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。
I have nine counters, each bearing one of the nine digits, 1, 2, 3, 4, 5, 6, 7, 8 and 9. I arranged them on the table in two groups, as shown in the illustration, so as to form two multiplication sums, and found that both sums gave the same product. You will find that 158 multiplied by 23 is 3,634, and that 79 multiplied by 46 is also 3,634. Now, the puzzle I propose is to rearrange the counters so as to get as large a product as possible. What is the best way of placing them? Remember both groups must multiply to the same amount, and there must be three counters multiplied by two in one case, and two multiplied by two counters in the other, just as at present.
我有九个计数器，每个计数器上都有九个数字中的一个：1、2、3、4、5、6、7、8和9。我将它们分成两组放在桌子上，如图所示，以形成两个乘法和，发现两个和得到相同的乘积。你会发现158乘以23是3,634，79乘以46也是3,634。现在，我提出的难题是重新排列计数器，以获得尽可能大的产品。放置它们的最佳方式是什么？请记住，两组必须乘以相同的数量，并且在一种情况下必须有三个计数器乘以二，在另一种情况下必须有两个计数器乘以两个计数器，就像目前一样。
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 51 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。