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Powering up the digits IV (Posted on 2011-04-21) Difficulty: 3 of 5
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.

No Solution Yet Submitted by K Sengupta    
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Solution solution | Comment 1 of 2
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

  Posted by Dej Mar on 2011-04-21 18:42:08
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