(This task is taken from a   Project Euler   problem.)

(All numbers herein are expressed in base ten.)


27   =   128   and   7   is
the first power of   2   whose leading decimal digits are   12.

The next power of   2   whose leading decimal digits
are   12   is   80,

280   =   1208925819614629174706176.



Define        p(L,n)      to be the  nth-smallest
value of    j    such that the base ten representation
of    2j    begins with the digits of    L .

    So   p(12, 1) =  7    and
         p(12, 2) = 80



You are also given that:

         p(123, 45)   =   12710



Task

   find: 
     p(12, 1)  
     p(12, 2)  
     p(123, 45)  
     p(123, 12345)  
     p(123, 678910)  
   display the results here, on this page.


F p(=l, n, pwr = 2)
   l = Int(abs(l))
   V digitcount = floor(log(l, 10))
   V log10pwr = log(pwr, 10)
   V (raised, found) = (-1, 0)
   L found < n
      raised++
      V firstdigits = floor(10 ^ (fract(log10pwr * raised) + digitcount))
      I firstdigits == l
         found++
   R raised
 
L(l, n) [(12, 1), (12, 2), (123, 45), (123, 12345), (123, 678910)]
   print(‘p(’l‘, ’n‘) = ’p(l, n))