Find multiples of each number in array using java

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Given an array of numbers. For each number in array, find the multiples that exist in the array.
I was supposed to get
Array - [2, 3, 6, 10]
Output-
Multiples of 2 - 6,10
Multiples of 3 - 6
Multiples of 6 - No multiples
Multiples of 10 - No multiples
Instead I am getting
Multiples of3is:3
No multiples

What I have tried:

import java.util.*;
import java.io.*;
import java.lang.*;
public class multiplesarrays
{
    public static void main(String[] args)
    { 
        Scanner sc=new Scanner(System.in);
        
        System.out.println("Enter the no: of elements in array");
        int n=sc.nextInt();
        int arr[]=new int[100];
        for(int i=0; i<n; i++)
        {
            arr[i]=sc.nextInt();
        }
        System.out.println("Multiples of each element of array is:");
       for(int i=0; i<n; i++)
       {
       for(int j=0; j<n; j++)
       {
           if(arr[j++]%arr[i++]==0)
           {
              
               System.out.println("Multiples of"+arr[i++]+"is:"+arr[j++]);
               break;
           }
           else
           {
               System.out.println("No multiples");
               break;
           }
       }
       }
    }
}

Posted 23-Jun-23 7:54am

Abhinav Kishore M

Updated 23-Jun-23 9:11am

v2

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Patrice T · Answer 1 · 2023-06-23T09:11:00

Quote:
Find multiples of each number in array using java

Never use postincrement on variables that control the loops. Unless you know exactly what you do.

Java

for(int i=0; i<n; i++) // Here is the only ++ you need on i
{
for(int j=0; j<n; j++) // Here is the only ++ you need on j
{
    if(arr[j++]%arr[i++]==0) // no ++ here
    {

        System.out.println("Multiples of"+arr[i++]+"is:"+arr[j++]); // no ++ here
        break;
    }
    else
    {
        System.out.println("No multiples");
        break;
    }
}
}

Other errors too.
Edit:
As for the errors, run your code and look at the results you should understand rather quickly what is wrong.
Hint: you should not test a value against itself.