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Program to check for Peterson number

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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.     

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

C++




// C++ program to determine whether the number is
// Peterson number or not
#include <iostream>
using namespace std;
  
// To quickly find factorial of digits
int fact[10]
    = { 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 == num);
}
  
// Driver Program
int main()
{
    int n = 145;
    if (peterson(n))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}


Java




//checks whether a number entered by user is peterson number or not
import java.util.*;
class GFG
{
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
      //taking input from the user 
      System.out.println("Enter the number");
        int num=sc.nextInt();
        int temp=num;//storing the number in a temporary variable
        int f=1,sum=0;
        while(num!=0)//running while loop until number becomes zero
        {
            f=1;
          //extracting last digit of the number 
          //and storing in r
            int r=num%10;
          //for loop to find the factorial of a digit
            for(int i=1;i<=r;i++)
            {
                f=f*i;
            }
            sum=sum+f;//adding the factotial of the digits
            num=num/10;
        }
      //checking if the sum of the factorial of digits 
      //is equal to the number or not
        if(sum==temp)
        System.out.println("PETERSON NUMBER");
        else
        System.out.println("NOT PETERSON NUMBER");
    }
    }


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[10] { 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




<?php
// PHP program to determine
// whether the number is
// Peterson number or not
  
// To quickly find 
// factorial of digits
$fact =array (1, 1, 2, 6, 24, 120, 720, 
                  5040, 40320, 362880);
  
// Function to check if 
// a number is Peterson 
// or not
function peterson($n)
{
    $num = $n; $sum = 0;
  
    // stores the sum of factorials of 
    // each digit of the number.
    while ($n > 0)
    {
        $digit = $n % 10;
        $n = $n / 10;     
    }
  
    // Condition check for
    // a number to be a
    // Peterson Number
    return ($sum == $num);
}
  
    // Driver Code
    $n = 145;
    if (peterson($n))
        echo "Yes";
    else
        echo"No";
      
// This code is contributed by ajit
?>


Javascript




<script>
  
// Javascript program to determine whether
// the number is Peterson number or not
  
// To quickly find factorial of digits
let fact = [ 1, 1, 2, 6, 24, 120, 
             720, 5040, 40320, 362880 ];
  
// Function to check if a number is
// Peterson or not
function peterson(n)
{
    let num = n, sum = 0;
  
    // stores the sum of factorials of
    // each digit of the number.
    while (n > 0)
    {
        let digit = n % 10;
        sum += fact[digit];
        n = parseInt(n / 10);
    }
  
    // Condition check for a number to
    // be a Peterson Number
    return (sum == num);
}
  
// Driver code
let n = 145;
if (peterson(n))
    document.write("Yes");
else
    document.write("No");
      
// This code is contributed by souravmahato348
  
</script>


Output

Yes

Time Complexity:  log10(n)
Auxiliary Space: O(1)



Last Updated : 17 Feb, 2023
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