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。
Four brothers—named John, William, Charles, and Thomas—had each a money-box. The boxes were all given to them on the same day, and they at once put what money they had into them; only, as the boxes were not very large, they first changed the money into as few coins as possible. After they had done this, they told one another how much money they had saved, and it was found that if John had had 2 s. more in his box than at present, if William had had 2 s. less, if Charles had had twice as much, and if Thomas had had half as much, they would all have had exactly the same amount. Now, when I add that all four boxes together contained 45 s., and that there were only six coins in all in them, it becomes an entertaining puzzle to discover just what coins were in each box.
四兄弟——约翰、威廉、查尔斯和托马斯——每人都有一个存钱罐。箱子都是当天给他们的，他们立刻就把身上的钱放进去。只是箱子不是很大，他们先把钱换成了尽可能少的硬币。做完这件事后，他们互相告诉对方自己存了多少钱，结果发现如果约翰有 2 s。如果威廉有 2 秒的话，他的盒子里的东西比现在还要多。少一点，如果查尔斯有两倍，如果托马斯有一半，他们都会有完全相同的数量。现在，当我添加所有四个盒子总共包含 45 枚硬币，并且里面总共只有 6 个硬币时，发现每个盒子里到底有哪些硬币就变成了一个有趣的谜题。
提示 1
先把每一步允许做什么写成状态表。
提示 2
找一个不会随操作改变的量，或把对象分成互斥类别。
提示 3
检查构造是否覆盖全部对象，而不是只给出一个漂亮例子。
完整解答
固定主人面向门口，就消除了旋转重复。先数其余人任意排列，再减去两位指定朝圣者相邻的情形。相邻时把这两人看成一个双人块，块内部有 2 种顺序；最后用总数减去坏排列。
Dudeney 的趣题常把难点藏在“看起来可以试”的地方。别急着猜答案；先把图、表或状态画出来，再问哪些限制一直没有变。这也是它和 Carroll 逻辑题互补的地方：一个拆句子，一个拆结构。