题面摘自 Henry E. Dudeney 公版文本；英文为古腾堡原文整理，中文为本站自译，提示、解答骨架和闲谈保留本站原创结构。
On Friday, April 1, 1898, three new clocks were all set going precisely at the same time—twelve noon. At noon on the following day it was found that clock A had kept perfect time, that clock B had gained exactly one minute, and that clock C had lost exactly one minute. Now, supposing that the clocks B and C had not been regulated, but all three allowed to go on as they had begun, and that they maintained the same rates of progress without stopping, on what date and at what time of day would all three pairs of hands again point at the same moment at twelve o'clock?
1898 年 4 月 1 日星期五，三个新时钟全部设定在同一时间——中午十二点。第二天中午，发现时钟 A 的时间准确，时钟 B 快了一分钟，时钟 C 慢了一分钟。现在，假设时钟 B 和 C 没有被调节，但所有三个时钟都允许按照它们开始时的方式继续运行，并且它们保持相同的进度而不停止，那么在一天中的什么日期和时间，所有三对指针将再次指向同一时刻十二点？
提示 1
先说出现象：哪些量会变，哪些约束不会变。
提示 2
找守恒量、相似关系、平衡条件或不变量，不急着代公式。
提示 3
把物理图景或谜题结构翻成一个最小方程组，再处理边界情况。
完整解答
解题主线是先把 Dudeney 谜题 34 的条件整理成一个稳定模型，再选择最少的变量。第一步确认约束，第二步写出关键关系，第三步检查特殊情形。这里给的是原创解法骨架；若要核对原始题面，请回到公版来源。
这类题最怕一上来套公式。先把图景或语言条件说清楚，答案通常会少绕很多路。