Verified answer

Solution:

The given program is written in Java.

import java.util.*;

public class Array  {

   public static void main(String s[])  {

       Scanner sc=new Scanner(System.in);

       int ELEMENTS=15,i;

       int numbers[]=new int[ELEMENTS];

       int digits[]=new int[ELEMENTS];

       for(i=0;i<ELEMENTS;i++){

           System.out.print("a["+i+"] = ");

           numbers[i]=sc.nextInt();

           digits[i]=numbers[i]%10;

       }

       System.out.println("Given Array: "+Arrays.toString(numbers));

       System.out.println("Last Digit of each numbers in the array - "+Arrays.toString(digits));

       System.out.println("Required Numbers are - ");

       for(int x:numbers){

           if(ItemPresent(digits,startingDigit(x))){

               System.out.print(x+"  ");

           }

       }

   }

   static int startingDigit(int x){

       for(;x>9;x/=10){}

       return x;

   }

   static boolean ItemPresent(int a[],int x){

       for(int y:a){

           if(x==y)

               return true;

       }

       return false;

   }

}

Explanation:

• Array numbers[] stores the elements and digit[] stores the unit digits of each element of the array numbers[].
• startingDigit() returns the starting digit of any number and ItemPresent() checks whether an element is present in array or not.
• A loop iterates through the elements of array numbers[]. Now, if the starting digit of the number starts with the element kept in array digits[], then it displays the number.

See the attachment for output.