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Reciprocal Period Poser (Posted on 2016-10-29) Difficulty: 3 of 5
N is a duodecimal positive integer ≤ BB.
Find the value of N such that 1/N has the maximum period of its digits after the duodecimal point.

No Solution Yet Submitted by K Sengupta    
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Solution computer solution Comment 2 of 2 |
The longest period is 138 based on N = B7, which is 139 in decimal, based upon the following results:

 138   139   11   7 
 0   1   0   5   2   1   10   9   5   11   5   9   4   11   0   7   3   0   3   1   3   6   5   8   4   5   10   5   4   2   9   1   9   9   0   9   3   10   7   5   1   1   5   7   4   0   8   3   5   5   3   2   3   11   7   10   3   3   4   4   10   0   2   0   10   4   3   9   6   11   10   11   6   9   10   1   2   6   0   6   2   7   0   11   4   8   11   8   10   8   5   6   3   7   6   1   6   7   9   2   10   2   2   11   2   8   1   4   6   10   10   6   4   7   11   3   8   6   6   8   9   8   0   4   1   8   8   7   7   1   11   9   11   1   7   8   2   5  

 0   1   0   5   2   1   10   9   5   11   5   9   4   11   0   7   3   0   3   1   3   6   5   8   4   5   10   5   4   2   9   1   9   9   0   9   3   10   7   5   1   1   5   7   4   0   8   3   5   5   3   2   3   11   7   10   3   3   4   4   10   0   2   0   10   4   3   9   6   11   10   11   6   9   10   1   2   6   0   6   2   7   0   11   4   8   11   8   10   8   5   6   3   7   6   1   6   7   9   2   10   2   2   11   2   8   1   4   6   10   10   6   4   7   11   3   8   6   6   8   9   8   0   4   1   8   8   7   7   1   11   9   11   1   7   8   2   5  

The decimal representation of the first 276 duodecimal digits is shown, with a blank line between the repetitions of the 138 digits.

    5   point 255
    6   kill "recipper.txt":open "recipper.txt" for output as #2
   10   for P1=0 to 11
   20   for P2=0 to 11
   30     N=12*P1+P2:n0=n
   40     Ct=1:numrtr=11
   50     if n>0 and n@2>0 and n@3>0 then
   60       :while numrtr @ n >0
   70         :inc ct
   80         :numrtr=12*numrtr+11
   85       :wend
   90       :if ct>mx then mx=ct:mxn=n:mxp1=p1:mxp2=p2
  100    next
  110    next
  120    print mx,mxn,mxp1,mxp2
  121    print #2, mx,mxn,mxp1,mxp2
  130    f=1/mxn
  140    for i=1 to 2*mx
  150      f=f*12
  159      print int(f);" ";
  160      print #2, int(f);" ";
  161      if i=mx then print:print
  162      if i=mx then print #2,:print #2,
  165      f=f-int(f)
  170    next
  180    print: print #2,
  190    close #2

  Posted by Charlie on 2016-10-29 10:49:24
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