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。
Mr. Morgan G. Bloomgarten, the millionaire, known in the States as the Clam King, had, for his sins, more money than he knew what to do with. It bored him. So he determined to persecute some of his poor but happy friends with it. They had never done him any harm, but he resolved to inoculate them with the "source of all evil." He therefore proposed to distribute a million dollars among them and watch them go rapidly to the bad. But he was a man of strange fancies and superstitions, and it was an inviolable rule with him never to make a gift that was not either one dollar or some power of seven—such as 7, 49, 343, 2,401, which numbers of dollars are produced by simply multiplying sevens together. Another rule of his was that he would never give more than six persons exactly the same sum. Now, how was he to distribute the 1,000,000 dollars? You may distribute the money among as many people as you like, under the conditions given.
百万富翁摩根·G·布卢姆加滕先生在美国被称为“蛤王”,由于他的罪孽,他拥有的钱多得不知道该怎么花。这让他感到无聊。于是他决定用它来迫害他一些贫穷但快乐的朋友。他们从未对他造成任何伤害,但他决心给他们接种“万恶之源”。因此,他提议向他们分发一百万美元,然后看着他们迅速走向恶化。但他是一个有着奇怪的幻想和迷信的人,对他来说,永远不要送出不是一美元或七的幂的礼物——比如7、49、343、2,401,这些美元的数字是简单地将七相乘得出的,这是一条不可违反的规则。他的另一条规则是,他永远不会给超过六个人完全相同的金额。现在,他该如何分配这100万呢?您可以在给定的条件下将钱分配给任意数量的人。
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 - 16,第三人得 x + 19。三人总和给出 4x + 19 - 16 = 87,所以 x = 21。三份分别是 26、21、40 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。