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mc1-013 · MATHEMATICS EXTENSION

林德纸草问题79Problem 79 of the Rhind papyrus

数学学习拓展 约 10 分钟阅读 中文English

内容概述

  • 数学故事
  • 数学史 / 古埃及数学 / 趣味算题
  • 叙事 + 题目

故事介绍古埃及数学重要文献林德(阿默斯)纸草卷,重点讲述第79题:一组以“房子、猫、鼠、麦穗、量斗”为名的数字,其实是7的前五次幂及其和。历史学家康托认为它可能是中世纪及后世“去罗马路上的七个老妇人”“St. Ives 儿歌”等连锁乘法趣题的古老源头。

中文正文

古埃及数学的主要来源是林德(Rhind)-阿默斯(Ahmes)纸草卷,一本兼有实用手册性质的数学教科书,包括85个数学问题,是阿默斯在大约公元前1650年用写经文的草书体从一部古书抄录下来的。纸草卷原由英国的埃及学专家林德(A. Henry Rhind)在埃及买下,后来归了一家英国博物馆。

尽管林德纸草卷里的多数问题都不难破译和解释,但有一个问题,问题79,却不那么肯定。问题出现在一组奇怪的数字里,我们转录在下面:

| 财物 | | |------|---| | 房子 | 7 | | 猫 | 49 | | 鼠 | 343 | | 麦穗 | 2401 | | 量斗 | 16807 | | | 19607 |

很容易看出前五个数是7的幂,然后是它们的和。因为这一点,乍看起来作者似乎在给这些幂次引入代表性的名词,如房子代表1次,猫代表2次,等等。

然而,历史学家康托(Moritz Cantor)在1907年提出一个更合理也有趣的解释。他从这个问题发现了一个流行在中世纪的问题的源头,那是1202年斐波那契(Leonardo Fibonacci)在他的《算经》(Liber abaci)中提出的。那本书有很多问题,其中的一个说:“去罗马的路上走着七个老妇人。每个妇人有七匹骡子,每匹骡子驮着七个麻袋,每个麻袋装着七块大饼,每块大饼旁放着七把小刀,每把小刀套着七重鞘。妇人、骡子、麻袋、大饼、小刀和刀鞘,去罗马的路上共有多少东西?”同一个问题还有我们更熟悉的形式,就是后来唱的那支古老的英国儿歌:

As I was going to St. Ives

I met a man with seven wives;

Every wife had seven sacks;

Every sack had seven cats;

Every cat had seven kits.

Kits, cats, sacks, and wives,

How many were going to St. Ives?

据康托的解释,林德纸草卷的那个原始问题后来也许演变成下面的形式:“一笔财富七匹马,一马背上七只猫,一猫抓住七只鼠,一鼠偷吃七棵穗,一穗产出七斗粮。马猫鼠穗和稻粮,东西一共是多少?”

看来,这个问题流传到今天,仍然是一个疑难。当阿默斯抄录的时候,它已经很古老了,而在斐波那契把它写入《算经》时,它差不多已经3000年了。750多年后的今天,我们还在把它传唱给我们的孩子。我们不禁想知道,那古老的英国儿歌会不会也出现在古埃及的问题里?尽管它完全可能是盎格鲁-撒克逊人的杰作。

不时出现在我们今天杂志上的许多疑难,都能找到它在中世纪的影子。现在几乎不可能确定它们还能追溯到多远的过去。

英文正文

13° Problem 79 of the Rhind papyrus. One of our chief primary sources concerning the mathematics of ancient Egypt is the Rhind, or Ahmes, papyrus, a mathematical text partaking of the nature of a practical handbook and consisting of eighty-five problems copied about 1650 B.C. in hieratic writing by the scribe Ahmes from an earlier work. The papyrus was purchased in Egypt by the English Egyptologist A. Henry Rhind and then later acquired by the British Museum.

Although little difficulty was encountered in deciphering and then interpreting most of the problems in the Rhind papyrus, there is one problem, Problem Number 79, for which the interpretation is not so certain. In this problem occurs the following curious set of data, here transcribed:

| Estate | | |--------|---| | Houses | 7 | | Cats | 49 | | Mice | 343 | | Heads of wheat | 2401 | | Hekat measures | 16807 | | | 19607 |

One easily recognizes the numbers as the first five powers of seven, along with their sum. Because of this it was at first thought that perhaps the writer was here introducing the symbolic terminology houses, cats, and so on, for first power, second power, and so on.

A more plausible and interesting explanation, however, was given by the historian Moritz Cantor in 1907. He saw in this problem an ancient forerunner of a problem that was popular in the Middle Ages, and which was given by Leonardo Fibonacci in 1202 in his Liber abaci. Among the many problems occurring in this work is the following: "There are seven old women on the road to Rome. Each woman has seven mules; each mule carries seven sacks; each sack contains seven loaves; with each loaf are seven knives; and each knife is in seven sheaths. Women, mules, sacks, loaves, knives, and sheaths, how many are there in all on the road to Rome?" As a later and more familiar version of the same problem we have the old English children's rhyme:

As I was going to St. Ives

I met a man with seven wives;

Every wife had seven sacks;

Every sack had seven cats;

Every cat had seven kits.

Kits, cats, sacks, and wives,

How many were going to St. Ives?

According to Cantor's interpretation, the original problem in the Rhind papyrus might then be formulated somewhat as follows: "An estate consisted of seven houses; each house had seven cats; each cat ate seven mice; each mouse ate seven heads of wheat; and each head of wheat was capable of yielding seven hekat measures of grain. Houses, cats, mice, heads of wheat, and hekat measures of grain, how many of these in all were in the estate?"

Here, then, may be a problem that has been preserved as part of the puzzle lore of the world. It was apparently already old when Ahmes copied it, and older by close to three thousand years when Fibonacci incorporated a version of it in his Liber abaci. More than seven hundred and fifty years later we are reading another variant of it to our children. One cannot help wondering if a surprise twist such as occurs in the old English rhyme may also have occurred in the ancient Egyptian problem, though, in all likelihood, this twist was an Anglo-Saxon contribution.

There are many puzzle problems popping up every now and then in our present-day magazines that have medieval counterparts. How much further back some of them go is now almost impossible to determine.

相关题目

  • 题目:

- 中文转述:房子 7,猫 49,鼠 343,麦穗 2401,量斗 16807,求和为 19607。 - 英文转述:Houses 7, Cats 49, Mice 343, Heads of wheat 2401, Hekat measures 16807, sum 19607. - 康托重构:一笔财富七匹马,一马背上七只猫,一猫抓住七只鼠,一鼠偷吃七棵穗,一穗产出七斗粮;求总数。

  • 提示:无(原书未给出提示。)
  • 解答或证明:无(原书未给出完整解答,仅说明 19607 是前五个 7 的幂之和;具体计算过程为教学加工补充。)

我们可以从故事中思考什么

  • 林德纸草第79题中的数字为什么恰好是 7 的幂?
  • 康托的解释为什么比“房子=一次幂”的说法更合理?
  • St. Ives 儿歌的“陷阱”在哪里?真正“去 St. Ives”的可能只有谁?
  • 一个数学问题为什么能在不同文明、不同时代反复出现?

核心知识点讲解:幂与等比数列求和

选择理由

故事中的核心数据 7、49、343、2401、16807 恰好是 7 的连续正整数次幂,而 19607 是它们的和。这是小学到初中阶段理解“幂”和“等比数列求和”最直观的历史素材之一。

概念介绍

幂(power)表示同一个数连续相乘的结果。7¹ = 7,7² = 7 × 7 = 49,7³ = 7 × 7 × 7 = 343,依此类推。等比数列是指每一项与前一项的比值(公比)都相等的数列。等比数列求和就是把所有项加起来。

具体内容

  • 7¹ = 7
  • 7² = 49
  • 7³ = 343
  • 7⁴ = 2401
  • 7⁵ = 16807
  • 7 + 49 + 343 + 2401 + 16807 = 19607

用等比数列求和公式:S₅ = 7(7⁵ − 1)/(7 − 1) = 7 × 16806 / 6 = 19607。

应用场景

  • 计算细胞分裂、利息复利、连锁反应等问题。
  • 解决古代趣题中“每个主体又包含若干个同类主体”的层级计数问题。

与生活的联系

  • 社交媒体转发:如果一个人发给7个人,每个人再发给7个人,层级传播的数量就是 7 的幂。
  • 病毒传播模型:早期阶段的感染者数量常常按等比数列增长。

主要思想方法与延伸讨论

主要思想方法

  • 归纳与模式识别:从具体数字中识别出 7 的幂规律。
  • 历史比较法:把古埃及问题与中世纪、近代儿歌进行对比,追溯问题传承。
  • 层级计数法:用乘法和加法解决“每个对象内部又包含多个对象”的问题。

延伸讨论

  • 如果用 5 代替 7,构造一个“五层”连锁问题,结果会是什么?
  • St. Ives 儿歌真正想问的是“有多少人/物在去 St. Ives 的路上”,还是一个语言陷阱?
  • 古代没有现代指数符号,阿默斯如何表达“7 的五次幂”?