1. Introduction Methods 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. 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. Taeler Porter and Scott Gildemeyer Advisor: Dr. Abdollah Khodkar 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. Further information tporter1@my.westga.edutporter1@my.westga.edu – Taeler Porter email sgildem1@my.westga.edusgildem1@my.westga.edu – Scott Gildemeyer email akhodkar@westga.eduakhodkar@westga.edu – Abdollah Khodkar email Two and Three Dimensional Silver Cubes Department of Mathematics, 12 31 1547 3126 6735 2413 13261110 41 395 5268 11 845723 973 18 794261 2 dimensional order 6 2 dimensional order 2 2 dimensional order 4 752 246 321 314 537 263 643 124 475 3 dimensional order 3 1397 4279 5613 65810 2479 3197 65 8 5631 8 13 856 1324 7942 865 8 31 9742 3124 3 dimensional order 4 n d-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. 4 2 – (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. 4 2 – 4 = multiple of 3 16 – 4 = 12 The repetition of 4 is a possibility. 2n – 1, 3n – 2 These are the two equations used to figure out the highest consecutive number in the order n. Results 12345 6110117 7101132 87 129 9116813 103274 1295108 864312 1211413 25 611 51316 9111223 133798 2124310 741129 48101 561312 12953 685114 43712 1361189 27184 110613 5102137 312495 3 dimensional order 5 1238910 4519 8 716 89 111213123 1213114141 13111216115 6719108 312 89 14589 12131115161 131112312 1112131414 5141089 16789 2319 8 1311121414 11121311516 121311231 141516123 1516144111 16141513112 123567 481675 1019756 16151412131 161415312 1415161411 9101675 312756 148567 1614151114 14151611213 151614231 814756 1910567 231675 3 dimensional order 6 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 even for 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. 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.