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A repunit number (Posted on 2011-12-26) Difficulty: 2 of 5
3 integers form an arithmetic progression with d=2 i.e. a, a+2, a+4.
The sum of their squares is a 4-digit number divisible by 1111.
List all possible triplets.

No Solution Yet Submitted by Ady TZIDON    
Rating: 4.0000 (1 votes)

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Question is there a mistake in the puzzle? | Comment 1 of 5

The relevent subset of such progressions follows, together with the sum of the squares of their members:

-22 -20 -18    1208
-21 -19 -17    1091
-20 -18 -16    980
-19 -17 -15    875
-18 -16 -14    776
-17 -15 -13    683
-16 -14 -12    596
-15 -13 -11    515
-14 -12 -10    440
-13 -11 -9     371
-12 -10 -8     308
-11 -9 -7      251
-10 -8 -6      200
-9 -7 -5       155
-8 -6 -4       116
-7 -5 -3       83
-6 -4 -2       56
-5 -3 -1       35
-4 -2  0       20
-3 -1  1       11
-2  0  2       8
-1  1  3       11
 0  2  4       20
 1  3  5       35
 2  4  6       56
 3  5  7       83
 4  6  8       116
 5  7  9       155
 6  8  10      200
 7  9  11      251
 8  10  12     308
 9  11  13     371
 10  12  14    440
 11  13  15    515
 12  14  16    596
 13  15  17    683
 14  16  18    776
 15  17  19    875
 16  18  20    980
 17  19  21    1091
 18  20  22    1208
 19  21  23    1331

No sum seems to be a rep-digit.


  Posted by Charlie on 2011-12-26 10:59:43
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