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 2N Digit Numbers (Posted on 2009-05-09)
A positive integer contains each of the base 2N digits from 0 to 2N - 1 exactly once such that the successive pairs of digits from left to right are divisible in turn by 2,3,....,2N. That is, the two digit base 2N number constituted by the ith digit from the left and the (i+1)th digit from the left is divisible by (i+1), for all i = 1,2,3,....,2N-1.

For example, considering the octal number 16743250, we observe that the octal number 32 which is formed by the fifth digit and the sixth digit is not divisible by 6. Therefore, the octal number 16743250 does not satisfy this property.

For which positive integer value(s) of N apart from 5, with 2 ≤ N ≤ 12, do there exist at least one base 2N number that satisfies this property?

Note: Think of this problem as an extension of Ten-Digit Numbers.

 No Solution Yet Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(2): computer solution | Comment 4 of 5 |
(In reply to re: computer solution by Daniel)

Thanks; I've placed a caveat at the beginning of my post.
 Posted by Charlie on 2009-05-10 13:01:17

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