Program to check if a number is divisible by sum of its digits
Given an integer N, the task is to check whether the number is divisible by the sum of its digits or not. If divisible, then print “YES” else print “NO”.
Examples:
Input: N = 12
Output: YES
Explanation:
As sum of digits of 12 = 1 + 2 = 3
and 12 is divisible by 3
So the output is YES
Input: N = 123
Output: NO
Approach: The idea to solve the problem is to extract the digits of the number and add them. Then check if the number is divisible by the sum of its digit. If it is divisible then print YES otherwise print NO.
Below is the implementation of above approach:
Implementation:
C++
// CPP implementation of above approach #include <bits/stdc++.h> using namespace std; // Function to check // if the given number is divisible // by sum of its digits string isDivisible(long long int n) { long long int temp = n; // Find sum of digits int sum = 0; while (n) { int k = n % 10; sum += k; n /= 10; } // check if sum of digits divides n if (temp % sum == 0) return "YES"; return "NO"; } // Driver Code int main() { long long int n = 123; cout << isDivisible(n); return 0; } |
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Java
// Java implementation of above approach class GFG { // Function to check if the // given number is divisible // by sum of its digits static String isDivisible(long n) { long temp = n; // Find sum of digits int sum = 0; while (n != 0) { int k = (int) n % 10; sum += k; n /= 10; } // check if sum of digits divides n if (temp % sum == 0) return "YES"; return "NO"; } // Driver Code public static void main(String []args) { long n = 123; System.out.println(isDivisible(n)); } } // This code is contributed by Ryuga |
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Python3
# Python 3 implementation of above approach # Function to check if the given number # is divisible by sum of its digits def isDivisible(n): temp = n # Find sum of digits sum = 0; while (n): k = n % 10; sum += k; n /= 10; # check if sum of digits divides n if (temp % sum == 0): return "YES"; return "NO"; # Driver Code n = 123; print(isDivisible(n)); # This code is contributed by # Akanksha Rai |
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C#
// C# implementation of above approach using System; class GFG { // Function to check if the // given number is divisible // by sum of its digits static String isDivisible(long n) { long temp = n; // Find sum of digits int sum = 0; while (n != 0) { int k = (int) n % 10; sum += k; n /= 10; } // check if sum of digits divides n if (temp % sum == 0) return "YES"; return "NO"; } // Driver Code public static void Main() { long n = 123; Console.WriteLine(isDivisible(n)); } } // This code is contributed by anuj_67.. |
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PHP
<?php // PHP implementation of above approach // Function to check if the given number // is divisible by sum of its digits function isDivisible($n) { $temp = $n; // Find sum of digits $sum = 0; while ($n) { $k = $n % 10; $sum += $k; $n = (int)($n / 10); } // check if sum of digits divides n if ($temp % $sum == 0) return "YES"; return "NO"; } // Driver Code $n = 123; print(isDivisible($n)); // This code is contributed // by chandan_jnu ?> |
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Output:
NO
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