Introduction

1. Two and Three Dimensional Silver Cubes Taeler Porter and Scott GildemeyerAdvisor: Dr. Abdollah Khodkar Department of Mathematics, Conclusions and Future Work We have learned to correctly use the formulas to help find possible numbers to place in the silver cells. Using these formulas we found different possibilities of two and three dimensional silver cubes. From our research we have found the three dimensional order of 7, found by Kevin Ventullo, which was the goal for this research. Currently we are working on finding the algorithm for the 7 x 7. The next step is to find the three dimensional order eleven and its algorithm. Introduction Results An n × n matrix A is said to be silver if, for every i = 1, 2, . . . , n, each symbol in {1, 2, . . . , 2n − 1} appears either in the ith row or the ith column of A. A problem of the 38th International Mathematical Olympiad in 1997 introduced this definition and asked to prove that no silver matrix of order 1997 exists. In [2] the motivation behind this problem as well as a solution is presented: a silver matrix of order n exists if and only if n = 1 or n is evenfor two dimensional silver matrix. The next step was to find three dimensional silver matrix. All were found up to the 7 x 7. It was an open problem and we were chosen to do the research to find the 3 dimensional matrix of order 7. 2 dimensional order 2 2 dimensional order 4 2 dimensional order 6 3 dimensional order 3 Methods Literature cited [1] F. J. MacWilliams and N. J. A. Sloane, The theory of error-correcting codes II, North-Holland Publishing Co., Amsterdam, 1977. [2] M. Mahdian and E. S. Mahmoodian, The roots of an IMO97 problem, Bull. Inst. Combin. Appl. 28 (2000), 48–54. [3] P. J. Wan, Near-optimal conflict-free channel set assignments for an optical clusterbased hypercube network, J. Comb. Optim., 1 (1997), pp. 179–186. 2n – 1 , 3n – 2 These are the two equations used to figure out the highest consecutive number in the order n. 3 dimensional order 4 • Example • 2 dimensional order of 4.2(4) – 1= 7 so the cube would go 1 – 7. • 3 dimensional order of 4. 3(4) – 2 = 10 so the cube would go 1 – 10. 3 dimensional order 5 Acknowledgments We would like to acknowledge the National Science Foundation STEP grant #DUE-0336571 for the opportunity to research. Also, the GEMS Summer Fellows Program for the experience of research and skills. nd-1 _ (# of repetition)= multiple of d This is the equation to find possibilities of silvers for the dimension d. If the equation is true, then there is a possibility that the number of repetitions will work in order to create the silver cube of order n. • Example • 2 dimensional order of 4 with numbers repeating 3 times. 42 – (3) = multiple of 2 • 16 – 3 = 13 • The repetition of 3 is not a possibility. • 3 dimensional order of 4 with numbers repeating 4 times. 42 – 4 = multiple of 3 • 16 – 4 = 12 • The repetition of 4 is a possibility. Further information tporter1@my.westga.edu – Taeler Porter email sgildem1@my.westga.edu – Scott Gildemeyer email akhodkar@westga.edu – Abdollah Khodkar email 3 dimensional order 6