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Evil powers (Posted on 2006-08-31) Difficulty: 4 of 5
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 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  1.234567892751309448479852377911957

Break 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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