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。
"This is queer," said McCrank to his friend. "Twopence added to twopence is fourpence, and twopence multiplied by twopence is also fourpence." Of course, he was wrong in thinking you can multiply money by money. The multiplier must be regarded as an abstract number. It is true that two feet multiplied by two feet will make four square feet. Similarly, two pence multiplied by two pence will produce four square pence! And it will perplex the reader to say what a "square penny" is. But we will assume for the purposes of our puzzle that twopence multiplied by twopence is fourpence. Now, what two amounts of money will produce the next smallest possible result, the same in both cases, when added or multiplied in this manner? The two amounts need not be alike, but they must be those that can be paid in current coins of the realm.
“这很奇怪，”麦克兰克对他的朋友说。 “两便士加两便士就是四便士，两便士乘以两便士也是四便士。”当然，他认为金钱可以使金钱倍增的想法是错误的。乘数必须被视为一个抽象数字。确实，两英尺乘以两英尺等于四平方英尺。同样，两便士乘以两便士将产生四平方便士！读者会很困惑什么是“方便士”。但为了解谜，我们假设两便士乘以两便士等于四便士。现在，当以这种方式相加或相乘时，哪两笔钱会产生下一个最小的可能结果，在两种情况下都相同？两个金额不必相同，但必须是可以用该领域现行硬币支付的金额。
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。