Check if a given number divides the sum of the factorials of its digits
Given an integer N, the task is to check whether N divides the sum of the factorials of its digits.
Examples:
Input: N = 19
Output: Yes
1! + 9! = 1 + 362880 = 362881 which is divisible by 19.Input: N = 20
Output: No
0! + 2! = 1 + 4 = 5 which is not divisible by 20.
Approach: First, store the factorials of all the digits from 0 to 9 in an array. And, for the given number N check if it divides the sum of the factorials of its digits.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Function that returns true if n divides// the sum of the factorials of its digitsbool isPossible(int n){ // To store factorials of digits int fac[10]; fac[0] = fac[1] = 1; for (int i = 2; i < 10; i++) fac[i] = fac[i - 1] * i; // To store sum of the factorials // of the digits int sum = 0; // Store copy of the given number int x = n; // Store sum of the factorials // of the digits while (x) { sum += fac[x % 10]; x /= 10; } // If it is divisible if (sum % n == 0) return true; return false;} // Driver codeint main(){ int n = 19; if (isPossible(n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation of the approach class GFG { // Function that returns true if n divides // the sum of the factorials of its digits static boolean isPossible(int n) { // To store factorials of digits int fac[] = new int[10]; fac[0] = fac[1] = 1; for (int i = 2; i < 10; i++) fac[i] = fac[i - 1] * i; // To store sum of the factorials // of the digits int sum = 0; // Store copy of the given number int x = n; // Store sum of the factorials // of the digits while (x != 0) { sum += fac[x % 10]; x /= 10; } // If it is divisible if (sum % n == 0) return true; return false; } // Driver code public static void main (String[] args) { int n = 19; if (isPossible(n)) System.out.println("Yes"); else System.out.println("No"); } } // This code is contributed by Ryuga |
Python3
# Python 3 implementation of the approach # Function that returns true if n divides# the sum of the factorials of its digitsdef isPossible(n): # To store factorials of digits fac = [0 for i in range(10)] fac[0] = 1 fac[1] = 1 for i in range(2, 10, 1): fac[i] = fac[i - 1] * i # To store sum of the factorials # of the digits sum = 0 # Store copy of the given number x = n # Store sum of the factorials # of the digits while (x): sum += fac[x % 10] x = int(x / 10) # If it is divisible if (sum % n == 0): return True return False # Driver codeif __name__ == '__main__': n = 19 if (isPossible(n)): print("Yes") else: print("No") # This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the approach using System;class GFG { // Function that returns true if n divides // the sum of the factorials of its digits static bool isPossible(int n) { // To store factorials of digits int[] fac = new int[10]; fac[0] = fac[1] = 1; for (int i = 2; i < 10; i++) fac[i] = fac[i - 1] * i; // To store sum of the factorials // of the digits int sum = 0; // Store copy of the given number int x = n; // Store sum of the factorials // of the digits while (x != 0) { sum += fac[x % 10]; x /= 10; } // If it is divisible if (sum % n == 0) return true; return false; } // Driver code public static void Main () { int n = 19; if (isPossible(n)) Console.WriteLine("Yes"); else Console.WriteLine("No"); } } // This code is contributed by Code_Mech. |
PHP
<?php// PHP implementation of the approach // Function that returns true if n divides// the sum of the factorials of its digitsfunction isPossible($n){ // To store factorials of digits $fac = array(); $fac[0] = $fac[1] = 1; for ($i = 2; $i < 10; $i++) $fac[$i] = $fac[$i - 1] * $i; // To store sum of the factorials // of the digits $sum = 0; // Store copy of the given number $x = $n; // Store sum of the factorials // of the digits while ($x) { $sum += $fac[$x % 10]; $x /= 10; } // If it is divisible if ($sum % $n == 0) return true; return false;} // Driver code$n = 19; if (isPossible($n)) echo "Yes";else echo "No"; // This code is contributed by Akanksha Rai?> |
Output:
Yes
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