水手分椰子类型题简易通解公式和推导中英对照

1、 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

2、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 i

3、s 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 th

4、is 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 cou

5、ntry.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

6、 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,

7、six days of effort, finally figured out all kinds of questions of this kind of simple general solution formula:y=andb/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,

8、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=andb/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 sailor of coconut - five monkeys sub peach a question, and through the use equation, the solution obt

9、ained。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 o

10、f simple general solution formula:y=andb/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”

11、,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

12、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 s

13、ystem, now its publication as follows, to share with you: HYPERLINK /showpic.html l url=/orignal/a1494e13td858d3a4a032&690 t _blank 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-1db/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：

14、（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-1db/c y The total number is to be assigned coco

15、nuta each time you want to assign the number of copies of (non-zero natural number) n Coconut assigned the total number of timescafter each allocation, to take away part of the daftereachallocation.Pick up the part after, theremainingpart. m (a/d) of the greatest common divisorNotes: (1) In the abov

16、e 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

17、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=andb/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=an-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=an-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=an-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。 2. General solution formula three special forms:(1) When (a / d) of the Convention, only the number

18、of m = 1, the general solution formula can be abbreviated; y = a n-db / c (2) When the formula m and c are equal to 1, the general solution of equation can be written as a special simplified form: y = a n-db (3) When the formula m, c and b are equal to 1, the general solution of equation can be writ

19、ten as a special simplified form y = an-d In the Sailor of coconut in this topic: Since (c = 1, b = 1) and thus it happens to belong to the above, y = a nd type, we can see Sailor of the coconut, the subject, in this general solution formula where is The most simple one calculate the type.二，公式的推导Two

20、, Formula Derivation设，第6次,5个水手一起分椰子时,看到的数量为(last, five seaman with distribution of coconut,see the coconut number)ax+b,x为最后一次分a份后,每份的个数 (X,for the sixth time, distribution of coconut, coconut each a number)。 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)：(axb)a/db=a2x/dba/db 第

21、4个水手分椰子时,看到的椰子数为（Fourth seaman divided coconut, see the coconut number）：(a2x/dab/db)a/db=a3x/d2b(a/d)2ba/db 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/d3b(a/d)3b(a/d)2ba/db 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

22、 coconut, see the coconut number)： y=a6x/d5(a/d)5(a/d)4(a/d)3(a/d)2(a/d)1b, 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 ofbrackets,is apublic(d/a) as thenumber cases,According to the geometric progression recursion formula and generalize:y=anx/dn-1(a/d)n-11(d/a)n/(1d/a)/b=anx/dn-1(a/d)n-1d/aab/c=

23、anx/dn-1ban/cdn-1db/c=(canxban)/cdn-1db/c=an(cxb)/cdn-1db/c=an(xb/c)/dn-1db/c算式推导这里时，出现了两种情况（1）当上式中的a(a/d)n-1部分, 若(a/d)无公约数时,则an与dn-1互质, 故上式可进一步写成：y=an(xb/c)/dn-1-db/c从上式可看出：根据题意dn-1必然是正整数，当(b/c)也为正整数，则(x+b/c)/dn-1必可取得最小自然数1, 或1 的任意整倍数, 通常在计算时，为了简便，一般取最小自然数1, 则上述方程可简写成，简易公式：y=andb/c,这个公式可看作是所有这种类型题目

24、的通解，但不一定是最小解Formula deduced this step, there were two situations (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(xb/c)/dn-1-db/c From the above equation can be seen: According to the meani

25、ng of the questions dn-1 must be a positive integer, when (b / c) is also positive integer, then (x + b / c) / dn-1 will be available to the smallest natural number 1, or 1 any integer multiples, usually in the calculation, for simplicity, and generally the smallest natural number 1, the above equat

26、ion can be abbreviated as, simple formula: y = an-db / c, this formula can be seen as all the general solution of this type of problem, but not necessarily a minimal solution （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=andb/c更小的解，现在我们接着y=an(xb/c)/dn-1-db/c，这一步继续求证，设m为(a/b)的最大公约数，则有： ya(a/m)/(d/m)n-1(xb/c)-db

27、/ca(a/m)n-1(xb/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(xb/c)/dn-1-db/c this step and continue projections, if the set: m of (a/b) of the greatest common divisor, then there:

28、y a(a/m)/(d/m)n-1(xb/c)-db/c a(a/m)n-1(xb/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-1db/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ya(a/m）n-1db/candb/c.也就是说：后者实质上是前者特殊形式，而ya(a/m）n-1db/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 According to the first case above, followed by the same

29、 token, can be obtained: y=a(a/m）n-1db/c Clearly, if we put one, is seen as a (a / d) divisor , then, when (a / d) of the divisor is only 1, then ya(a/m）n-1db/candb/cThat is: the latter in essence, is a special form of the former, and ya(a/m）n-1db/candb/c.not only a sailor assigned coconut, this typ

30、e of problem solving generic formula , is also intended to meet the requirements of questions, solving all problems of this type, the general solution of equation minimal solution（3）在通解公式:y=a(a/m）n-1db/c,里的db/c中，只要出现（b/c）不为正整数，通解公式便无解，现求证如下;现在我们对推导过程中的最后一步ya(a/m)n-1(xb/c)/(d/m)n-1 -db/c。进行进一步分析， (3)

31、 the general solution of equationy=a(a/m）n-1db/c Lane, db/c, as long as appears (b/c), this part is not a positive integer, the general solution formula has no solution, now Prove that the following We turn now to the last step in the derivation, ya(a/m)n-1(xb/c)/(d/m)n-1 -db/c. For further analysis

32、, 在上式中若要便公式ya(a/m)n-1(xb/c)/(d/m)n-1 -db/c。有解，则公式中的(xb/c)/ (d/m)n-1应为整数。现设：（xb/c)/(d/m)n-1=k，(k为正整数)，则有x= k(d/m)n-1b/c， 显然，若b/c不为整数时，则x不为整数，而x若不为整数时，则说明通解公式已无求解的意义；也就是说：当b/c不为正整数时，此时的简易通解公式y=a(a/m）n-1db/c也必定无解,反之则必定有解。 In the above formula, to make ya(a/m)n-1(xb/c)/(d/m)n-1 -db/c. A solution, the fo

33、rmula of the (xb/c)/ (d/m)n-1应为整数。 should be an integer. Let now: ：（xb/c)/(d/m)n-1=k, (k is a positive integer), then x= k(d/m)n-1b/c，Obviously, if b/c is not an integer, then x, is not an integer, and x, if not an integer, then the solution through the solution formula has no significance; that: wh

34、en b / c is not positive integer, then the general solution of a simple equationy=a(a/m）n-1db/c will also no solution, and vice versa must be solvable. 三，公式的验算现在用上述公式来求解,本人在上月博客中12、15、16日所出的三道此种类型题目 Now with the formula to solve, I blog in April, 12, 15, 16, the three such topics 例一：在九猴分桃中(Exam1: “n

35、ine monkeys peach) a=9, n=10, b=8, d=7, c=2 根据通解公式有(According to the general formula)： y= 91087/2=3486784373又如,十六水手分椰子中(Exam2:“16 sailors allot coconuts)：a=16, n=11, b=12, d=13, c=3 根据通解公式有：(According to the general formula)： y=161112同样，可得二十三海盗分珠宝的解为： (Exam 3: as the same ,“23 pirates part jewelry” according to the general formu