harold 的下一根是 $1 + h \ge 6$ ，Ulysses 不可能刷它（否则需 $u = 1$ 会从第 $3$ 根起刷所有尖桩），所以必须由 Tanya 刷，迫使 $t = 2$ 。接着第 $5$ 根若要被刷就需 $u = 2$ ，但这样 Tanya 和 Ulysses 合起来会覆盖从第 $2$ 根起的所有尖桩， Harold 的第 $1 + h$ 根会被刷两次。因此 $h = 3$ 或 $h = 4$ 。 $h = 1$ ，

若 $h = 3$ ，Harold 刷 $1, 4, 7, \ldots$ 。Ulysses 不能刷第 $5$ 根（那会使 $u = 2$ ，并重复刷 $7$ ），所以 Tanya 刷它： $t = 3$ ，覆盖 $2, 5, 8, \ldots$ 。剩下的正好是 $3, 6, 9, \ldots$ ，所以 $u = 3$ ，得到 $333$ 。若 $h = 4$ ，Harold 刷 $1, 5, 9, \ldots$ ；第 $4$ 根再次迫使 $t = 2$ ，剩余的尖桩 $3, 7, 11, \ldots$ 迫使 $u = 4$ ，得到 $424$ 。

所有可刷数之和为 $333 + 424 = 757$ 。

The three progressions $\{1, 1 + h, \ldots\},$ $\{2, 2 + t, \ldots\},$ $\{3, 3 + u, \ldots\}$ must partition the positive integers. If $h = 1,$ Harold paints every picket and overlaps the other two painters. If $h = 2,$ Harold paints picket $3,$ which Ulysses also paints, so $h \ge 3.$ If $h \ge 5,$ consider picket $4:$ Harold's next picket is $1 + h \ge 6,$ and Ulysses cannot paint it (that would need $u = 1,$ repainting everything from $3$ on), so Tanya must, forcing $t = 2.$ Then picket $5$ is unpainted unless $u = 2,$ but then Tanya and Ulysses together cover every picket from $2$ on, and Harold's picket $1 + h$ is painted twice. So $h = 3$ or $h = 4.$

If $h = 3,$ Harold paints $1, 4, 7, \ldots.$ Ulysses cannot paint picket $5$ (then $u = 2$ and he would repaint $7$ ), so Tanya does: $t = 3,$ covering $2, 5, 8, \ldots.$ What remains is exactly $3, 6, 9, \ldots,$ so $u = 3,$ giving $333.$ If $h = 4,$ Harold paints $1, 5, 9, \ldots;$ picket $4$ again forces $t = 2,$ and the leftover pickets $3, 7, 11, \ldots$ force $u = 4,$ giving $424.$

The sum of the paintable integers is $333 + 424 = 757.$

2002 AIME I 第 9 题

答案：757

难度评级：2840

解答：

其他年份的第 9 题