1. Write an application that inputs five numbers, each between 10 and 100, inclusive. As each number is read, display it only if it’s not a duplicate of a number already read. Provide for the "worst case," in which all five numbers are different. Use the smallest possible array to solve this problem. Display the complete set of unique values input after the user enters each new value. This is a sample run of your program: Enter·an·integer·between·10·and·100 :100↵ This·is·the·first·time·100·has·been ·entered↵ Enter·an·integer·between·10·and·100 :100↵ Enter·an·integer·between·10·and·100 :10↵ This·is·the·first·time·10·has·been· entered↵ Enter·an·integer·between·10·and·100 :20↵ This·is·the·first·time·20·has·been· entered↵ Enter·an·integer·between·10·and·100 :20↵ The·complete·set·of·unique·values·e ntered·is:↵ Unique·Value·1:·is·100↵ Unique·Value·2:·is·10↵ Unique·Value·3:·is·20

Check the explanation

Explanation:

import java.util.Scanner;

public class DuplicateElimination {

   public static void main(String[] args) {

       Scanner scanner = new Scanner(System.in);

       int arr[] = new int[5];

       int size = 0, n;

       for(int i = 0;i<5;i++){

           System.out.print("Enter an integer between 10 and 100:");

           n = scanner.nextInt();

           if(size==0 || elemanVarmı(arr,size,n)){

               System.out.println("This is the first time "+n+" has been entered");

               arr[size] = n;

               size++;

           }

       }

       System.out.println("The complete set of unique values entered is");

       for(int i = 0;i<size;i++){

           System.out.println("Unique Value "+(i+1)+": is "+arr[i]);

       }

   }

   private static boolean elemanVarmı(int dizi[], int size, int n)

   {

       for (int i = 0; i < size; i++)

       {

           if(dizi[i]==n)

           {

               return false;

           }

       }

       return true;

   }

}

i've had the same problem. i generally just ignore it, as a ton of these "verified answers" are completely wrong.