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Permute 2 Power (Posted on 2009-11-18) Difficulty: 2 of 5
Prove that there does not exist any positive integer N which is a power of 2 such that the digits of N (in the base ten representation) can be permuted to form a different power of 2. It is known that neither N nor any of the permutations of the digits of N can contain any leading zero.

No Solution Yet Submitted by K Sengupta    
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Solution solution (spoiler) Comment 1 of 1

When a number is doubled, its digital root doubles mod 9, and this is applicable to powers of 2:

   1 1 
   2 2
   4 4
   8 8
  16 7
  32 5
  64 1
 128 2
 256 4
 512 8
1024 7

One can see how the digital root is related to the idea of "casting out 9's".

The thing to note here is that for powers of 2, the digital root has a cycle length of 6: as you go from one power of 2 to the next, you advance 1/6 of the way in the cycle.

If two powers of 2 contain the same number of digits, the larger could be 2, 4 or 8 times the smaller; that is, it can be only 3 powers of 2 higher, which is insufficient to advance to the repetition of the same digital root, which requires advancing six powers of 2. So the two can't have the same digits, otherwise they'd have the same digital root.

BTW, without the restriction against leading zeros, its probably also true, though I don't see a proof. The following program lists the non-zero digits used for powers of 2 up to 2000, in ascending order, each power on a separate record (line) of the file:

    5   kill "pow2dig.txt"
   10   open "pow2dig.txt" for output as #2
   20   P=8
   30   for Pwr=4 to 2000
   40     P=P*2
   50     S=str(P)
   51     for I=1 to len(S)
   52       if mid(S,I,1)="0" then mid(S,I,1)=" "
   53     next
   54     S=cutspc(S)
   60     repeat
   70       Done=1
   80       for I=1 to len(S)-1
   90         if mid(S,I,1)>mid(S,I+1,1) then
  100             :H=mid(S,I,1):mid(S,I,1)=mid(S,I+1,1):mid(S,I+1,1)=H
  110             :Done=0
  120       next
  130     until Done
  140     print #2,S;" ";Pwr
  150    next

such that, for example, the resulting file shows

122579  21

to represent 2^21 = 2097152

This file was sorted so that any powers of 2 using the same non-zero digits would be on successive records of the file, and no such matches were found in the portion of the records preceding the space before the power itself.

  Posted by Charlie on 2009-11-18 15:15:20
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