Using the digits 0-9, each only once, construct A+B=C.
arithmetic/digits/equations/ten.digits.sum.product
Problem
Using the digits 0-9, each only once, construct A+B=C. Construct AxB=C the same way. Try each of these excluding 0. - Using the digits 0-9, construct A+B=C so that each digit is only used once. - Using the digits 0-9, construct AxB=C so that each digit is used only once. - Try each of the above, using excluding 0. ... the original inspiration was "The Moscow Puzzles" where an incomplete answer to the first question is given. -- Harold Cataquet <100341.1300@compuserve.com>, March 1996
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Reading: each listed digit is used exactly once in total, numbers have no leading zeros, and A + B and B + A (A x B and B x A) count as one solution. All solutions were found by a C++ program that tries every arrangement of the digits and every split into A, B, C. 1. A + B = C with the digits 0-9: 84 solutions. C always has 4 digits, and A, B have 2 + 4 or 3 + 3 digits. For example 246 + 789 = 1035 and 26 + 4987 = 5013. All of them: 26 + 4987 = 5013 74 + 5938 = 6012 289 + 746 = 1035 439 + 587 = 1026 27 + 4986 = 5013 75 + 1968 = 2043 289 + 764 = 1053 452 + 637 = 1089 34 + 5978 = 6012 76 + 1958 = 2034 324 + 765 = 1089 457 + 632 = 1089 34 + 5987 = 6021 78 + 1956 = 2034 325 + 764 = 1089 473 + 589 = 1062 37 + 5984 = 6021 78 + 1965 = 2043 342 + 756 = 1098 473 + 625 = 1098 38 + 5974 = 6012 78 + 5934 = 6012 346 + 752 = 1098 475 + 623 = 1098 43 + 5978 = 6021 78 + 5943 = 6021 347 + 859 = 1206 476 + 829 = 1305 47 + 2968 = 3015 84 + 2967 = 3051 349 + 857 = 1206 479 + 583 = 1062 48 + 2967 = 3015 84 + 5937 = 6021 352 + 746 = 1098 479 + 826 = 1305 48 + 5973 = 6021 86 + 1957 = 2043 356 + 742 = 1098 483 + 579 = 1062 56 + 1978 = 2034 86 + 4927 = 5013 357 + 849 = 1206 487 + 539 = 1026 56 + 1987 = 2043 87 + 1956 = 2043 359 + 847 = 1206 489 + 537 = 1026 57 + 1986 = 2043 87 + 2964 = 3051 364 + 725 = 1089 489 + 573 = 1062 58 + 1976 = 2034 87 + 4926 = 5013 365 + 724 = 1089 624 + 879 = 1503 64 + 2987 = 3051 87 + 5934 = 6021 423 + 675 = 1098 629 + 874 = 1503 65 + 1978 = 2043 246 + 789 = 1035 425 + 673 = 1098 674 + 829 = 1503 67 + 2948 = 3015 249 + 786 = 1035 426 + 879 = 1305 679 + 824 = 1503 67 + 2984 = 3051 264 + 789 = 1053 429 + 876 = 1305 743 + 859 = 1602 68 + 1975 = 2043 269 + 784 = 1053 432 + 657 = 1089 749 + 853 = 1602 68 + 2947 = 3015 284 + 769 = 1053 437 + 589 = 1026 753 + 849 = 1602 73 + 5948 = 6021 286 + 749 = 1035 437 + 652 = 1089 759 + 843 = 1602 Why C is a multiple of 9: the digit sums of A, B and C add to 0 + 1 + ... + 9 = 45, and A + B = C gives digitsum(A) + digitsum(B) = digitsum(C) mod 9, so 2 * digitsum(C) = 45 = 0 mod 9. Indeed the 17 values of C (1026, 1035, 1053, 1062, 1089, 1098, 1206, 1305, 1503, 1602, 2034, 2043, 3015, 3051, 5013, 6012, 6021) all have digit sum 9 or 18. 2. A x B = C with the digits 0-9: 22 solutions, all 1 x 4 = 5 or 2 x 3 = 5 digits: 3 x 5694 = 17082 6 x 5817 = 34902 45 x 396 = 17820 3 x 6819 = 20457 7 x 3094 = 21658 46 x 715 = 32890 3 x 6918 = 20754 7 x 4093 = 28651 52 x 367 = 19084 3 x 8169 = 24507 7 x 9304 = 65128 54 x 297 = 16038 3 x 9168 = 27504 7 x 9403 = 65821 63 x 927 = 58401 4 x 3907 = 15628 27 x 594 = 16038 78 x 345 = 26910 4 x 7039 = 28156 36 x 495 = 17820 4 x 9127 = 36508 39 x 402 = 15678 3. A + B = C with the digits 1-9 (no 0): 168 solutions, all of the form 3 + 3 = 3 digits, e.g. 124 + 659 = 783, 125 + 739 = 864, 127 + 359 = 486. By the same digit-sum argument C is a multiple of 9; there are 31 different values of C, from 459 to 981, each with 4 or 8 solutions (e.g. 783 = 124 + 659 = 129 + 654 = 154 + 629 = 159 + 624 and four more). 4. A x B = C with the digits 1-9: 9 solutions: 4 x 1738 = 6952 18 x 297 = 5346 39 x 186 = 7254 4 x 1963 = 7852 27 x 198 = 5346 42 x 138 = 5796 12 x 483 = 5796 28 x 157 = 4396 48 x 159 = 7632 (5346 and 5796 each arise twice.) -- Claude (rec.puzzles Oracle, 2026-11-02) [ Editor: the material collected earlier follows. ] [ Editor: the proposer offered to send his answers, which are not in the collected material. ]