Check whether sum of digits at odd places of a number is divisible by K

Given two integers ‘N’ and ‘K’, the task is to find the sum of digits of ‘N’ at its odd places (right to left) and check whether the sum is divisible by ‘K’. If it is divisible, output YES, otherwise output NO.

Examples:

Input: N = 4325, K = 4 
Output: YES 
Since, 3 + 5 = 8, which is divisible by 4.

Input: N = 1209, K = 3 
Output: NO  

Approach:

• Find the sum of the digits of ‘N’ at odd places (right to left).
• Then check the divisibility of the sum by taking its modulo with ‘K’.
• If it is divisible, then output ‘YES’, otherwise output ‘NO’.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// function that checks the
// divisibility of the sum
// of the digits at odd places
// of the given number
bool SumDivisible(int n, int k)
{
    int sum = 0, position = 1;
    while (n > 0) {

        // if position is odd
        if (position % 2 == 1)
            sum += n % 10;
        n = n / 10;
        position++;
    }

    if (sum % k == 0)
        return true;
    return false;
}

// Driver code
int main()
{
    int n = 592452;
    int k = 3;

    if (SumDivisible(n, k))
        cout << "YES";
    else
        cout << "NO";
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class solution
{

// function that checks the
// divisibility of the sum
// of the digits at odd places
// of the given number
static boolean SumDivisible(int n, int k)
{
    int sum = 0, position = 1;
    while (n > 0) {

        // if position is odd
        if (position % 2 == 1)
            sum += n % 10;
        n = n / 10;
        position++;
    }

    if (sum % k == 0)
        return true;
    return false;
}

// Driver code
public static void main(String arr[])
{
    int n = 592452;
    int k = 3;

    if (SumDivisible(n, k))
        System.out.println("YES");
    else
        System.out.println("NO");

}
}
//This code is contributed by Surendra_Gangwar

Python 3

# Python 3 implementation of the approach

# function that checks the divisibility
# of the sum of the digits at odd places
# of the given number
def SumDivisible(n, k):

    sum = 0
    position = 1
    while (n > 0) :

        # if position is odd
        if (position % 2 == 1):
            sum += n % 10
        n = n // 10
        position += 1

    if (sum % k == 0):
        return True
    return False

# Driver code
if __name__ =="__main__":
    n = 592452
    k = 3

    if (SumDivisible(n, k)):
        print("YES")
    else:
        print("NO")

# This code is contributed
# by ChitraNayal

C#

// C# implementation of the approach
using System;

class GFG
{
// function that checks the
// divisibility of the sum
// of the digits at odd places
// of the given number
static bool SumDivisible(int n, int k)
{
    int sum = 0, position = 1;
    while (n > 0)
    {

        // if position is odd
        if (position % 2 == 1)
            sum += n % 10;
        n = n / 10;
        position++;
    }

    if (sum % k == 0)
        return true;
    return false;
}

// Driver code
static public void Main ()
{
    int n = 592452;
    int k = 3;

    if (SumDivisible(n, k))
        Console.WriteLine("YES");
    else
        Console.WriteLine("NO");
}
}

// This code is contributed by Sachin

PHP

<?php
// PHP implementation of the approach

// function that checks the divisibility
// of the sum of the digits at odd places
// of the given number
function SumDivisible($n, $k)
{
    $sum = 0;
    $position = 1;
    while ($n > 0)
    {

        // if position is odd
        if ($position % 2 == 1)
            $sum += $n % 10;
        $n = (int)$n / 10;
        $position++;
    }

    if ($sum % $k == 0)
        return true;
    return false;
}

// Driver code
$n = 592452;
$k = 3;

if (SumDivisible($n, $k))
    echo "YES";
else
    echo "NO";

// This code is contributed
// by Sach_Code
?>

Javascript

<script>

// JavaScript implementation of the approach

// function that checks the
// divisibility of the sum
// of the digits at odd places
// of the given number
function SumDivisible(n, k)
{
    let sum = 0, position = 1;
    while (n > 0) {

        // if position is odd
        if (position % 2 == 1)
            sum += n % 10;
        n = Math.floor(n / 10);
        position++;
    }

    if (sum % k == 0)
        return true;
    return false;
}

// Driver code

    let n = 592452;
    let k = 3;

    if (SumDivisible(n, k))
        document.write("YES");
    else
        document.write("NO");


// This code is contributed by Surbhi Tyagi.

</script>

Output:

YES

Time Complexity: O(logn)

Auxiliary Space: (1)

Method #2:Using string() method:

1. Convert the integer to a string, then traverse the string and find the sum of all odd indices by storing it in sum.
2. If the sum is divisible by k, then return True else False.

Below is the implementation:

C++

// C++ implementation of the
// above approach
#include <bits/stdc++.h>
using namespace std;

bool sumDivisible(int n, int k)
{
    int sum = 0;

    // Converting integer to string
    string num = to_string(n);
    int i;

    // Traversing the string
    for (i = 0; i < num.size(); i++) {
        if (i % 2 != 0) {
            sum = sum + (num[i] - '0');
        }
    }

    if (sum % k == 0) {
        return true;
    }
    else {
        return false;
    }
}

// Driver code
int main()
{
    int n = 592452;
    int k = 3;
    if (sumDivisible(n, k)) {
        cout << ("YES");
    }
    else {
        cout << ("NO");
    }
    return 0;
}

// This code is contributed by gauravrajput1

Java

// Java implementation of the
// above approach
import java.io.*;
class GFG
{
static boolean sumDivisible(int n, int k)
{
    int sum = 0;

    // Converting integer to string
    String num = Integer.toString(n);
    int i;

    // Traversing the string
    for (i = 0; i < num.length(); i++) {
        if (i % 2 != 0) {
            sum = sum + (num.charAt(i) - '0');
        }
    }

    if (sum % k == 0) {
        return true;
    }
    else {
        return false;
    }
}

// Driver code
public static void main(String[] args)
{
    int n = 592452;
    int k = 3;
    if (sumDivisible(n, k)) {
        System.out.println("YES");
    }
    else {
        System.out.println("NO");
    }
}
}

// This code is contributed by shivanisinghss2110

Python3

# Python3 implementation of the
# above approach

def sumDivisible(n, k):
    sum = 0

    # Converting integer to string
    num = str(n)

    # Traversing the string
    for i in range(len(num)):
        if(i % 2 != 0):
            sum = sum+int(num[i])

    if sum % k == 0:
        return True
    return False


# Driver code
n = 592452
k = 3
if sumDivisible(n, k) == True:
    print("YES")
else:
    print("NO")

# This code is contributed by vikkycirus

C#

// C# implementation of the
// above approach
using System;
class GFG
{
static bool sumDivisible(int n, int k)
{
    int sum = 0;

    // Converting integer to string
    string num = n.ToString();
    int i;

    // Traversing the string
    for (i = 0; i < num.Length; i++) {
        if (i % 2 != 0) {
            sum = sum + (num[i] - '0');
        }
    }

    if (sum % k == 0) {
        return true;
    }
    else {
        return false;
    }
}

// Driver code
static public void Main ()
{
    int n = 592452;
    int k = 3;
    if (sumDivisible(n, k)) {
        Console.Write("YES");
    }
    else {
        Console.Write("NO");
    }
}
}

// This code is contributed by shivanisinghss2110

Javascript

<script>

// javascript implementation of the
// above approach

function sumDivisible(n, k){
    var sum = 0

    // Converting integer to string
    var num = n.toString()

    // Traversing the string
    for(var i=0 ; i < num.length ; i++) {
        if(i % 2 != 0) {
            sum = sum + Number(num[i])
        }
    }

    if (sum % k == 0){
        return true;
    }
    else{
    return false;
    }
}

// Driver code
var n = 592452
var k = 3
if(sumDivisible(n, k)){
    document.write("YES");
}
else{
    document.write("NO");
}





</script>

Output:

Yes

Time Complexity: O(d), where d is the number of digits in n.

Auxiliary Space: O(d), where d is the number of digits in n.