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.Subsequently, after the introduction of the American Mathematical Science Popularization master Gardiner * Martin, the subject has been more widely spread
     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“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, 对此题给出过一个答案为(-4)巧妙的特解。近几十年来, 在一些后来者的不断努力下，该问题的求解方法研究，也逐步的有了一些进展。但严格的来说：到目前为止所取得的成果，基本上还是局限于“水手分椰子”（五猴分桃）这一个具体题目上,离简便而又全面的求解所有这种类型的题目，还存在着较大的距离.

Has widely circulated on the "sailor split coconut", played an important role, famous modern mathematical logic scientist Huai Dehai, Use high order difference equation theory, general solution and special relationship, For"seaman divided coconut"question,has given an answer to (-4) clever solution.
     In recent decades, some of the newcomers, but also constantly on this issue,  did everything possible to explore，But strictly speaking: present the results obtained, the basic is confined to:" seaman divided coconut" (Five monkeys divided peaches) such a specific topic, From comprehensive, complete and convenient, solving all of this type of problem, There are still relatively large distance  

 

本人曾于1979年,  在月刊《中国青年》看到中国版的“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核心求解基本公式：y=aⁿ－db/c. 但是，由于当时自己在乡下，信息闭塞,也没把这个“公式”很当一回事。

          Iwas 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=aⁿ－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、三十年的热门数学话题；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”类型题的基本核心解题公式,y=aⁿ－db/c，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  

    通解公式 (1), y=a(a/m)^n-1-db/c,(用于b/c为正整数)

 通解公式 (2), y=[ka(a/m)^n-1-db]/c,(用于b/c不为正整数)

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

   （a）在上试公式中，按照这种类型题题意的要求；y、a、b、c、d、m、都为正整数,n大于等于2(反之无意义）

 （b）对于公式(1)，若b/c为正整数,则通解公必定会有解。若b/c不是一个正整数,则用通解公式(2)求解,
    (c）对于公式(2)，若b/m为正整数,则通解公必定会有解。若b/m不是一个正整数,则例题本身无解,也为通解公式(2)无解。
   （d）解公式（2）中的k，可通过求k公式: k=(fc+b)(m/d)^n-1而得到, (试中k大于等于1, f 是能使k取整的自然数, 在一般情况下, k会小于=c),然后再将k值代进公式2, 便可很容易的直接得到解。

 

     一， Sailors assigned coconut problem, simple General Solution Formula:

    那么，第5个水手分椰子时, 看到的椰子数为  (Fifth  divided  coconut, see the coconut number)： (ax＋b)a/d＋b=a2x/d＋ba/d＋b  

     第4个水手分椰子时,看到的椰子数为（Fourth  seaman  divided coconut, see the coconut number）： (a2x/d＋ab/d＋b)a/d＋b=a3x/d2＋b(a/d)2＋ba/d＋b  

    同样有，第3个水手分椰子时, 看到的椰子数为 （Equally third seaman  divided coconut, see the coconut number )： a4x/d3＋b(a/d)3＋b(a/d)2＋ba/d＋b 

    然后, 再一路往后推，第1个水手分椰子时, 看到的椰子数为 (Then, as before  pushing back, first seaman divided coconut, see the coconut number)：

     y=a6x/d5＋[(a/d)5＋(a/d)4＋(a/d)3＋(a/d)2＋(a/d)＋1]b,                                    

     上式中的括号内是一个公比为(d/a)的等比数例, 根据等比数例递推公式有 （n the type of

brackets,is a'public'(d/a) as the'number cases',According to the geometric progression recursion formula and generalize:

 y=anx/dn-1＋{(a/d)n-1[1－(d/a)n/(1－d/a)]}/b

 =anx/dn-1＋[(a/d)n-1－d/a]ab/c

 =anx/dn-1＋ban/cdn-1－db/c

 =(canx＋ban)/cdn-1－db/c

 =an(cx＋b)/cdn-1－db/c

 =an(x＋b/c)/dn-1－db/c

 从而得到求解的基本方程.y=aⁿ(x+b/c)/d^n-1-db/c,并由此可申引出,一个及简易公式和二个通解公式,现分析如下:

   Thereby obtaining the basic equation solving. y=aⁿ(x+b/c)/d^n-1-db/c And thus obtain a simple formula, and two-pass solution formula is analyzed as follows:  

     (1)当上式中的a(a/d)^n-1部分, 若(a/d)无公约数时,则aⁿ与d^n-1互质, 故上式可进一步写成：y=aⁿ[(x+b/c)/d^n-1]-db/c

 从上式可看出：根据题意d^n-1必然是正整数,当(b/c)也为正整数，则(x+b/c)/d^n-1 必可取得最小自然数1, 或1 的任意整倍数, 即y=kaⁿ－db/c,故(b/c)为正整数时,公式必定有解,

 通常在计算时为了简便，k一般取最小自然数1, 则上述方程可简写成，简易公式：y=aⁿ－db/c,这个公式可看作是这种类型题目其中的一个通解，但不一定是最小解。

    (1) When the above formula,a(a/d)^n-1 section, if (a/d) no divisor, then the an and dn-1 are relatively prime, the above formula can be further written as:y=aⁿ[(x＋b/c)/d^n-1]-db/c   

    be seen from the above equation: According to these questions is intended, ((b/c)must be a positive integer when((b/c)is also a positive integer,then y=kaⁿ－db/c,inevitable.get the smallest natural number1, or any multiple of 1,ie,y=aⁿ－db/c,so when((b/c)when the a positive integer,the formula is bound to answers.

    (2) 若出现(a/d)有公约数这种情况时，此时y值，还会有比公式，y=aⁿ－db/c更小的解，现在我们接着  y=aⁿ(x＋b/c)/d^n-1-db/c，这一步继续求证，设m为(a/b)的最大公约数，则有：

    y＝a[(a/m)/(d/m)]^n-1(x＋b/c)-db/c

     ＝a(a/m)^n-1(x＋b/c)/(d/m)^n-1 -db/c。

    (2) If there is (a/d) a common divisor, in this case, y value is also there will be a ratio of General Solution smaller solution, now we then,y=aⁿ(x＋b/c)/d^n-1-db/c this step and continue projections, if the set: m of (a/b) of the greatest common divisor, then there: 

    y＝a[(a/m)/(d/m)]^n-1(x＋b/c)-db/c

     ＝a(a/m)^n-1(x＋b/c)/(d/m)^n-1 -db/

 

    根据上面第一种情况后面的同样道理，可得到：y=a(a/m）^n-1－db/c

    显然，如果我们把 1也看做是（a/d）的公约数，那么当（a/d）的公约数只有 1时，则y＝a(a/m）^n-1－db/c＝aⁿ－db/c 

    也就是说：后者实质上是前者特殊形式，而y＝a(a/m)^n-1－db/c,不仅是“五猴分桃”这种类型题的通解公式，同时也是符合题意要求的，求解所有的此种类型题的最小解的通解公式,它的解集为：y＝ka(a/m)^n-1－db/c(k为任意整数)

     According to the first case above, followed by the same token, can be obtained: y=a(a/m）^n-1－db/c

    Clearly, if we put one, is seen as a (a/d) divisor , then, when (a/d) of the divisor is only 1, then y＝a(a/m)^n-1－db/c＝aⁿ－db/c  

   

    In other words: the latter essence, it Is the a special form of the former, while the y=a(a/m）^n-1－db/c Not just the "Sailor assigned coconut" type general solution formula. Meanwhile, It is also consistent with the requirements of the subject, to find the smallest answer, the general solution.

（3）若b/c不为正整数，则可用通解公式(2),y=[ka(a/m)^n-1-db]/c,来求解，其公式推导如下：

    对于基础方程: y=k(a/m)^n-1(x+b/c)/(d/m)^n-1-db/c。可将其中的 "a(a/m)^n-1(x+b/c)/(d/m)^n-1" 部分的分子和分母,同时剩以c,得到y=k(a/m)^n-1(x+b/c)c/c(d/m)^n-1-db/c,记为h,并使得(x＋b/c)c=k(d/m)^n-1 (k正整数)这样式h,便变成了y=k(a/m)^n-1/c-db/c,进一步变形得到即通解公式2;y=[k(a/m)^n-1-db]/c.(即通解公式2).

 (3) If b/c is not a positive integer,the available general solution formula 2; y=[ka(a/m)^n-1－db]/c to solve, the formula is derived as follows:

 For the base equation: y=k(a/m)^n-1(x+b/c)/(d/m)^n-1-db/c。We can Put them, "a(a/m)^n-1(x+b/c)/(d/m)^n-1" of this part of the numerator and denominator, and multiplying c, to obtain y=k(a/m)^n-1(x+b/c)c/c(d/m)^n-1-db/c,, denoted by h, and such that (x＋b/c)c=k(d/m)^n-1 (k positive integer), where h, then becomesy=k(a/m)^n-1/c-db/c, further deformation to obtain y=[k(a/m)^n-1-db]/c(ie, the general solution of equation 2)

 

     (4)关于公式2,求k公式的推导; 设x=[k(d/m)^n-1-b]/c, 则有cx+b=k(d/m)^n-1所以有k=cx+b/(d/m)^n-1，最后得到:k=(fc+b)(m/d)^n-1 (k为正整数，f为能使k取整数的自然数)

     (4) 2,the derivation of a formula of k's formula; Let x=[k(dm)^n-1-b]/c, there cx+b=k(d/m)^n-1

there : k=cx+b/(d/m)^n-1, the last obtained: k = (fc+b) (m d)^n-1 (1.k is a positive integer, 2.f that make k get natural number integer)

 

     三，后记

  本文在解决“五猴分桃问题”简易计算方法这个问题上，跳出了单个、局部考虑问题的思路, 从这个问题的分的规律来寻找整体解决方案，得到了简易通解公式(1):y=a(a/m)^n-1－db/c。和它的姊妹公式(2):y=[ka(a/m)^n-1－db]/c。在通解公式里，　

 1，由于a和n可为任意数,且其它影响计算结果的因素又都可以是变量; 因此 这个公式穷极了,求解这类问题的深度和广度。

 2、各个变量基本上都没有相关的系数计算，且基本上都是单独出现,另外公式有解无解的条件也简单明了， 因此这个通解公式应又是求解这种类型题的最简计算公式。

 3通解公式所得到的解，又是符合这类题目的求最小解的要求， 因此可以说，这个提出来已半个多世纪的数学问题 ，已经得到较园满的解决 

 

 Third, Postscript

 In this paper, the simple calculation method to solve the "five monkey sub peach problem," this issue, out of a single, partial consider the idea, from the point of law of this issue to find the overall solution, get a simple solution formula (1): y=a(a/m)^n-1-db/c. And its sister formula (2): y = [ka(a/m)^n-1-db]/c.In general solution formula, the

 1, due to a and n can be any number, and other factors also affect the calculation results can be variable; therefore a very poor this formula, the depth and breadth of solving such problems.

 2, each variable is essentially no correlation coefficient calculation, and there are basically alone, while no solution equation solvability conditions are simple and clear, so the general solution formula should also be the easiest to solve this type of problem formula .

 3-way solution formula obtained solution, and is consistent with the purpose of seeking such practice requires a minimum solution, it can be said that the proposed mathematical problem to have over half a century, has been relatively garden full of resolve

 

 

    引用本文中的公式及公式的推导请注明来源, 否则依法追究侵权责任。 

    Cite this article equation and Formula Derivation, please indicate the source, otherwise shall be held tort liability.

    

                                     本文作者：中国湖南省祁阳县陈小刚 2012年 5月  9日

     Author: Xiao Gang-chen (Qiyang county 426100, Hunan province, China)

                                中译英文：中国湖南省祁阳县唐文军 2012年 5月 21日

     Translator: Tang Wen-jun(University of South China, Hunan province , China)            

                                        公式发表时间：2012年4月21日（见本人微博）

                                       Formula Posted: 21 April 2012 (see my microblogging)

                                                   本文修改时期；2014年 1月 18日

                                      This paper modify period; January 18, 2014