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2N Digit Numbers (Posted on 2009-05-09) Difficulty: 2 of 5
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    
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Hints/Tips re: computer solution NEEDS CORRECTION | Comment 2 of 5 |
(In reply to computer solution by Charlie)

      4   1230         NO    12   23    30
4   3210               OK    32 21  20
6   143250           OK     14 43 32 25 50
6   543210           NO      54 43 32  21   10
8   32541670        NO    32 25 54  41 16    67 70

..........

and so on .......

Please read the problem carefuly and debug your program.

Edited on May 10, 2009, 7:22 am
  Posted by Ady TZIDON on 2009-05-10 07:07:57

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