數學

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.

令Sn 為{1,2,…,n}任意排列所成的集合，π ? Sn 為其中的一個元素，我們記π = (π(1), π(2),…, π(n))。今給定π ? Sn ，若對所有i,1? i ? n，都有π (π (i)) = i 時，我們稱π 為involution。假設π ? Sn ，並給定σ ? Sm (m ? n)，當π 中任取m 項，其大小關係的順序都和σ 不同，我們稱π 避開σ，或稱π 是一個σ-avoiding 排列。在這篇報告中，我們主要分析了2143-avoiding involution，1234-avoiding involution，和3412-avoiding involution 中的一些統計量，給出了十數個結果與幾個猜想。Let Sn be the set of permutations on {1,2,…,n} and π ? Sn be an element in Sn. Denote π as π = (π(1), π(2),…, π(n)). We say that π is an involution if π(π(i)) = i for every i, 1? i ? n. Given π ? Sn and σ ? Sm (m ? n) , we say that π avoids σ (or π is an σ-avoiding permutation) if π does not contain any m-term subsequence in the order of σ. In this paper, we discuss some classic statistics on 2143-avoiding involutions, 1234-avoiding involutions and 3412-avoiding involutions. We get many new results in this field and give some interesting conjectures.

In this study, we mainly explore the geometric construction in 3D. By conducting some problems about constructing circles, we define the PLC construction in 2D as constructing a circle, either passing through a given point (P), tangent to a given line (L) or tangent to a given circle (C). Besides, we aim to discuss the properties of the PLC construction and the relations between each other. We discover if we find a plane satisfying certain conditions in space, the properties in the PLC construction can apply to such a plane. Furthermore, we extend the properties in PLC to the PES construction in 3D, defined as constructing a sphere, either passing through a given point (P), tangent to a given plane (E) or tangent to a given sphere (S). Also we discuss the relations among them.這個研究主要在探討3D 的尺規實作。藉由歸納某些有關作圓的題目，我們定義2D 中的PLC作圖─作圓，過已知點（P）、切已知線（L）、切已知圓（C）。並探討PLC 作圖的性質及彼此的關聯性。而我們發現：在空間中只要找到滿足特定條件的平面，則2D 幾何作圖性質在該平面仍能沿用。此外，運用PLC 作圖性質，我們進一步推廣到空間中的PES 作圖─作球，過已知點（P）、切已知面（E）、切已知球（S），並探討各個類型間的關聯性。

高斯曾經提出八皇后問題：八個皇后在8 × 8 的棋盤上有幾種放法可以使任意兩皇后不會互相攻擊？我們在原來的問題上加上一些條件，改變皇后攻擊規則，使得皇后失去一條對角線的攻擊方向，稱之為「跛腳皇后」。我們稱一個在棋盤上放置最多跛腳皇后使其不互相攻擊的放法為好放法；研究跛腳皇后放置在各類棋盤上其好放法的個數和性質。我們分別在六種棋盤上做討論：（1） 在平面n x n 棋盤上，我們證明了其好放法與完美極致史考倫型數列之間的對應關係，並歸納出相關的性質和定理。（2） 在平面m x n 棋盤上，我們固定一邊長度n，做出n 較小時好放法數的通式；我們也將其好放法對應至廣義史考倫。（3） 在環面n x n 棋盤上，我們說明了其好放法與完全剩餘系排列之間的對應關係，並歸納出相關的性質和定理。（4） 在環面m x n 棋盤上，我們固定gcd（m,n），做出gcd（m,n）較小時好放法數的通式。（5） 在柱面n x n 棋盤上，我們證明其與環面n x n 棋盤等價，說明其好放法具有和環面n x n 棋盤好放法相同的性質和定理。（6） 在柱面m x n 棋盤上，分成左右柱面以及上下柱面來做討論。我們歸納出相關性質和定理；並固定一邊長度n，做出n 較小時好放法數的通式。Gauss had researched about putting eight queens on the chessboard on the way that doesn’t make any queen attacks another one. Thus, we added some rules on the question: the queen loses one diagonal attacking-way and become the “lame queen”. We call a way that doesn’t make any lame queen attacks another one “a good way”. We have been investigating the amount and properties of good ways based on six kinds of chessboard: （1）We found the correspondences between the “good way” on n × n plane chessboard and the Perfect extremal Skolem-type sequence, and concluded some associated properties and theorems. （2） On m× n plane chessboard, we fixed the length n of one side of the chessboard, and accomplished the amount of good ways when n is small. We also correspond the good way to the Generalized Skolem.（3）We found the correspondences between the “good way” on n × n torus chessboard and the arrayal of complete residue system, and concluded some associated properties and theorems.（4）On m× n torus chessboard, we fixed the gcd(m,n) （greatest common divisor of m and n）, and accomplished the amount of good ways when gcd(m,n) is small.（5）On n × n cylinder chessboard, we proved that this kind of chessboard is equal to torus chessboard. So the good ways, characters, and theorems on cylinder chessboard are the same as on the torus one.（6）On m× n cylinder chessboard, we separate it into two cases: left-right cylinder chessboard and up-down cylinder chessboard. We concluded some associated properties and theorems, and we also fixed the length n of one side of the chessboard and accomplished the amount of\r the good ways when n is small.

給定下面範例：\r 058823529411764705882 +235294117647058823532\r =0588235294117647058823529411764705882353，\r 其等式結果與質數17 的倒數結果(1/17)有某種關聯(卻沒有一個決定性的證據)，意即\r 1/17=0.0588235294117647=\r 0.058823529411764705882352941176470588235294117647...... ( Len(17) =16 )\r \r 曾經在下列網站上發現過幾組數字(挑戰試題)，引起我們極大的興趣。\r http://www.domino.research.ibm.com/Comm/wwwr_pondernsf/challenges/March2000.html\r http://www.math.smsu.edu/~les/POW08_96.html\r \r \r The two examples that I have are 0588 2+23532=05882353 and 058823529411764705882+23529411764705882353 2=0588235294117647058823529411764705882353 These were found by the Canadian professor Alf van der Poorten, and he gave a talk on these identities in December at the west coast number theory conference. He was unspecific as to exactly where these identities were coming from, but they are connected with reciprocals of primes:1/17 = 0.0588235294117647= 0.058823529411764705882352941176470588235294117647 ΛΛ ( Len(17) = 16 ) Though not mentioning how to obtain these equations, Prof. Poorten demonstrated the relationship between the above examples and the reciprocal of the prime numbers 17 (1/17 ) without a definitive proof.

在這個科展中我們要研究兩個非常有趣的問題：\r 停車場問題 與 彈硬幣遊戲.\r 停車場問題是這樣的：在一條單行道上有n個車位，編號從1到n。現在有n個司機排成一排要進入停車。但是每個司機都有怪癖，各自有最想要停的位子。他們依序將車子開進單行道，如果想要停的位子是空的，當然停在這個位子。但是如果不巧那個位子已經被停了，不得已只好找下一個空位，姑且停之。但是如果往下找都沒有空位，由於是單行道，司機就只好開走不停了。\r 比如說，如果現在有五輛車，司機的喜好分別是(3,1,2,5,2)。則五輛車都可以順利停車。但是司機的喜好如果是(3,1,4,5,4)，有些車就無法停車了。\r 彈硬幣遊戲是這樣的：考慮圓內接正n+1邊形，任意兩點都連線。這正n+1邊形中有一個頂點P是特殊的，每個頂點上一開始都放有一些硬幣(各點硬幣數可以不同)。如果P以外的某個頂點上的硬幣數n個，我們可以對這個頂點進行操作：一次操作是指將這個頂點上的硬幣各分一個給每個其他頂點。點P只在其他點都無法操作時操作。我們不理會頂點P上的錢數，因此這個遊戲可以無限地玩下去。

如果n-1 個正四面體在n 個面的多面體棋盤中翻轉(如：正八面體、正二十面體、五邊雙菱錐及三角錐台棋盤)，則正四面體相互之間會受到很多限制。本研究為探討n-1 個正四面體在n 個面的多面體棋盤中，利用其翻轉的特性，翻轉至相鄰空格，進而完成：1. 每個正四面體的底面及側面均能「同色共面」。2. 在滿足以上的兩個條件下做「數字排序」。3. 探討每個遊戲在各階段是否有解的情形。最後，將遊戲推廣至正八面體在多面體棋盤中的翻轉。If n-1 pyramids can turn in the chessboard of n planes pyramids (ex: 8 planes pyramid, 20 planes pyramid, 5 sides bi rhombus pyramid and triangular pyramid chessboard) there are lost of restrictions between these pyramids. This research is discussing about the n-1 pyramids in the chessboard of n planes multi pyramid, which have the peculiarity of turn that can turn to nearby space. We found that: 1. Each pyramid’s button plane and aide plane can be “same plane same color”. 2. Matching about 2 conditions then can do “order by numbers”. 3. Search for answers of each levels in each game. Finally, the game will be propagated to eight planes pyramid can turn on the chessboard of multi pyramid.

At the website “MathLinks EveryOne,” we found a problem “Snakes on a chessboard,” which was raised by Prof. Richard Stanley. The following is the problem. A snake on the m n chessboard is a nonempty subset S of the squares of the board with the following property: Start at one of the squares and continue walking one step up or to the right, stopping at any time. The squares visited are the squares of the snake. Prove that the total number of ways to cover an m × n chessboard with disjoint snakes is a product of Fibonacci numbers. We call the total number of ways to cover a chessboard with disjoint snakes “the snake-covering number.” This problem hasn’t been solved since it was posted on September 18, 2004, so it aroused our interest to study it. First, we used the way in which we added each block to the chessboard, and therefore we discovered some regulations about the snake-covering number of the1 × n , 2 × n and 3 × n chessboard. Through “recursive relation” and “mathematical induction”, we proved the general term of the snake-covering number of the1 × n , 2 × n and 3 × n chessboard. In the following study, we found a key method in which we added a group of blocks to the chessboard. Finally, we proved the general term of the snake-covering number of the m × n chessboard. Also, we discovered the way to figure out the snake-covering number of the nonrectangular chessboard.在網站“ MathLinks EveryOne ”中，我們找到了一個有趣的問題“棋然上的蛇” ( Snakes on a chessboard ) ，這個問題是由教授 Richard Stanley 所提出。問題如下：在m x n棋盤形格子上，蛇由任意一格出發，但蛇的走法只能往右 → ，往上↑，或停住 ‧ 若此蛇已停住，將由另一條蛇來走，且不同蛇走過的格子不可重疊”證明：將 m × n 棋盤形格子完全覆蓋的總方法數為費氐（ Fibonacci ）數列某些項的乘積。我們將把棋盤形格子完全覆蓋的所有方法數稱之為“蛇填充數” 由於這個問題自從 2004年 9 月 18 日被登在網站上後，還沒有人提出解答，於是引發了我們研究的興趣。首先，我們使用了將一個一個格子加到棋盤上的方法，並發現了 l × n 、 2 x n、 3 × n 棋盤形格子蛇填充數的一些規律。我們使用遞迴關係及數學歸納法來證明 l x n 、 2 x n ， 3 × n 棋盤形格子蛇填充數的一般項。在接下來的研究中我們發現一個特別的方法，一次增加數個方塊 ‧ 最後我們證明了，m x n, ，棋然形格子的蛇填充數的一般項 ‧ 而且，我們也找到如何求出不規則棋盤形格子的蛇填充數。

Generalized Petersen graph P(n,k)，定義為n 為不小於2 的整數以及1≤ k ≤ n−1，有頂點{ u0, u1, . . . , un−1, v0 , v1 , . . . , vn−1 }，及路徑{ uiui+1 , uivi , vivi+k：1≤ i ≤ n−1 }。在 [2] 中，我們可以知道P(n,5) 是Hamiltonian 等價於當n≠11。 在這一篇報告中，我們證明當generalized Petersen graph P(n,5) 是hyper Hamiltonian（一種Hamiltonian graph 再去掉任何一點後，仍然是Hamiltonian graph）的充要條件是n 為不等於11 的奇數且n ≥ 7。 The generalized Petersen graph P(n,k), n ≥ 2 and 1≤ k ≤ n−1, has vertex-set { u0, u1, . . . , un−1, v0 , v1 , . . . , vn−1 } and edge-set { uiui+1 , uivi , vivi+k：1≤ i ≤ n−1 with subscripts reduced modulo n}. And we can know that P(n,5) is Hamiltonian if and only if n≠11 from [2].In this paper it is proved that generalized Petersen graph P(n,5) is Hyper Hamiltonian (A Hamiltonian graph can still be a Hamiltonian graph when any one of the nodes fault) if and only if n is odd and n≠11.