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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 | Comment 2 of 5 |
For simplicity call the numbers a-2, a, a+2
Then the sum of squares becomes 3a^2 + 8

We are seeking a solution to 3a^2 + 8 = 1111n where n is a single digit.
Solving for a:
a = √((1111n-8)/3)
and trying the 9 possibilities of n finds the integer solution
n=5, a=43
So the triplets are
(41,43,45) and (-45,-43,-41)

  Posted by Jer on 2011-12-26 11:25:28
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