(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.


 
public class FirstPowerOfTwo {
 
    public static void main(String[] args) {
        runTest(12, 1);
        runTest(12, 2);
        runTest(123, 45);
        runTest(123, 12345);
        runTest(123, 678910);
    }
 
    private static void runTest(int l, int n) {
        System.out.printf("p(%d, %d) = %,d%n", l, n, p(l, n));
    }
 
    public static int p(int l, int n) {
        int test = 0;
        double log = Math.log(2) / Math.log(10);
        int factor = 1;
        int loop = l;
        while ( loop > 10 ) {
            factor *= 10;
            loop /= 10;
        }
        while ( n > 0) {
            test++;
            int val = (int) (factor * Math.pow(10, test * log % 1));
            if ( val == l ) {
                n--;
            }
        }
        return test;
    }
 
}
 