(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.


// float*int --> float and int*float --> int
fcn p(L,nth){   // 2^j = <L><digits>
   var [const] ln10=(10.0).log(), ld10=(2.0).log() / ln10;
   digits := (10).pow(L.numDigits - 1);
   foreach i in ([1..]){
      z:=ld10*i;
      if(L == ( ln10 * (z - z.toInt()) ).exp()*digits and (nth-=1) <= 0)
	 return(i);
   }
}