Program to check for Peterson number

A number is said to be a Peterson number if the sum of factorials of each digit of the number is equal to the number itself.

Example:

Input : n = 145
Output = Yes
Explanation:
145 = 5! + 4! + 1!
= 120 + 24 +1
= 145

Input  : n = 55
Output : No
Explanation: 5! + 5!
= 120 + 120
= 240
Since 55 is not equal to 240
It is not a Peterson number.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We will pick each digit (Starting from the last digit) of given number and find its factorial. And add all factorials. Finally we check if sum of factorials is equal to number or not.

C++

 // C++ program to determine whether the number is // Peterson number or not #include using namespace std;    // To quickly find factorial of digits int fact = {1, 1, 2, 6, 24, 120, 720,                       5040, 40320, 362880};    // Function to check if a number is Peterson  // or not bool peterson(int n) {     int num = n, sum = 0;        // stores the sum of factorials of      // each digit of the number.     while (n > 0) {         int digit = n % 10;         sum += fact[digit];         n = n / 10;             }        // Condition check for a number to      // be a Peterson Number     return (sum == n); }    // Driver Program int main() {     int n = 145;     if (peterson(n))        cout << "Yes";     else        cout << "No";     return 0; }

Java

 // java program to determine whether the // number is Peterson number or not import java.io.*;    public class GFG {        // To quickly find factorial of digits     static int []fact = new int[]{1, 1, 2,                     6, 24, 120, 720, 5040,                             40320, 362880};            // Function to check if a number is     // Peterson or not     static boolean peterson(int n)     {         int num = n;         int sum = 0;                // stores the sum of factorials of          // each digit of the number.         while (n > 0)         {             int digit = n % 10;             sum += fact[digit];             n = n / 10;          }                // Condition check for a number to          // be a Peterson Number         return (sum == num);     }            // Driver Program     static public void main (String[] args)     {         int n = 145;                    if (peterson(n))             System.out.println("Yes");         else             System.out.println("No");     } }    // This code is contributed by vt_m.

Python3

 # Python3 code to determine whether the # number is Peterson number or not    # To quickly find factorial of digits fact = [1, 1, 2, 6, 24, 120, 720,         5040, 40320, 362880]    # Function to check if a number # is Peterson or not def peterson (n):     num = n     sum = 0            # stores the sum of factorials of      # each digit of the number.     while n > 0:         digit = int(n % 10)         sum += fact[digit]         n = int(n / 10)            # Condition check for a number     # to be a Peterson Number     return (sum == num)    # Driver Code n = 145 print("Yes" if peterson(n) else "No")    # This code is contributed by "Sharad_Bhardwaj"..

C#

 // C# program to determine whether the // number is Peterson number or not using System;    public class GFG {        // To quickly find factorial of digits     static int []fact = new int{1, 1, 2,                       6, 24, 120, 720, 5040,                              40320, 362880};            // Function to check if a number is     // Peterson or not     static bool peterson(int n)     {         int num = n;         int sum = 0;                // stores the sum of factorials of          // each digit of the number.         while (n > 0)         {             int digit = n % 10;             sum += fact[digit];             n = n / 10;          }                // Condition check for a number to          // be a Peterson Number         return (sum == num);     }            // Driver Program     static public void Main ()     {         int n = 145;                    if (peterson(n))             Console.WriteLine("Yes");         else             Console.WriteLine("No");     } }    // This code is contributed by vt_m.

PHP

 0)     {         \$digit = \$n % 10;         \$n = \$n / 10;          }        // Condition check for     // a number to be a     // Peterson Number     return (\$sum == \$n); }        // Driver Code     \$n = 145;     if (peterson(\$n))         echo "Yes";     else         echo"No";        // This code is contributed by ajit ?>

Output:

Yes

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Improved By : vt_m, jit_t

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