Dudeney 谜题 49

A man had in his office three cupboards, each containing nine lockers, as shown in the diagram. He told his clerk to place a different one-figure number on each locker of cupboard A, and to do the same in the case of B, and of C. As we are here allowed to call nought a digit, and he was not prohibited from using nought as a number, he clearly had the option of omitting any one of ten digits from each cupboard. Now, the employer did not say the lockers were to be numbered in any numerical order, and he was surprised to find, when the work was done, that the figures had apparently been mixed up indiscriminately. Calling upon his clerk for an explanation, the eccentric lad stated that the notion had occurred to him so to arrange the figures that in each case they formed a simple addition sum, the two upper rows of figures producing the sum in the lowest row. But the most surprising point was this: that he had so arranged them that the addition in A gave the smallest possible sum, that the addition in C gave the largest possible sum, and that all the nine digits in the three totals were different. The puzzle is to show how this could be done. No decimals are allowed and the nought may not appear in the hundreds place.

一名男子的办公室里有三个柜子，每个柜子里有九个储物柜，如图所示。他告诉他的职员在 A 柜子的每个储物柜上放置一个不同的一位数字，对 B 柜子和 C 柜子也做同样的事情。由于我们在这里可以将“0”称为数字，并且他也没有被禁止使用“0”作为数字，因此他显然可以选择省略每个柜子中十个数字中的任何一个。现在，雇主并没有说储物柜要按任何数字顺序编号，当工作完成时，他惊讶地发现这些数字显然被不加区别地混淆了。这位古怪的小伙子请他的职员做出解释，他说他想到了这样一个想法，即排列这些数字，使它们在每种情况下形成一个简单的加法总和，上面两行数字在最低行中产生总和。但最令人惊讶的一点是：他这样排列它们，使得A中的加法得到尽可能小的和，C中的加法得到尽可能大的和，并且三个总数中的九个数字都不同。这个谜题是为了展示如何做到这一点。不允许有小数，并且百位上不能出现零。