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Canterbury Puzzle 6 · Merchant’s Coins 6

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。

The Canterbury Puzzles 1907 6 arithmetic

Perhaps no puzzle of the whole collection caused more jollity or was found more entertaining than that produced by the Host of the "Tabard," who accompanied the party all the way. He called the pilgrims together and spoke as follows: "My merry masters all, now that it be my turn to give your brains a twist, I will show ye a little piece of craft that will try your wits to their full bent. And yet methinks it is but a simple matter when the doing of it is made clear. Here be a cask of fine London ale, and in my hands do I hold two measures—one of five pints, and the other of three pints. Pray show how it is possible for me to put a true pint into each of the measures." Of course, no other vessel or article is to be used, and no marking of the measures is allowed. It is a knotty little problem and a fascinating one. A good many persons to-day will find it by no means an easy task. Yet it can be done.

也许整个系列中没有哪个谜题比全程陪伴聚会的“战袍”主持人制作的谜题更令人欢乐或更有趣。他把朝圣者们召集在一起,说道:“各位快乐的主人,现在轮到我来给你们开动脑筋了,我将向你们展示一件小手艺,它将充分考验你们的智慧。然而,我认为,当弄清楚它的做法时,这只是一件简单的事情。这是一桶上等的伦敦啤酒,我手里拿着两份量——一份五品脱,另一份三品脱。请展示我怎么可能在每一项措施中都投入真正的品脱。”当然,不得使用其他容器或物品,也不允许标注措施。这是一个棘手的小问题,也是一个令人着迷的问题。今天,很多人会发现这绝不是一件容易的事。但这是可以做到的。

提示 1

先列小例,不要凭直觉猜最大或最小。

提示 2

看奇偶、整除、余数和总量是否已经锁住选择。

提示 3

最后用上下界排除所有漏网情形。

完整解答

设第二人得 x 枚,则第一人得 2x - 6,第三人得 x + 9。三人总和给出 4x + 9 - 6 = 47,所以 x = 11。三份分别是 16、11、20 枚。

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。