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 numberbool 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 codeint main(){ int n = 592452; int k = 3; if (SumDivisible(n, k)) cout << "YES"; else cout << "NO"; return 0;} |
Java
// Java implementation of the approachimport java.util.*;class solution{// function that checks the// divisibility of the sum// of the digits at odd places// of the given numberstatic 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 codepublic 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 numberdef 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 codeif __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 numberfunction 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(log10n), since each time the value of n is reduced to n/10.
Auxiliary Space: (1), since no extra space has been taken.
Method #2:Using string() method:
- Convert the integer to a string, then traverse the string and find the sum of all odd indices by storing it in sum.
- 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 codeint 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 approachimport 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 codepublic 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 approachdef 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 coden = 592452k = 3if sumDivisible(n, k) == True: print("YES")else: print("NO")# This code is contributed by vikkycirus |
C#
// C# implementation of the// above approachusing 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 codestatic 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 approachfunction 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 codevar n = 592452var k = 3if(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.
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