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. |
Javascript
<script>// JavaScript implementation of the approach// Function that returns true if n divides // the sum of the factorials of its digits function isPossible(n) { // To store factorials of digits var fac = new Array(10); fac[0] = fac[1] = 1; for(var i = 2; i < 10; i++) fac[i] = fac[i - 1] * i; // To store sum of the factorials // of the digits var sum = 0; // Store copy of the given number var x = n; // Store sum of the factorials // of the digits while (x != 0) { sum += fac[x % 10]; x = parseInt(x / 10); } // If it is divisible if (sum % n == 0) return true; return false; } // Driver Codevar n = 19; if (isPossible(n)) document.write("Yes"); else document.write("No"); // This code is contributed by Khushboogoyal499</script> |
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
Time Complexity: O(logn)
Auxiliary Space: O(1)
Approach 2:
Here’s another approach to check if a number n divides the sum of the factorials of its digits:
- Traverse the digits of the given number from right to left.
- Initialize a variable ‘sum’ to 0.
- For each digit, calculate its factorial and add it to ‘sum’.
- If at any point ‘sum’ becomes greater than or equal to ‘n’, check if it is divisible by ‘n’. If yes, return true.
- If all digits have been traversed and ‘sum’ is divisible by ‘n’, return true. Otherwise, return false.
Here’s the implementation of this approach in C++, Java and Python.
C++
#include <iostream>using namespace std;// Function that returns true if n divides// the sum of the factorials of its digitsbool isPossible(int n){ int sum = 0; int x = n; while (x > 0) { int digit = x % 10; int fact = 1; for (int i = 2; i <= digit; i++) { fact *= i; } sum += fact; if (sum >= n && sum % n == 0) { return true; } x /= 10; } return (sum % n == 0);}// Driver codeint main(){ int n = 19; if (isPossible(n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
import java.util.*;public class GFG { // Function that returns true if n divides the sum of // the factorials of its digits public static boolean isPossible(int n) { // Initialize sum to 0 int sum = 0; int x = n; while (x > 0) { int digit = x % 10; int fact = 1; for (int i = 2; i <= digit; i++) { fact *= i; // Calculate the factorial of the digit } sum += fact; // Add the factorial to the sum if (sum >= n && sum % n == 0) { return true; // If sum is divisible by n, return true } x /= 10; // Move to the next digit } return (sum % n == 0); // Return true if sum is divisible by n, otherwise false } // Driver code public static void main(String[] args) { int n = 19; if (isPossible(n)) { System.out.println("Yes"); } else { System.out.println("No"); } }} |
Python3
# Function that returns true if n divides# the sum of the factorials of its digitsdef is_possible(n): sum = 0 x = n while x > 0: digit = x % 10 fact = 1 for i in range(2, digit+1): fact *= i sum += fact if sum >= n and sum % n == 0: return True x //= 10 return sum % n == 0# Driver codedef main(): n = 19 if is_possible(n): print("Yes") else: print("No")if __name__ == '__main__': main() |
C#
using System;public class GFG{ // Function that returns true if n divides // the sum of the factorials of its digits public static bool IsPossible(int n) { int sum = 0; int x = n; while (x > 0) { int digit = x % 10; int fact = 1; // Calculate factorial of the current digit for (int i = 2; i <= digit; i++) { fact *= i; } // Add the factorial to the sum sum += fact; // Check if the sum is greater than or equal to n and divisible by n if (sum >= n && sum % n == 0) { return true; } // Move to the next digit x /= 10; } // Check if the sum is divisible by n return (sum % n == 0); } // Driver code public static void Main(string[] args) { int n = 19; if (IsPossible(n)) Console.WriteLine("Yes"); else Console.WriteLine("No"); }} |
Javascript
<script>// JavaScript code of the above approach// Function that returns true if n divides// the sum of the factorials of its digitsfunction isPossible(n) { let sum = 0; let x = n; while (x > 0) { let digit = x % 10; let fact = 1; for (let i = 2; i <= digit; i++) { fact *= i; } sum += fact; if (sum >= n && sum % n === 0) { return true; } x = Math.floor(x / 10); } return sum % n === 0;}// Driver codelet n = 19;if (isPossible(n)) { document.write("Yes");} else { document.write("No");}// This code is contributed by Susobhan Akhuli</script> |
Output
Yes
Time Complexity: O(logn)
Auxiliary Space: O(1)
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