Determine the probability that for a base 11 perfect square P chosen at random between 1,000,000,000 (base 11) and A,AAA,AAA,AAA (base 11) inclusively, the five digit number formed by the last five digits of P (reading left to right) is precisely one more than the number formed by the first five digits.

list
   10   dim Dig(10)
   20   Strt=-int(-sqrt(11^9))
   30   Fin=int(sqrt(11^10-1))
   40   for Root=Strt to Fin
   50      Sq=Root*Root
   60      for I=0 to 9
   70         Dig(I)=Sq @ 11
   80         Sq=Sq\11
   90      next I
  100      N1=0
  110      for I=9 to 5 step -1
  120         N1=N1*11+Dig(I)
  130      next I
  140      N2=0
  150      for I=4 to 0 step -1
  160         N2=N2*11+Dig(I)
  170      next I
  180      if N2=N1+1 then print Root*Root
  185        :for I=9 to 0 step -1:print Dig(I);:if I=5 then print "   ";:endif
  186        :next:print
  190   next Root
  200   print:print Fin-Strt+1
OK
run
 2881864489
 1  2  4  9  8     1  2  4  9  9
 6484275625
 2  8  2  8  2     2  8  2  8  3
 6484597729
 2  8  2  8  4     2  8  2  8  5
 11528102161
 4  9  8  6  3     4  9  8  6  4
 18012055681
 7  7  0  3  3     7  7  0  3  4

 112492
OK

shows that 5 of the 112,492 perfect squares in the given range satisfy the condition, so the probability is 5/112,492.

The table printed shows the decimal representation of the square, followed by the base-11 representation. The middle of the latter is spaced farther than the other digits to highlight the two separate parts asked for by the puzzle.

Posted by Charlie on 2010-09-10 14:16:29