數學

本文由‘‘分數7/17是否能表示成兩個相異的埃及分數之和’’這個問題出發，藉由簡單數論的性質以及反證法，得到一個真分數可表示成兩個相異埃及分數之和的定理檢驗法(定理1)。有了這個基礎，我們進ㄧ步推廣定理1 的結果，做出了嶄新的結果(定理2、定理3) 。此定理分別可以用來檢驗真分數表示成三個、四個相異埃及分數之和的存在性; 至於將真分數表示為5 項、6 項….k 項相異埃及分數之和的部分尚在嘗試。利用定理1、2，我們寫了兩個Matlab 軟體工具的電腦程式，使得我們可以檢驗任意真分數是否可以表示成兩項及三項的和，並可把所有的解列出來; 最後我們研究的是一個有關埃及分數的猜想(Erdos-Strauss Conjecture)問題，當分子為4，且分母為4k、4k+2、4k+3 時，猜想皆成立。對於分母為4k+1 而言，當k 為3r+1、3r+2 猜想亦成立，k=3r 且r 為奇數時也是成立的，因此目前需解決的問題只剩分母為24t+1 的情況了。值得一提的是，我們用Matlab 的程式檢驗出當分母為1014 至1014 +240000 之內的正整數時，猜想都是成立的，這已經超越了已知文獻的結果。This paper begins with the question: ‘‘Is 7/17 able to be the sum of two different Egyptian fractions?’’ to discuss the problem of Egyptian fractions. According to the complete division properties and the counter-evidence method, we get a back-check theorem which is about a true fraction can be the sum of two different Egyptian fractions (see theorem 1). Using the same method we obtain a new back-check theorem that is a fraction can be the sum of three or four different Egyptian fractions (thereom2, thereom3). Similarly, we can follow the same procedure to get the rule that a fraction can be the sum of five or six …or even more different Egyptian fractions. By the theorem1 and 2, we propose two programs written vie the Matlab software to examine that any true fraction can be the sum of two items and three items or not. Finally we focus on the Erdos-Straus Conjecture, which related about true fractions can be divided by three different Egyptian fractions. The conjecture is when the denominator is 4k, 4k+2, or 4k+3, the problem mentioned above can be solved. As for the denominator is 4k+1, then the conjecture also can be solved, as k equals to 3r+1 or 3r+2. Also, k being 3r and r is an odd number, the conjecture is satisfied. As for the case of r equals to even number, the problem has not been solved. But it is worth to mention here that we use Matlab software to examine the conjecture is agreeable as the denominator is between 1014to 1014+ 240000. This is beyond the results from the literatures.

Two soldiers walk on a checkerboard. They can only walk one step once a time and two directions, front and left, are decided randomly. The gunshot is the column and row where a soldier is located, and one will die if he enters the gunshot area of the other. To treat the probability of winning, we first study the cases of 1×n, 2×n, 3×n, and 4×n rectangles iterately. Then we establish a general form of the probability of winning in a general n×k rectangle by using recurrence technique and generating function, respectively. Finally, we extend to the general n×m×k cuboid case to obtain the first soldier’s probability of winning.在一個長方形的棋盤中，兩士兵行走，每一次只走一步，而且上和左兩個方向是隨機的，射程範圍是所在的此行和此列，而進入他人射程範圍則死亡。探討其獲勝機率，從1×n 、2×n、3×n、4×n 矩形的情形逐步研究，並分別運用遞迴式的技巧及生成函數，導出 n×k 矩形中先走士兵獲勝機率的一般式。更進一步地，我們也獲得了n×m×k 立體空間先走士兵的獲勝機率。

對4 × 4 方格表中計數問題的二個解題方法(1..解方程式的方法， 2.分割圖形的方法)作分析和研究後，首先我推廣分割圖形的方法來証明 : “好的n × n 方格表” 存在若且惟若n 為偶數。同時証明這種“好的n × n 方格表”內所有n2 個數的總和f(n) 為n(n+2)/4。當討論一般的n×m 方格表時，發現分割圖形的方法盲點，無法繼續推廣來証明。再經過深入分析與推廣解方程式的方法，藉由n×m 變數方格表，我們終於找到構造所有“好的n × m方格表”的方法。同時計算“好的n × m 方格表” (n≦m)內所有mn 個數的總和f(n,m), n≦7和証明好的nxm 方格表會有2(n+1)行一個循環的現象。We first studied two solution methods (1.solving equations,2.dissecting diagrams.) for calculations on 4x4 checkboard. Using the method of dissecting diagrams, we proved that``good nxn checkboard'' exists if and only if n is even. Furthermore, the sum f(n) of those n2 numbers in a ``good'' nxn checkboard is equal to n(n+2)/4.In studying the more general nx m checkboards, we found that the method of dissecting diagrams does not work, However, by extending the method of solving equations, and by considering nx m variable checkboards, we obtained a way of obtaining all ``good nxm checkboards.'' By way of computing the sum f(n,m) (n≦7) of those mn numbers in a ``good nxm checkboards,'' periodicity in every 2(n+1) rows is observed.

本研究是藉由數學手法探討；如何由一疊36 張四種花色的撲克牌中，尋找出保證可猜中最多張花色的方法。研究過程是以在適當的猜牌時機，以鴿籠原理、邏輯推理、二進位、分析與歸納……等數學原理與方法，搭配巧妙的策略運用而達到目的。猜牌方法：先約定好猜牌規則，助手將36 張牌背圖樣相同但非對稱的撲克牌，以旋轉牌背的方向傳達訊息。在本研究中得出利用數學原理與方法可「經由巧妙的猜牌方法保證可以猜中26 張花色」，並提供後續研究者利用本研究之結果繼續深入探討與研究。 The study is mathematically based with reasonable explanations behind it. We are to correctly guess as many cards as possible from a deck of 36 cards, with random numbers and four different suits. We will apply mathematical methods, such as pigeonhole principle, logic inference, binary system, and analytical reduction, upon right timing. Using careful arrangement of the principles and reasoning, we can reach our ultimate goal. To state guessing: Conference between the guesser and the assistant about the guessing rules, the assistant will have 36 cards with the same exact pattern on the back but not symmetrical. The pattern of the cards will be different when rotated 180o. The only communication between the two is by rotating cards. In the process we will obtain mathematical theory and methods assuring 26 cards correctly guessed, and the study is for further and deeper discussion.

(一) 本研究首先導出ΔABC等分面積線移動所包絡出的曲線方程式，其圖形是由等分面積線段PQ(其中P、Q皆在ΔABC的周界上)的中點所構成，具有3 條曲線段(分別為3 條雙曲線之一部分)的封閉曲線，形成內文所謂的「包絡區」。利用包絡區的區隔，我們找出：1.當P 點在包絡區內，則有3 條等分面積線。2.當P 點在包絡區周界上，則有2 條等分面積線。3.當P 點曲線段的端點或在包絡區外，則有1 條等分面積線。(二) 以三角形的研究當基礎，擴展到凸n 邊形(不包含點對稱圖形)，我們發現：等分面積線數量之分布，仍然與包絡區息息相關，且1.凸2m +1邊形最多有2m +1條等分面積線。2.凸2m邊形，必發生內文所謂的「換軌」。因此，最多只有2m ?1條等分面積線。3.包絡曲線所分割出的區域，於相同區域其等分面積線數量相同，且相鄰兩區域數量差兩條。(三) 若凸n邊形有k個「換軌點」，則此n邊形過定點等分面積線至多有n ? k 條。(四) 若凸n 邊形為點對稱圖形(如正偶數邊形、平行四邊形)，則所有等分面積線皆過中心點。1) Our study got a curve equation of bisectors of a triangle. When a bisector is moving, we get three curves. They’re constructed by the midpoints of PQ. The three parts of the three curves make a closed curve which we called “the Envelope Area”. We found out:\r 1. When Point P is in the Envelope Area, we can get 3 bisectors. 2. When Point P is on the curves of the Envelope Area, we can get 2 bisectors. 3. When Point P is outside of the Envelope Area, we can get only 1 bisector. 2) Based on our study of triangles, we found that in Convex polygons(not including Point Symmetry Convex polygons), the distribution of bisectors is related to the Envelope Area. 1. We can get at most 2m +1 bisectors in a 2m +1 Convex polygon. 2. We can get at most 2m ?1 bisectors in a 2m Convex polygon, and the bisectors on the curves will “Change the Track”. 3. Envelope curve will divide a Convex polygon into several areas. The same area has the same numbers of bisectors, and the near areas have less or more 2 bisectors. 3) If a Convex polygon has k points to change the track, it will have at most n – k bisectors.\r 4) In a Point Symmetry Convex polygon (ex. Regular 2m convex polygons and parallelograms), all the bisectors will come through the center point.

正方形和長方形是每一個人都非常熟悉的圖形，但其中卻隱藏了非常多奇妙的“數學之謎”。 所謂「完美長方形」是：在一個長方形中 (長、寬不等)，能否分割出最少大小相異的正方形。 這個研究中，首先用「草圖」的解題方法研究完美長方形，接下來利用「平面圖形」的解題方法可簡化計算的過程，最後利用「對偶關係」證明出：完美長方形的最少階數為 9 階。 進而，我們將這個問題擴展至三維空間，思索在一個長方體中（長、寬、高都不等長） ，能否仿照二維空間，分割出最少大小相異的正方體，而完成這個研究。 Square and the rectangle are figures that everyone knows well very much, but what a wonderful " mystery of mathematics " is hidden among them. What is called the perfect rectangle is whether in a rectangle (its length and width is different ) could cut apart two squares as the least difference in size . In this research, the solution approach of "the sketch map" is used to study the perfect rectangle at first, then the solution approach of "the level figure" to simplify the complicatied calculation of the solving course , and "the dual relation" is finally used to prove 9 orders are the least orders for a perfect rectangle . And then, we expand this question to three-dimensional space, considering in a cuboid (its length, width, and height is different) whether could follow the two-dimensional space model to cut apart two squares as the least difference in size, and finish this research.

在2003 年台灣國際科展之作品說明書「Concatenating Squares」中【9】與2004第三屆旺宏科學獎「SA3-119 ：與特殊型質數之倒數關聯的兩平方總和的整數分解」成果報告書中【10】，就已有令人驚訝的結果。在2005 第四屆旺宏科學獎「SA4-298 ：分和累乘再現數產生的方法及其性質探討之推廣與應用」成果報告書中【11】，更是以逆向思維進行研究推廣，創造出許多新穎且引人注意的美麗數式，由原創性的觀點來進行逆向思維的研究。正由於這些再現數充滿奇巧，且與不定方程（代數）、簡單的數整除性分析（數論）以及同餘式理論都有所關聯，我們要發展前所未有的同步關聯與研究分歧， 且是最新的發現，更以豐富的想像力、創造力與推理能力提出令人耳目一新的重要結果，得到許多從未見過世面的美麗數式，回眸觀賞時內心充滿了數學之美！;Number which when chopped into two（three）parts, added / subtracted and squared（cubed） result in the same number. Consider an n -digit number k , square it and add / subtract the right n or n .1 digits. If the resultant sum is k , then k is called a Kaprekar number. The set of n -Kaprekar integers is in one-to-one correspondence with the set of unitary divisors of 10 n m1. If instead we work in binary, it turns out that every even perfect number is n - Kaprekar for some n . We wish to find a general pattern for numbers these numbers cubed or 3-D Kaprekar. We also investigate some 3-D Kaprekar of special forms. In addition, some results relating to the properties of the Kaprekar numbers also presented. This study indicates the “interesting” and “pragmatic” natures of the research project. We have developed the original results based upon his initiatives and has thus created a new horizon through the research project. This is the proudest achievement for this study. Generalization of Some Curiously Fascinating Integer Sequences：Various Recurrent Numbers！

This study was to explore the nature of two basic constitutes of the regular pentagon,With these two constitutes, the regular pentagon could be multiplied into any times. We used four multiplication methods (m2 = 2m1 + n1 、n2 = m1 + n1 、m2= k2m1 、n2= k2n1、a2 = a1 + 1、a2 = a1 + ) to show how the regular pentagon could enlarge and to verify that the enlarged regular pentagons derived from computer did exist. By integrating these four multiplication methods, we were able to arrange regular pentagon of any length of side, and evidenced the equation was ( If the side length of a regular pentagon is a form of m,n is the number of A,B respectively ) We further proved that the first multiplication method could be developed into a new modified method, which could divide a regular pentagon with a given side length into a combination of A and B. But only when the x and y of side length of a regular pentagon could be divided by a natural number, k, and made x/k into an item of the Fibonacci Sequence and y/k a successive item. When we tried to verify if any regular pentagon could be constituted by other smaller regular pentagons, we also found that it was un-dividable only if the length of pentagon side were ( the number of A, B were the 2n and 2n-1 item of Lucas Sequence). Otherwise, any regular pentagon might be able to be constituted by other smaller regular pentagons. 本研究是以正五邊形的兩個基本組成元素(B)作為討論對象，利用此二元素可以將正五邊形做任意倍數的放大。我們共使用4種繁殖法則(m2 = 2m1 + n1 、n2 = m1 + n1 、m2= k2m1 、n2= k2n1、a2 = a1 + 1、a2 = a1 + ) 來說明正五邊形的放大情形，並利用此4 種繁殖法驗證電腦運算出的放大圖形確實存在。利用這4 種繁殖法則的改良與整合，已達到能排出任意邊長之正五邊形的目標，並能計算並證明出其通式為。 (若正五邊形的邊長為形式,m、n代表、的個數) 更特別的是，我們能用第一繁殖法反推出一種方法，將給定邊長的正五邊形利用簡單的切割方式分成由A、B 組合成的形式，但只有正五邊形邊長之x、y 值可同除以任一自然數k 而使 x/k 為費波那契數列之一項且 y/k 為其後一項者才可以使用。 將此想法推廣至一個正五邊形能否由比他小的其他五邊形組合而成時，我們也發現當正五邊形之邊長為時（其A、B 個數為盧卡斯數列之第2n,2n-1 項），不可分解，否則應該皆可將一個正五邊形分解成比它小的其他五邊形組合（我們也可以利用這些質形檢驗出其他正五邊形是否也為質形）。但其分解形式，不只一種，而我們推測只用兩種較小的正五邊形就能達成，我們期待能找出一或多種分解方法，能將正五邊形分解成標準的分解形式。

In this work I make the analysis of the possibility of the existence of the number system with non-natural base & its investigation. The question examined in my work is totally opened:\r ‧ making the list of new characteristics, rules of the translation of the numbers, and also rules of the simple calculating operations, checked the operations of subtraction & division;\r ‧ checked the Euclidean algorithm, its characteristics by means of estimating the coefficients;\r ‧ found the practical appliance of new method in compiling & solving of the tasks.\r Investigation I’ve suggested stipulates for independence of new system & its appliance in type of tasks, that is beyond the course of school program & also beyond the whole system of school education.

Purpose of the Research:\r 1) To determine whether a poor understanding and inability of Grade 7 and 8 learners to apply the BODMAS principle in mathematics, influences scores obtained in a mathematics test.\r 2) To determine whether scores obtained in the given mathematics test can be improved with a BODMAS learning tool.\r Procedures:\r 1. Get the educators opinion on mathematics in schools. Send a total of 50 questionnaires to four schools.\r 2. Determine what percentage of a mathematical test/examination requires the application of BODMAS\r 3. Do a pre-test at two schools, a total of 370 grade 7 and 8 learners.\r 4. Design a BODMAS learning tool and verify it with three educators.\r 5. Implement the tool at the two schools.\r 6. Do a post-test at the two schools.\r 7. Get all the educators who were at the implementation session to evaluate the session.\r 8. Investigate two other schools, by sending 270 pre-tests to those two schools, to determine whether applying the BODMAS principle correctly is also a problem for learners in those schools.\r 9. Implement the BODMAS learning tool into the intermediate phase syllabus.\r Data:\r 1. Of the 41 educators in the sample, 52% think the standard of maths in their schools is average.\r 2. 38.9% of a grade 8 mathematics examination paper and 46% of grade 8 mathematics tests contains questions that are BODMAS related.\r 3. The learners achieved an overall average of 22.57% in the pre-test\r 4. The educators evaluated the BODMAS learning tool as very good as it is.\r 5. Learners and educators enjoyed the implementation session of the BODMAS learning tool.\r 6. In the post test learners did much better, the overall average increased by 21.00% to 43.57%.\r 7. Educators were positive about the way in which the tool was explained.\r 8. The learners in the other two schools also struggled with applying the BODMAS principle.\r 9. A second pilot study is being done in four primary schools by the Department of Education for the implementation in the Free State mathematics 2013 syllabus. \r Conclusion:\r My hypothesis is supported. \r 1) A poor understanding and inability of Grade 7 and 8 learners to apply the BODMAS principle in mathematics, influenced scores obtained in a mathematics test.\r 2) Scores obtained in the given mathematics test were improved with a BODMAS learning tool.