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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: is there a mistake in the puzzle? | Comment 3 of 5 |
(In reply to is there a mistake in the puzzle? by Charlie)

Why did you stop short of 9999, indeed you stopped short of 1111.  If the number is to be divisible by 1111 it must be in this higher range.

  Posted by Jer on 2011-12-26 11:29:12

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