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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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Solution computer solution | Comment 1 of 5

The following is based on an incorrect reading of the problem. See subsequent posts:


  100      cls
  300      for Bse=2 to 24
  400       for S=1 to Bse-1
  500       Value=S
  600       Ss$=mid("123456789abcdefghijklmnopqrstuvwxyz",S,1)
  700       gosub *Addon
  800       next
  900      next
 1000      end
 1100 
 1200    *Addon
 1201   local NewDig
 1300       for NewDig=0 to Bse-1
 1400       Nd$=mid("0123456789abcdefghijklmnopqrstuvwxyz",NewDig+1,1)
 1500       if instr(Ss$,Nd$)=0 then
 1600       :Value=Value*Bse+NewDig
 1700       :Ss$=Ss$+Nd$
 1800       :
 1900       :Tst=Value-int(Value/len(Ss$))*len(Ss$)
 2000       :if Tst=0 then
 2100       ::if len(Ss$)=Bse then
 2200       ::if Bse<>PBase then
 2300       ::print Bse,Ss$
 2400       ::endif
 2500       ::PBase=Bse
 2600       ::else
 2700       ::if Bse<>PBase then
 2800       ::gosub *Addon
 2900       ::endif
 3000       ::endif
 3100       ::endif
 3200       :
 3300       :Ss$=left(Ss$,len(Ss$)-1)
 3400       :Value=(Value-NewDig)/Bse
 3500       :endif
 3600       next
 3700      return
 3800 
finds only
base    number in that base (as an example)
 2      10
 4      1230
 6      143250
 8      32541670
 10     3816547290
 14     9c3a5476b812d0 (where a represents 10, b represents 11, etc.)
 

so the bases in question are all of 2, 4, 6, 8 and 10, but not 12. Up through base 24, the highest base in which this happens is base 14.
 
But if the restriction against showing more than one result per base is removed, and the decimal equivalents of the subtotals are shown, the results are:

2   10           2
4   1230         6  27   108
4   3210        14  57   228
6   143250      10  63   380   2285   13710
6   543210      34  207  1244  7465   44790
8   32541670    26  213  1708  13665  109326  874615   6996920
8   52347610    42  339  2716  21735  173886  1391089  11128712
8   56743210    46  375  3004  24035  192282  1538257  12306056
10  3816547290  38  381  3816  38165  381654  3816547  38165472 381654729 3816547290
14  9c3a5476b812d0 138 1935 27100 379405 5311674 74363443 1041088208 14575234923 204053288930
                                 2856746045021 39994444630296 559922224824157 7838911147538198
 

                               
The puzzle only asks for even bases, but I'm not sure why this can happen only in even bases.                                

Edited on May 10, 2009, 1:00 pm
  Posted by Charlie on 2009-05-09 18:11:55

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