1、 水手分椰子類型題簡易通解公式及推導(dǎo) （中英對照修改版）Sailors assigned coconut problem, simple General Solution Formula and derivation of the (Bilingual modified version) 中國湖南省祁陽縣 陳小剛引言,“水手分椰子”是趣味數(shù)學(xué)題”水手、猴子和椰子”的習(xí)慣簡稱，在中國被改為（五猴分桃）這是一道世界著名的趣味數(shù)學(xué)題，于1926年,首先刊登在美國星期六晚郵報上，據(jù)說,最早是由偉大物理學(xué)家狄拉克提出來的，這一貌似簡單的問題曾困擾住了他，為了獲得簡便的計算方法，他把問題提交給當(dāng)時的一些數(shù)

2、學(xué)家，有意思的是，竟然也沒有得到滿意的結(jié)果，隨后，在經(jīng)過美國數(shù)學(xué)科普大師馬丁*加德納的介紹后，該題得到了更為廣泛的流傳。1979年，諾貝爾物理學(xué)獎獲得者，李政道博士在“中國科技大學(xué)”講學(xué)時，特地提到此題；自此以后，研究該題的簡易計算方法迅速風(fēng)靡國內(nèi)。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.曾對“五水手分椰子”的廣泛流傳起過重要作用的, 著名現(xiàn)代數(shù)理邏輯學(xué)家懷德海, 曾用高階差分方程理論的通解和特解的關(guān)系, 對“水手分椰子”一題,給出過一個答案為(-4)的巧妙特解。近十多年來, 在后來者的不斷努力下，一些比較簡便的方法也逐步出現(xiàn)。但嚴(yán)格的來說：目前所取得的成果，其本上還是局限于“水手分椰子”(或五猴分桃)這一個具體題目，離全面徹底而又簡捷地求解所有這種類型的題目，還有著較大的距離。 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, 在月刊中國青年看到中國式的水手分椰子，(五猴分桃)一題, 并通過用不定方程求得其解。當(dāng)時,

8、本人覺得就題論題意義己不大。同時在非常繁復(fù)的計算過程中, 隱隱略略覺得這種類型題好象能找到某種規(guī)律。于是通過五、六天的努力,終于演算出所有這種類題型的簡捷的通解公式：y=andb/c. 但是，由于當(dāng)時自己在鄉(xiāng)下，信息閉塞,也沒把這個“通解公式”很當(dāng)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. 一幌三十多年又過去了，近段時間, 因較空閑，經(jīng)常上上網(wǎng),于是驚呀發(fā)現(xiàn):尋找“五猴分桃”類型題的簡易計算方法，竟是一個具有較深背景的，已討論了二、三十年的熱門話題；而且至今仍未找到完美解決辦法。于是自己邊回想、邊演算，終于又重新推導(dǎo)出了“五猴分桃”類型題的“通解公式”

11、,并通過進(jìn)一步分析，得到了這類問題的完美求解體系，現(xiàn)將其發(fā)表如下，與大家共同分享：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，水手分椰子類型題簡易通解公式及特殊形式: 1.水手分椰子問題的簡易通解公式y(tǒng)=a(a/m)n-1db/c其中: y被分的椰子的總個數(shù) a每次分的份數(shù), （可為任意數(shù)） n 總共分的次數(shù)（可為任意數(shù)） b每次分a份后的余數(shù). c 每次分a份后拿走的份數(shù), d 每次分a份后拿走c份后,剩下再分的份數(shù). m (a/d)的最大公約數(shù) 注：

14、（1）在上述公式中，按照這種類型題題意的要求；y、a、b、c、d、n、m都為正整數(shù)， （2）當(dāng)b/c不為正整數(shù)時，題目本身無解；若b/c為正整數(shù)時，則題目必定有解（后面會有論述）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、種特殊形式: (1)當(dāng)出現(xiàn)（a/d)的公約數(shù)只有m =1時，通解公式可簡化為；y=andb/c (2)當(dāng)式中的m和c都等于1時,通解公式可寫成特殊簡化形式:y=an-db (3)當(dāng)式中的m，c和b都等于1時,通解公式可寫成特殊簡化形式y(tǒng)=an-d 在五猴分桃一題中：由于(c=1，b=1）因而它正好屬于公上面y=an-d的類型，由此可見五猴分桃一題，在這個簡易通解公式里，是計算最為簡單的一個類型。 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，公式的推導(dǎo)Two

20、, Formula Derivation設(shè)，第6次,5個水手一起分椰子時,看到的數(shù)量為(last, five seaman with distribution of coconut,see the coconut number)ax+b,x為最后一次分a份后,每份的個數(shù) (X,for the sixth time, distribution of coconut, coconut each a number)。 那么，第5個水手分椰子時, 看到的椰子數(shù)為 (Fifth divided coconut, see the coconut number)：(axb)a/db=a2x/dba/db 第

21、4個水手分椰子時,看到的椰子數(shù)為（Fourth seaman divided coconut, see the coconut number）：(a2x/dab/db)a/db=a3x/d2b(a/d)2ba/db 同樣有，第3個水手分椰子時, 看到的椰子數(shù)為 （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個水手分椰子時, 看到的椰子數(shù)為(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, 上式中的括號內(nèi)是一個公比為(d/a)的等比數(shù)例,根據(jù)等比數(shù)例遞推公式有（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算式推導(dǎo)這里時，出現(xiàn)了兩種情況（1）當(dāng)上式中的a(a/d)n-1部分, 若(a/d)無公約數(shù)時,則an與dn-1互質(zhì), 故上式可進(jìn)一步寫成：y=an(xb/c)/dn-1-db/c從上式可看出：根據(jù)題意dn-1必然是正整數(shù)，當(dāng)(b/c)也為正整數(shù)，則(x+b/c)/dn-1必可取得最小自然數(shù)1, 或1 的任意整倍數(shù), 通常在計算時，為了簡便，一般取最小自然數(shù)1, 則上述方程可簡寫成，簡易公式：y=andb/c,這個公式可看作是所有這種類型題目

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）若出現(xiàn)(a/d)有公約數(shù)這種情況時，此時y值，還會有比公式，y=andb/c更小的解，現(xiàn)在我們接著y=an(xb/c)/dn-1-db/c，這一步繼續(xù)求證，設(shè)m為(a/b)的最大公約數(shù)，則有： 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/ 根據(jù)上面第一種情況后面的同樣道理，可得到：y=a(a/m）n-1db/c顯然，如果我們把 1也看做是（a/d）的公約數(shù)，那么當(dāng)（a/d）的公約數(shù)只有 1時，則ya(a/m）n-1db/candb/c.也就是說：后者實質(zhì)上是前者特殊形式，而ya(a/m）n-1db/c，不僅是“五猴分桃”這種類型題的通解公式，同時也是符合題意要求的，求解所有的此種類型題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，只要出現(xiàn)（b/c）不為正整數(shù)，通解公式便無解，現(xiàn)求證如下;現(xiàn)在我們對推導(dǎo)過程中的最后一步y(tǒng)a(a/m)n-1(xb/c)/(d/m)n-1 -db/c。進(jìn)行進(jìn)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y(tǒng)a(a/m)n-1(xb/c)/(d/m)n-1 -db/c。有解，則公式中的(xb/c)/ (d/m)n-1應(yīng)為整數(shù)?，F(xiàn)設(shè)：（xb/c)/(d/m)n-1=k，(k為正整數(shù))，則有x= k(d/m)n-1b/c， 顯然，若b/c不為整數(shù)時，則x不為整數(shù)，而x若不為整數(shù)時，則說明通解公式已無求解的意義；也就是說：當(dāng)b/c不為正整數(shù)時，此時的簡易通解公式y(tǒng)=a(a/m）n-1db/c也必定無解,反之則必定有解。 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應(yīng)為整數(shù)。 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，公式的驗算現(xiàn)在用上述公式來求解,本人在上月博客中12、15、16日所出的三道此種類型題目 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 根據(jù)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 根據(jù)通解公式有：(According to the general formula)： y=161112同樣，可得二十三海盜分珠寶的解為： (Exam 3: as the same ,“23 pirates part jewelry” according to the general formu