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

Given two integer ‘N’ and ‘K’, the task is to find the sum of digits of ‘N’ at it’s 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 it’s modulo with ‘K’.
  • If it is divisible then output ‘YES’, otherwise output ‘NO’.

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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

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Python 3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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Output:

YES


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