(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.


object FirstPowerOfTwo {
  def p(l: Int, n: Int): Int = {
    var n2 = n
    var test = 0
    val log = math.log(2) / math.log(10)
    var factor = 1
    var loop = l
    while (loop > 10) {
      factor *= 10
      loop /= 10
    }
    while (n2 > 0) {
      test += 1
      val value = (factor * math.pow(10, test * log % 1)).asInstanceOf[Int]
      if (value == l) {
        n2 -= 1
      }
    }
    test
  }
 
  def runTest(l: Int, n: Int): Unit = {
    printf("p(%d, %d) = %,d%n", l, n, p(l, n))
  }
 
  def main(args: Array[String]): Unit = {
    runTest(12, 1)
    runTest(12, 2)
    runTest(123, 45)
    runTest(123, 12345)
    runTest(123, 678910)
  }
}