Java help pls

Rearrange the following lines of code to produce an improved version of the intName method. One line is not useful. Mouse: Drag/drop Keyboard: Grab/release Spacebar (or Enter). Move Cancel Esc Unused IntegerName.java Load default template... import java.util.Scanner; if (part >= 100) { name = Helper.digitName (part / 100) + " hundred "; part = part % 100; } public class IntegerName { if (part >= 100) { = name + Helper.digitName (part / 100) + part = part % 100; } name hundred "; Turns a number into its English name. @param number a positive integer = 20) { name = name + Helper.tensName(part) + " "; part = part % 10; } Rearrange the following lines of code to produce an improved version of the intName method. One line is not useful. Mouse: Drag/drop Keyboard: Grab/release spacebar (or Enter). Move ID Cancel Esc Unused IntegerName.java V Load default template... I. { @return the English name of the number name = name + Helper.tens Name (part) + " part = part % 10; } return name; public static String intName(int number) { // The part that still needs to // be converted int part = number; // The return value String name = ""; else if (part >= 10) { name = name + Helper.teenName(part); part = 0; } if (part > 0) { name = name + Helper.digitName (part); } } public static void main (String [] args) { Scanner in = new Scanner(System.in); int numberToConvert = in.nextInt(); String strNum = intName (numberToConvert); System.out.println(strNum); } if (part > 0) { name = name + Helper.digitName(part); } if (part >= 1000) { name = Helper.digitName(part / 1000) + " thousand "; part = part % 1000; }