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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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Solution Solution (Spoiler) | Comment 4 of 5 |

The sum of the 3 squares has to be 1111, 2222, 3333, ..., or 9999.

If x is the central of the 3 numbers, then
x^2 + (x-2)^2 + (x+2)^2 = N, where N is one of the 9 numbers above.

3x^2 + 8 = N, so mod(N,3) = 2; so N can only be:
2222, or 5555, or 8888
Dividing by 3 and taking the square root to get an approximation of x leads to the following triplet.

(41, 43, 45)


  Posted by Larry on 2011-12-26 11:32:42
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