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The class of ‘55’ (Posted on 2021-02-27) Difficulty: 3 of 5
How many numbers comply with the two conditions:
1. It is a permutation of 5 consecutive digits.
2. It is divisible by 55.

See The Solution Submitted by Ady TZIDON    
Rating: 5.0000 (1 votes)

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Solution More if wraparound at 9,0 is considered consecutive. | Comment 2 of 4 |
(In reply to Let me count the ways (spoiler) by Steve Herman)

I found the same 20 as Steve Herman, 4 for each of the schemes below (the second is just the first with a,c and b,d swapped):
(a,c) (b,d) e 
(1,4) (2,3) 0 
(2,3) (1,4) 0 
(2,3) (4,6) 5 
(6,7) (3,4) 5
(4,6) (7,8) 5

But if strings such as 67890 and 89012 are considered consecutive, then there are 4 more groups, or 16 more possibilites:
(a,c) (b,d) e 
(6,9) (7,8) 0
(7,8) (6,9) 0
(2,8) (1,9) 0
(1,9) (2,8) 0

  Posted by Larry on 2021-02-27 10:55:54
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