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Power of 2 (Posted on 2016-05-27) Difficulty: 4 of 5
Find integers N for which the sum of the digits in 2N equals N.

No Solution Yet Submitted by Ady TZIDON    
Rating: 5.0000 (1 votes)

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Solution computer solution | Comment 5 of 6 |
   10   for N=1 to 3000
   20    if N=fnSod(2^N) then print N,2^N
   30   next
  999   end
 1000   fnSod(X)
 1010   local S$,Tot,I
 1015   Tot=0
 1020   S$=cutspc(str(X))
 1030   for I=1 to len(S$)
 1040     Tot=Tot+val(mid(S$,I,1))
 1050   next
 1060   return(Tot)

finds only

n     2^n
5    32
70   1180591620717411303424

Why no more?

There doesn't seem to be a specific reason ruling out further cases. The number of digits in the power should vary linearly with the power and should be about n multiplied by the common log of 2. The "typical" s.o.d. should also then be proportional to n. And the maximum possible sod would be 9*n*log(2), and as small as needed.

Perhaps it's just the great range and standard deviation in the sod that make it more and more unlikely to be a "hit" as the powers get larger.

  Posted by Charlie on 2016-05-27 15:12:52
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