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Powering up the digits IV (Posted on 2011-04-21) |
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Each of X1, X2 and X3 represents a nonzero digit of the 3-digit base M positive integer X1X2X3; where X1, X2 and X3 are not necessarily distinct.
Determine the possible positive integer values of M, with 7 ≤ M ≤ 206, such that this equation has at least one valid solution.
X1X2 + X2X3 + X3X1 = X1X2X3 ± 6
Note: X1X2X3 denotes the concatenation of the three digits.
solution
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M = {7, 8, 9, 11, 12, 14, 15, 17, 19, 20, 22, 23, 25, 28, 30, 62, 118, 126} 1267 = 12 + 26 + 61 -2 3257 = 32 + 25 + 53 3347 = 33 + 34 + 43 3437 = 34 + 43 + 33 +6 5137 = 51 + 13 + 35 +6 1348 = 13 + 34 + 41 +6 2538 = 25 + 53 + 32 +5 2728 = 27 + 72 + 22 +5 4428 = 44 + 42 + 24 +2 7328 = 73 + 32 + 27 -6 1539 = 15 + 53 + 31 2279 = 22 + 27 + 72 +6 23511 = 23 + 35 + 52 +4 24411 = 24 + 44 + 42 +2 33511 = 33 + 35 + 53 +6 61311 = 61 + 13 + 36 +4 16312 = 16 + 63 + 31 -1 22812 = 22 + 28 + 82 -4 13514 = 13 + 35 + 51 -6 14414 = 14 + 44 + 41 -5 29114 = 29 + 91 + 12 -3 12815 = 12 + 28 + 81 -2 36215 = 36 + 62 + 23 -6 25417 = 25 + 54 + 42 -6 36317 = 36 + 63 + 33 18319 = 18 + 83 + 31 32920 = 32 + 29 + 93 -1 24522 = 24 + 45 + 52 -4 19323 = 19 + 93 + 31 +6 52525 = 52 + 25 + 55 -2 37328 = 37 + 73 + 33 -6 14530 = 14 + 45 + 51 -5 14662 = 14 + 46 + 61 -5 755118 = 75 + 55 + 57 +6 147126 = 14 + 47 + 71 -5
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Posted by Dej Mar
on 2011-04-21 18:42:08 |
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