水手分椰子类型题简易通解公式及推导中英对照

水手分椰子类型题简易通解公式及推导（中英对照修改版）Sailors assigned coconut problem, simple General Solution Formula and derivation of the(Bilingual modified version)中国湖南省祁阳县陈小刚引言,“水手分椰子”是趣味数学题”水手、猴子和椰子”的习惯简称，在中国被改为（五猴分桃）这是一道世界著名的趣味数学题，于1926年,首先刊登在美国《星期六晚邮报》上，据说,最早是由伟大物理学家狄拉克提出来的，这一貌似简单的问题曾困扰住了他，为了获得简便的計算方法，他把问题提交给当时的一些数学家，有意思的是，竟然也没有得到满意的结果，随后，在经过美国数学科普大师马丁*加德纳的介绍后，该题得到了更为广泛的流传。

1979年，诺贝尔物理学奖获得者，李政道博士在“中国科技大学”讲学时，特地提到此题；自此以后，研究该题的简易计算方法迅速风靡国内。

Preface:"seaman divided coconut" is interesting mathematical topics "seaman, the monkey and the coconut" used simple name, (China was changed to five monkeys divided peach).This is a very famous interesting mathematical problem, first published in the United States" the Saturday evening post.It is said that the famouse physicist Dirac is the earliest man who brought out this problem,, this seemingly simple problems had plagued him. In order to obtain a simple method, he put this problem, give some mathematicians, interestingly, also did not get satisfactory result. On 1979 years, the Nobel Prize winner, Lee Dr. "China University of Technology" lecture, specially referred to this question; since then, to study the problems of simple calculation method, quickly swept the country..曾对“五水手分椰子”的广泛流传起过重要作用的, 著名现代数理逻辑学家怀德海, 曾用高阶差分方程理论的通解和特解的关系, 对“水手分椰子”一题,给出过一个答案为(-4)的巧妙特解。

近十多年来, 在后来者的不断努力下，一些比较简便的方法也逐步出现。

但严格的来说：目前所取得的成果，其本上还是局限于“水手分椰子”(或五猴分桃)这一个具体题目，离全面彻底而又简捷地求解所有这种类型的题目，还有着较大的距离。

I was in 1979, in the monthly "Chinese youth," and see the Chinese-style sailor of coconut - "five monkeys sub peach" a question, and through the use equation, the solution obtained。

At that time I felt that doing this particular subject, has little significance. Meanwhile in a very complex calculation process, feel slightly faint if this type of problem can find some regularity. So through five, six days of effort, finally figured out all kinds of questions of this kind of simple general solution formula:y=a n－db/c.However, because of their own in the country, lack of information, did not put the "general solution formula" very seriously.本人曾于1979年, 在月刊《中国青年》看到中国式的水手分椰子，(五猴分桃)一题, 并通过用不定方程求得其解。

当时, 本人觉得就题论题意义己不大。

同时在非常繁复的计算过程中, 隐隐略略觉得这种类型题好象能找到某种规律。

于是通过五、六天的努力,终于演算出所有这种类题型的简捷的通解公式：y=a n －db/c.但是，由于当时自己在乡下，信息闭塞,也没把这个“通解公式”很当一回事。

Iwas in 1979, in the monthly "Chinese youth," and see the Chinese-style sailorof coconut - "five monkeys sub peach" a question, and through the use equation, the solution obtained。

At that time I felt that doing this particular subject, has little significance. Meanwhile in a very complex calculation process, feel slightly faint if this type of problem can find some regularity. So through five, six days of effort, finally figured out all kinds of questions of this kind of simple general solution formula:y=a n－db/c.However, because of their own in the country, lack of information, did not put the "general solution formula" very seriously.一幌三十多年又过去了，近段时间, 因较空闲，经常上上网,于是惊呀发现:寻找“五猴分桃”类型题的简易计算方法，竟是一个具有较深背景的，已讨论了二、三十年的热门话题；而且至今仍未找到完美解决办法。

于是自己边回想、边演算，终于又重新推导出了“五猴分桃”类型题的“通解公式”,并通过进一步分析，得到了这类问题的完美求解体系，现将其发表如下，与大家共同分享：thirty years passed in a flash, Recently, due to relatively idle, often on the Internet, so surprised and found that: looking for "five minutes peach monkey" type questions simple calculation method was actually one with a darker background, has been discussed for two or three decades a hot topic; but has yet to find the perfect solution。

So he, while recall, while calculus, and finally deduces again "five monkeys of peaches" This type of title of "general problem-solving formula", and through further analysis, got it, the perfect kind of problem for solving system, now its publication as follows, to share with you:一，水手分椰子类型题简易通解公式及特殊形式:1.水手分椰子问题的简易通解公式 y=a(a/m)n-1－db/c其中:y ──被分的椰子的总个数a ──每次分的份数, （可为任意数）n ──总共分的次数（可为任意数）b ──每次分a份后的余数.c ──每次分a份后拿走的份数,d ──每次分a份后拿走c份后,剩下再分的份数.m —— (a/d)的最大公约数注：（1）在上述公式中，按照这种类型题题意的要求；y、a、b、c、d、n、m都为正整数，（2）当b/c不为正整数时，题目本身无解；若b/c为正整数时，则题目必定有解（后面会有论述）One, five monkey peach type of problem solving simple generic formula and special forms:1 five monkeys of peaches problem solving simple generic formula；y=a(a/m)n-1－db/cy ——The total number is to be assigned coconuta ——each time you want to assign the number of copies of (non-zero natural number)n ——Coconut assigned the total number of timesc ── after each allocation, to take away part of thed ── after each allocation.Pick up the part after,the remaining part.m —— (a/d) of the greatest common divisorNotes: (1) In the above formula, according to this type of problem title meaning requirements; y, a, b, c, d, n, m, are positive integers,(2)When the b / c is not a positive integer, the title itself is no solution; if b / c are positive integers, then the problem must be solvable (there will be discussed later)2.通解公式的三种特殊形式:(1)当出现（a/d)的公约数只有m =1时，通解公式可简化为；y=a n－db/c(2)当式中的m和c都等于1时, 通解公式可写成特殊简化形式: y=a n-db(3)当式中的 m，c和b都等于1时,通解公式可写成特殊简化形式y=a n-d在《五猴分桃》一题中：由于(c=1，b=1）因而它正好属于公上面y=a n-d 的类型，由此可见《五猴分桃》一题，在这个简易通解公式里，是计算最为简单的一个类型。