superprismatic Posted July 1, 2010 Report Share Posted July 1, 2010 Rearrange the rows of the following matrix so that each column of the resulting matrix is an arithmetic progression modulo 31 reading from top to bottom (or, equivalently, from bottom to top). 30 8 30 5 7 25 15 10 13 4 21 7 10 21 28 14 5 1 11 25 25 12 2 27 29 4 2 7 10 30 28 0 5 9 13 4 5 1 12 24 20 26 0 12 27 3 4 20 24 15 17 17 25 24 7 9 6 20 14 16 19 26 1 27 9 18 26 23 11 13 19 14 8 13 3 18 11 11 4 28 15 26 23 0 17 18 7 0 15 28 2 20 21 30 29 10 22 7 23 21 0 3 13 2 19 8 27 17 21 3 26 23 3 8 9 17 0 16 8 6 14 6 16 6 24 12 25 19 14 15 22 19 23 11 20 19 28 29 6 5 9 22 17 21 27 27 8 11 16 9 20 30 29 25 6 13 16 14 10 2 18 2 18 22 4 29 12 23 1 16 24 1 1 15 3 24 18 28 26 5 12 29 22 10 30 22 [/code] Quote Link to comment Share on other sites More sharing options...
0 bushindo Posted July 2, 2010 Report Share Posted July 2, 2010 (edited) Rearrange the rows of the following matrix so that each column of the resulting matrix is an arithmetic progression modulo 31 reading from top to bottom (or, equivalently, from bottom to top). 30 8 30 5 7 25 15 10 13 4 21 7 10 21 28 14 5 1 11 25 25 12 2 27 29 4 2 7 10 30 28 0 5 9 13 4 5 1 12 24 20 26 0 12 27 3 4 20 24 15 17 17 25 24 7 9 6 20 14 16 19 26 1 27 9 18 26 23 11 13 19 14 8 13 3 18 11 11 4 28 15 26 23 0 17 18 7 0 15 28 2 20 21 30 29 10 22 7 23 21 0 3 13 2 19 8 27 17 21 3 26 23 3 8 9 17 0 16 8 6 14 6 16 6 24 12 25 19 14 15 22 19 23 11 20 19 28 29 6 5 9 22 17 21 27 27 8 11 16 9 20 30 29 25 6 13 16 14 10 2 18 2 18 22 4 29 12 23 1 16 24 1 1 15 3 24 18 28 26 5 12 29 22 10 30 22 You can satisfy the OP by reordering the rows so that the first column goes 0,1,2,3,... In general, you can construct a solution by ordering the rows so that the first column looks like the vector (a, 2a, ..., 31*a ) mod 31, for a = 1, 2, ... 30. For instance, for a = 2, the vector is (2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31)' For a = 5, the first column should be (5 10 15 20 25 30 4 9 14 19 24 29 3 8 13 18 23 28 2 7 12 17 22 27 1 6 11 16 21 26 31)' Edited July 2, 2010 by bushindo Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 2, 2010 Report Share Posted July 2, 2010 (edited) Edited July 2, 2010 by parik Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 2, 2010 Report Share Posted July 2, 2010 0 4 15 23 28 26 0 3 7 22 21 23 0 4 9 5 13 28 0 6 8 9 16 17 27 3 4 12 0 20 0 7 15 17 18 28 1 5 10 14 21 28 1 24 1 15 1 16 1 9 18 19 26 27 1 20 12 5 24 26 2 2 10 14 16 18 2 10 20 21 29 30 2 8 13 17 19 27 2 11 12 25 25 27 2 4 7 10 29 30 3 5 18 24 26 28 3 3 8 21 26 23 3 8 11 11 13 18 4 7 13 10 15 21 4 12 18 22 23 29 5 6 17 9 21 22 5 7 8 25 30 30 6 6 12 14 16 24 6 13 20 25 29 30 6 7 14 9 16 20 8 9 11 16 27 27 10 12 22 22 29 30 11 13 14 19 23 26 11 19 20 23 28 29 14 15 19 25 19 22 15 17 17 24 24 25 155 279 403 527 651 775 5.00 9.00 13.00 17.00 21.00 25.00 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 2, 2010 Report Share Posted July 2, 2010 5 7 8 25 30 30 4 7 13 10 15 21 1 5 10 14 21 28 2 11 12 25 25 27 2 4 7 10 29 30 0 4 9 5 13 28 1 20 12 5 24 26 27 3 4 12 0 20 15 17 17 24 24 25 6 7 14 9 16 20 1 9 18 19 26 27 11 13 14 19 23 26 3 8 11 11 13 18 0 4 15 23 28 26 0 7 15 17 18 28 2 10 20 21 29 30 0 3 7 22 21 23 2 8 13 17 19 27 3 3 8 21 26 23 0 6 8 9 16 17 6 6 12 14 16 24 14 15 19 25 19 22 11 19 20 23 28 29 5 6 17 9 21 22 8 9 11 16 27 27 6 13 20 25 29 30 2 2 10 14 16 18 4 12 18 22 23 29 1 24 1 15 1 16 3 5 18 24 26 28 10 12 22 22 29 30 155 279 403 527 651 775 5.00 9.00 13.00 17.00 21.00 25.00 I'm sure there are multiple answers though Quote Link to comment Share on other sites More sharing options...
Question
superprismatic
Rearrange the rows of the following
matrix so that each column of the
resulting matrix is an arithmetic
progression modulo 31 reading from
top to bottom (or, equivalently,
from bottom to top).
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