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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.

See The Solution Submitted by Ady TZIDON    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): is there a mistake in the puzzle? Comment 5 of 5 |
(In reply to re: is there a mistake in the puzzle? by Jer)

I goofed.
  Posted by Charlie on 2011-12-26 14:54:09

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