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# Check if a given number divides the sum of the factorials of its digits

• Last Updated : 20 May, 2021

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 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 approachclass 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 approachusing 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.



## Javascript


Output:
Yes

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