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 Evil powers (Posted on 2006-08-31)
Is there a power of 666 such that its decimal notation starts with the digits 123456789?

One may replace "starts" by "ends" for a much easier challenge ;-)

 See The Solution Submitted by JLo Rating: 4.0000 (3 votes)

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 solution | Comment 3 of 8 |

Below, L666 is the common log of 666; we continue adding it to Lp, which is the log of the power of 666 that we are up to. To begin with 123456789, the power must have a logarithm whose mantissa is within certain bounds of the mantissa of Lp.  L123p has the mantissa which should not be exceeded; L123m has a mantissa that must be exceeded. Originally the lowering of this second mantissa was to allow for rounding errors; but the lowering is actually more than needed, but it does show a progression, apparently resulting from 666^40,894,182 beginning with a 1 followed by eight zeros and then non-zero digits.

`   10   point 7   20   L666=log(666)/log(10)   21   P=1:Lp=L666   30   L123=log(123456789)/log(10)   31   L123p=log(123456790)/log(10)-8   32   L123m=log(1234567881)/log(10)-9   35   while Found=0   40      P=P+1:Lp=Lp+L666   45     if Lp-int(Lp)>L123m and Lp-int(Lp)<L123p then   50          :print P;int(Lp);10^(Lp-int(Lp))   90   wend`
`run 6722542  18980924  1.234567881198084806530809835259084 47616724  134444593  1.234567883508729726271308676128429 88510906  249908262  1.234567885819374650336462342734267 129405088  365371931  1.234567888130019578726270843170721 170299270  480835600  1.234567890440664511440734185531911 211193452  596299269  1.234567892751309448479852377911957Break in 60`

So the first power of 666 that has the required characteristics is represented by the 5th line above: 666^170,299,270 = 1.23456789044066451144073 x 10^480,835,600. (I've dropped the last 10 shown digits from the precision, as those are untrustworthy due to the 9-digit power of 10 (characteristic).)

 Posted by Charlie on 2006-08-31 13:51:05

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