Handling long type argument in Scala

Question

object LPrimeFactor {
  def main(arg:Array[String]):Unit = {
  start(13195)
  start(600851475143)

  }

  def start(until:Long){

    var all_prime_fac:Array[Int] = Arra         


        

Answer 1

To create a Long literal you must add L to the end of it.

start(600851475143L)