Find permutation of n which is divisible by 3 but not divisible by 6

Namachila November 19, 2025

0 1 4 minutes read

Given an integer $n$ . The task is to find another integer which is permutation of n, divisible by 3 but not divisible by 6. Given that n is divisible by 6. If no such permutation is possible print -1.

Examples:

Input: n = 336
Output: 363

Input: n = 48
Output: -1

For a number to be divisible by 6, it must be divisible by 3 as well as 2, means every even integer divisible by 3 is divisible by 6. So, an integer which is divisible by 3 but not 6 is odd integer divisible by 3.
So, if integer n contains any odd integer then there exists a permutation which is divisible by 3 but not 6, else no such permutation exist.

Algorithm:

let LEN is length of integer (i.e. ceil(log10(n))).
iterate over LEN and check whether n is even or odd.
- if n is odd return n
- else right – rotate n once. and continue.
if LEN is over return -1

Below is the implementation of the above approach:

C++

// C++ program to find permutation of n
// which is divisible by 3 but not
// divisible by 6
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the permutation
int findPermutation(int n)
{
    // length of integer
    int len = ceil(log10(n));
 
    for (int i = 0; i < len; i++) {
        // if integer is even
        if (n % 2 != 0) {
            // return odd integer
            return n;
        }
        else {
            // rotate integer
            n = (n / 10) + (n % 10) * pow(10, len - i - 1);
            continue;
        }
    }
 
    // return -1 in case no required
    // permutation exists
    return -1;
}
 
// Driver Code
int main()
{
    int n = 132;
 
    cout << findPermutation(n);
 
    return 0;
}

Java

// Java program to find permutation 
// of n which is divisible by 3 
// but not divisible by 6
import java.lang.*;
import java.util.*;
 
class GFG
{
// Function to find the permutation
static int findPermutation(int n)
{
    // length of integer
    int len = (int)Math.ceil(Math.log10(n));
 
    for (int i = 0; i < len; i++) 
    {
        // if integer is even
        if (n % 2 != 0) 
        {
            // return odd integer
            return n;
        }
        else
        {
            // rotate integer
            n = (n / 10) + (n % 10) * 
                (int)Math.pow(10, len - i - 1);
            continue;
        }
    }
 
    // return -1 in case no required
    // permutation exists
    return -1;
}
 
// Driver Code
public static void main(String args[])
{
    int n = 132;
 
    System.out.println(findPermutation(n));
}
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

Python3

# Python3 program to find permutation 
# of n which is divisible by 3 but 
# not divisible by 6
from math import log10, ceil, pow
 
# Function to find the permutation
def findPermutation(n):
     
    # length of integer
    len = ceil(log10(n))
 
    for i in range(0, len, 1):
         
        # if integer is even
        if n % 2 != 0:
             
            # return odd integer
            return n
        else:
             
            # rotate integer
            n = ((n / 10) + (n % 10) *
                  pow(10, len - i - 1))
            continue
         
    # return -1 in case no required
    # permutation exists
    return -1
 
# Driver Code
if __name__ == '__main__':
    n = 132
 
    print(int(findPermutation(n)))
 
# This code is contributed 
# by Surendra_Gangwar

C#

// C# program to find permutation 
// of n which is divisible by 3 
// but not divisible by 6
using System;
 
class GFG
{
// Function to find the permutation
static int findPermutation(int n)
{
    // length of integer
    int len = (int)Math.Ceiling(Math.Log10(n));
 
    for (int i = 0; i < len; i++) 
    {
        // if integer is even
        if (n % 2 != 0) 
        {
            // return odd integer
            return n;
        }
        else
        {
            // rotate integer
            n = (n / 10) + (n % 10) * 
                (int)Math.Pow(10, len - i - 1);
            continue;
        }
    }
 
    // return -1 in case no required
    // permutation exists
    return -1;
}
 
// Driver Code
public static void Main()
{
    int n = 132;
 
    Console.WriteLine(findPermutation(n));
}
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

PHP

<?php
// PHP program to find permutation 
// of n which is divisible by 3 but
// not divisible by 6
 
// Function to find the permutation
function findPermutation($n)
{
    // length of integer
    $len = ceil(log10($n));
 
    for ($i = 0; $i < $len; $i++) 
    {
        // if integer is even
        if ($n % 2 != 0) 
        {
            // return odd integer
            return (int)$n;
        }
        else
        {
            // rotate integer
            $n = ($n / 10) + ($n % 10) *
                  pow(10, $len - $i - 1);
            continue;
        }
    }
 
    // return -1 in case no required
    // permutation exists
    return -1;
}
 
// Driver Code
$n = 132;
 
echo findPermutation($n);
 
// This code is contributed by mits
?>

Javascript

<script>
// java script  program to find permutation
// of n which is divisible by 3 but
// not divisible by 6
 
// Function to find the permutation
function findPermutation(n)
{
 
    // length of integer
    let len = Math.ceil(Math.log10(n));
 
    for (let i = 0; i < len; i++)
    {
     
        // if integer is even
        if (n % 2 != 0)
        {
         
            // return odd integer
            return parseInt(n);
        }
        else
        {
         
            // rotate integer
            n = (n / 10) + (n % 10) *
                Math.pow(10, len - i - 1);
            continue;
        }
    }
     
    // return -1 in case no required
    // permutation exists
    return -1;
}
 
// Driver Code
let n = 132;
 
document.write( findPermutation(n));
 
// This code is contributed by sravan kumar (vignan)
</script>

Output:

Time complexity: O(logn) for given input number n

Auxiliary space: O(1)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!