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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

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 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           24   1230         6  27   1084   3210        14  57   2286   143250      10  63   380   2285   137106   543210      34  207  1244  7465   447908   32541670    26  213  1708  13665  109326  874615   69969208   52347610    42  339  2716  21735  173886  1391089  111287128   56743210    46  375  3004  24035  192282  1538257  1230605610  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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