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。
Of the Merchant the poet writes, "Forsooth he was a worthy man withal." He was thoughtful, full of schemes, and a good manipulator of figures. "His reasons spake he eke full solemnly. Sounding away the increase of his winning." One morning, when they were on the road, the Knight and the Squire, who were riding beside him, reminded the Merchant that he had not yet propounded the puzzle that he owed the company. He thereupon said, "Be it so? Here then is a riddle in numbers that I will set before this merry company when next we do make a halt. There be thirty of us in all riding over the common this morn. Truly we may ride one and one, in what they do call the single file, or two and two, or three and three, or five and five, or six and six, or ten and ten, or fifteen and fifteen, or all thirty in a row. In no other way may we ride so that there be no lack of equal numbers in the rows. Now, a party of pilgrims were able thus to ride in as many as sixty-four different ways. Prithee tell me how many there must perforce have been in the company." The Merchant clearly required the smallest number of persons that could so ride in the sixty-four ways.
关于商人，诗人写道：“他确实是一个值得尊敬的人。”他深思熟虑，满脑子都是计谋，而且善于操纵数字。 “他的理由很严肃。他的胜率不断增加。”一天早上，当他们在路上时，他身边的骑士和乡绅提醒商人，他还没有提出他欠公司的谜题。他随即说道：“是这样吗？那么，当我们下次停下来时，我将在这群快乐的伙伴面前提出一个数字谜语。今天早上，我们总共有三十个人骑马穿过公地。实际上，我们可以骑一个和一个，他们所谓的单列，或二和二，或三和三，或五和五，或六和六，或十和十，或十五和十五，或全部三十人排成一排。除此之外，我们不能骑马，以便有现在，一队朝圣者能够以多达六十四种不同的方式骑行。请告诉我，这群人中一定有多少人。”商人明确要求能够乘坐六十四路的最少人数。
提示 1
先只保留连接关系，把多余长度和角度擦掉。
提示 2
数端点、奇偶穿越和连通块，看看有没有不变量。
提示 3
最后再把抽象关系放回原图，确认没有偷换路径。
完整解答
一笔画回到起点需要所有路口度数为偶数；若有四个奇度路口，就不可能每条路恰好走一次并回到原点。至少要改变两条端点关系，或增加一条连接两个奇度点的路，把奇度点数降到 0。
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。