BMAD 62 Report post Posted July 29, 2013 Classic magic number square problem but hopefully with a slight twist. If all the rows, columns, and diagonals add up to 33 using any number you desire. Is the center uniquely determined? Share this post Link to post Share on other sites

0 Rainman 14 Report post Posted July 29, 2013 Yes. Let's label the squares as follows: ABC DEF GHI Now if we add them row by row we get 3*33 = (A+B+C)+(D+E+F)+(G+H+I). However, if we add the four lines that go through the center, we get 4*33 = (A+E+I)+(B+E+H)+(C+E+G)+(D+E+F) = (A+B+C+D+E+F+G+H+I)+3*E. Subtract the first equation from the second to get 33=3*E, which fixes E at 11. Share this post Link to post Share on other sites

0 gavinksong 11 Report post Posted July 29, 2013 (edited) Here's a simple magic square that adds up to 15: 4 9 2 3 5 7 8 1 6 We can add (33 - 15) / 3 = 6 to each cell to raise the total up to 33, turning our original square into: 10 15 8 9 11 13 14 7 12 We also have the option of doubling each cell and then incrementing each cell by one to raise the sum to 33, and that gives us: 9 19 5 7 11 15 17 3 13 which I honestly thought would give us a different center, but I guess not... Are we allowed to have negative numbers or fractions? Edited July 29, 2013 by gavinksong Share this post Link to post Share on other sites

0 BMAD 62 Report post Posted July 29, 2013 Here's a simple magic square that adds up to 15: 4 9 2 3 5 7 8 1 6 We can add (33 - 15) / 3 = 6 to each cell to raise the total up to 33, turning our original square into: 10 15 8 9 11 13 14 7 12 We also have the option of doubling each cell and then incrementing each cell by one to raise the sum to 33, and that gives us: 9 19 5 7 11 15 17 3 13 which I honestly thought would give us a different center, but I guess not... Are we allowed to have negative numbers or fractions? yes. any type of real number (no complex) Share this post Link to post Share on other sites

Classic magic number square problem but hopefully with a slight twist.

If all the rows, columns, and diagonals add up to 33 using any number you desire. Is the center uniquely determined?

## Share this post

## Link to post

## Share on other sites