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++
// 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; } |
chevron_right
filter_none
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 |
chevron_right
filter_none
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 |
chevron_right
filter_none
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 |
chevron_right
filter_none
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 ?> |
chevron_right
filter_none
Output:
YES
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
