perplexus.info :: Numbers : The class of ‘55’

The class of ‘55’ (Posted on 2021-02-27)

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)

Comments: ( Back to comment list | You must be logged in to post comments.)

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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