Use the digits 1-9 once each to create two multiplication problems, each with the same solution. That is, (AxB) = (CxD), and all nine digits are used once to make the numbers (9 digits total in A, B, C, and D.) How many such pairs of problems are there?
Source: Based on a Martin Gardner puzzle.
Assuming that A > B, C > D and A > C, we get 59 solution pairs to the problem A x B = C x D. They are listed below: 134 x 29 = 67 x 58 138 x 27 = 69 x 54 146 x 29 = 73 x 58 158 x 23 = 79 x 46 158 x 32 = 79 x 64 174 x 23 = 69 x 58 174 x 32 = 96 x 58 186 x 27 = 93 x 54 259 x 18 = 74 x 63 459 x 8 = 136 x 27 476 x 9 = 153 x 28 532 x 14 = 98 x 76 534 x 9 = 267 x 18 538 x 7 = 269 x 14 546 x 9 = 273 x 18 584 x 12 = 96 x 73 586 x 7 = 293 x 14 638 x 7 = 154 x 29 654 x 9 = 327 x 18 658 x 7 = 329 x 14 759 x 8 = 132 x 46 782 x 9 = 153 x 46 897 x 4 = 156 x 23 984 x 7 = 123 x 56 1358 x 2 = 679 x 4 1372 x 4 = 98 x 56 1458 x 3 = 729 x 6 1538 x 2 = 769 x 4 1576 x 2 = 394 x 8 1584 x 3 = 792 x 6 1586 x 2 = 793 x 4 1728 x 3 = 96 x 54 1746 x 3 = 582 x 9 1749 x 2 = 583 x 6 1756 x 2 = 439 x 8 1824 x 3 = 96 x 57 1854 x 3 = 927 x 6 1869 x 2 = 534 x 7 1943 x 2 = 67 x 58 3451 x 2 = 986 x 7 3458 x 2 = 91 x 76 3672 x 1 = 459 x 8 3752 x 1 = 469 x 8 4296 x 1 = 537 x 8 4632 x 1 = 579 x 8 4736 x 1 = 592 x 8 4876 x 1 = 92 x 53 5392 x 1 = 674 x 8 5394 x 1 = 87 x 62 5432 x 1 = 679 x 8 5742 x 1 = 638 x 9 5823 x 1 = 647 x 9 5936 x 1 = 742 x 8 6352 x 1 = 794 x 8 7456 x 1 = 932 x 8 7524 x 1 = 836 x 9 7536 x 1 = 942 x 8 7624 x 1 = 953 x 8 7632 x 1 = 954 x 8