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 Power of 2 (Posted on 2016-05-27)
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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 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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