If we place 192 in the 1st row, 384 in the second row and 576 in the 3rd row, then we get a 3x3 square, exhausting all nonzero digits.
In this square 1st row + 2nd row = 3rd row.
Please provide all other arrangements, satisfying this condition.
Due to the commutatative property of addition, and as each 3digit number abc (or def) is equivalent to a*100 + b*10 + c (or d*100 + e*10 + f), there are a total of 8 * 42 = 366 ways for the digits in the square to be arranged
[e.g., for abc + def = sum, where a, b, c, d, e & f are distinct digits, there are:
(1) abc + def = sum (2)def + abc = sum
(3) aec + dbf = sum (4)dbf + aec = sum
(5) abf + dec = sum (6)dec + abf = sum
(7) aef + dbc = sum & (8)dbc + aef = sum]
In the following, only the 42 arrangements abc + def = sum are given:
ROW1+ROW2 = ROW3
173 + 286 = 459
173 + 295 = 468
127 + 359 = 486
127 + 368 = 495
162 + 387 = 549
128 + 439 = 567
218 + 349 = 567
182 + 394 = 576
216 + 378 = 594
152 + 487 = 639
251 + 397 = 648
218 + 439 = 657
182 + 493 = 675
281 + 394 = 675
215 + 478 = 693
143 + 586 = 729
142 + 596 = 738
124 + 659 = 783
214 + 569 = 783
134 + 658 = 792
243 + 576 = 819
352 + 467 = 819
142 + 695 = 837
241 + 596 = 837
317 + 529 = 846
125 + 739 = 864
271 + 593 = 864
214 + 659 = 873
234 + 657 = 891
324 + 567 = 891
243 + 675 = 918
342 + 576 = 918
341 + 586 = 927
152 + 784 = 936
162 + 783 = 945
317 + 628 = 945
216 + 738 = 954
271 + 683 = 954
215 + 748 = 963
314 + 658 = 972
235 + 746 = 981
324 + 657 = 981

Posted by Dej Mar
on 20110914 19:25:20 