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Count digits in a factorial

Counting the number of digits in a factorial of a given number is an interesting problem that involves calculating the factorial and determining its digit count. A factorial of a non-negative integer 'n', denoted as 'n!', is the product of all positive integers from 1 to 'n'. In this article, we will explore how to count the digits in a factorial using a C program.

Problem Statement

Given a positive integer 'n', the task is to find and count the number of digits in 'n!'.

Example

For example, if 'n' is 5, then the factorial of 5 is 5! = 120, which has 3 digits. Similarly, for 'n' as 12, the factorial is 479,001,600, which has 9 digits.

Idea to Solve

To count the digits in 'n!', we can calculate 'n!' and then determine the number of digits in the result. We can use the logarithmic property of the factorial to estimate the number of digits.

Pseudocode

function factorial_digits(number):
    If number < 0:
        digit = 0
    Else if number > 1:
        Calculate result = (number * log10(number / M_E) + log10(2 * M_PI * number) / 2.0)
        digit = floor(result) + 1
    Print "Factorial of [", number, "]"
    Print "Digit :", digit

function main():
    factorial_digits(5)
    factorial_digits(12)
    factorial_digits(15)
    factorial_digits(220)

Algorithm Explanation

  1. The factorial_digits function takes one parameter: 'number'.
  2. If 'number' is less than 0, it sets digit to 0 (since factorials of negative numbers are undefined).
  3. If 'number' is greater than 1, it calculates the result using the logarithmic property of the factorial formula.
  4. It calculates the number of digits in 'result' using the floor of the logarithm and adds 1.
  5. It prints the calculated factorial digit count for the given 'number'.

Code Solution

// C program
// Count digits in a factorial 
#include <stdio.h>
#include <math.h>

//Find digits of a factorial number
void factorial_digits(int number)
{
	int digit = 1;
	if (number < 0)
	{
		//When get negative number
		digit = 0;
	}
	else if (number > 1)
	{
		//Find digit
		double result = ((number * log10(number / M_E) + log10(2 * M_PI * number) / 2.0));
		digit = floor(result) + 1;
	}
	printf("\n [%d] Factorial digit : [%d]", number, digit);
}
int main()
{
	//Test case
	/*
	     Factorial of [5]
	     // 120
	     // Digit : 3
	 */
	factorial_digits(5);
	/*
	     Factorial of [12]
	     // 479,001,600
	     // Digit : 9
	 */
	factorial_digits(12);
	/*
	    Factorial of [15]
	    // 1307674368000
	    // Digit : 13
	*/
	factorial_digits(15);
	/*
	Factorial of [220]

	//  2.283860335 E+421

	2283860335914641457397265865
	115333727042973071546228701773
	634716126027692603024845877776
	549791921102945706558196074779
	575009550523224197049956176972
	302056587667226166060976323404
	977554732543013557133146825747
	553799450849523377065894531021
	055272516334278466875614904921
	365807833845853428557155180084
	957884822642989867003294551385
	992993862178352349027264696691
	854493614080000000000000000000
	000000000000000000000000000000
	0000
	// Digit : 422

	*/
	factorial_digits(220);
	return 0;
}

Output

 [5] Factorial digit : [3]
 [12] Factorial digit : [9]
 [15] Factorial digit : [13]
 [220] Factorial digit : [422]
// Java program 
// Count digits in a factorial 
class FactorialNo
{
	//Find digits of a factorial number
	public void factorial_digits(int number)
	{
		int digit = 1;
		if (number < 0)
		{
			//When get negative number
			digit = 0;
		}
		else if (number > 1)
		{
			//Find digit
			double result = ((number * Math.log10(number / Math.E) + Math.log10(2 * Math.PI * number) / 2.0));
			digit = (int)Math.floor(result) + 1;
		}
		System.out.print("\n [" + number + "] Factorial digit : [" + digit + "]");
	}
	public static void main(String[] args)
	{
		FactorialNo obj = new FactorialNo();
		//Test case
		obj.factorial_digits(5);
		obj.factorial_digits(12);
		obj.factorial_digits(15);
		obj.factorial_digits(220);
	}
}

Output

 [5] Factorial digit : [3]
 [12] Factorial digit : [9]
 [15] Factorial digit : [13]
 [220] Factorial digit : [422]
//Include header file
#include <iostream>
#include<math.h>
using namespace std;

// C++ program 
// Count digits in a factorial 

class FactorialNo
{
	public:
		//Find digits of a factorial number
		void factorial_digits(int number)
		{
			int digit = 1;
			if (number < 0)
			{
				//When get negative number
				digit = 0;
			}
			else if (number > 1)
			{
				//Find digit
				double result = ((number *log10(number / M_E) + log10(2  * M_PI * number) / 2.0));
				digit = (int) floor(result) + 1;
			}
			cout << "\n [" << number << "] Factorial digit : [" << digit << "]";
		}
};
int main()
{
	FactorialNo obj = FactorialNo();
	//Test case
	obj.factorial_digits(5);
	obj.factorial_digits(12);
	obj.factorial_digits(15);
	obj.factorial_digits(220);
	return 0;
}

Output

 [5] Factorial digit : [3]
 [12] Factorial digit : [9]
 [15] Factorial digit : [13]
 [220] Factorial digit : [422]
//Include namespace system
using System;
// C# program 
// Count digits in a factorial 
class FactorialNo
{
	//Find digits of a factorial number
	public void factorial_digits(int number)
	{
		int digit = 1;
		if (number < 0)
		{
			//When get negative number
			digit = 0;
		}
		else if (number > 1)
		{
			//Find digit
			double result = ((number * Math.Log10(number / Math.E) + Math.Log10(2 * Math.PI * number) / 2.0));
			digit = (int) Math.Floor(result) + 1;
		}
		Console.Write("\n [" + number + "] Factorial digit : [" + digit + "]");
	}
	public static void Main(String[] args)
	{
		FactorialNo obj = new FactorialNo();
		//Test case
		obj.factorial_digits(5);
		obj.factorial_digits(12);
		obj.factorial_digits(15);
		obj.factorial_digits(220);
	}
}

Output

 [5] Factorial digit : [3]
 [12] Factorial digit : [9]
 [15] Factorial digit : [13]
 [220] Factorial digit : [422]
<?php
// Php program 
// Count digits in a factorial 
class FactorialNo
{
	//Find digits of a factorial number
	public	function factorial_digits($number)
	{
		$digit = 1;
		if ($number < 0)
		{
			//When get negative number
			$digit = 0;
		}
		else if ($number > 1)
		{
			//Find digit
			$result = (($number * log10(intval($number / M_E)) + intval(log10(2 * M_PI * $number) / 2.0)));
			$digit = (floor($result)) + 1;
		}
		echo "\n [". $number ."] Factorial digit : [". $digit ."]";
	}
}

function main()
{
	$obj = new FactorialNo();
	//Test case
	$obj->factorial_digits(5);
	$obj->factorial_digits(12);
	$obj->factorial_digits(15);
	$obj->factorial_digits(220);
}
main();

Output

 [5] Factorial digit : [1]
 [12] Factorial digit : [8]
 [15] Factorial digit : [11]
 [220] Factorial digit : [420]
// Node Js program 
// Count digits in a factorial 
class FactorialNo
{
	//Find digits of a factorial number
	factorial_digits(number)
	{
		var digit = 1;
		if (number < 0)
		{
			//When get negative number
			digit = 0;
		}
		else if (number > 1)
		{
			//Find digit
			var result = ((number * Math.log10(parseInt(number / Math.E)) + parseInt(Math.log10(2 * Math.PI * number) / 2.0)));
			digit = (Math.floor(result)) + 1;
		}
		process.stdout.write("\n [" + number + "] Factorial digit : [" + digit + "]");
	}
}

function main()
{
	var obj = new FactorialNo();
	//Test case
	obj.factorial_digits(5);
	obj.factorial_digits(12);
	obj.factorial_digits(15);
	obj.factorial_digits(220);
}
main();

Output

 [5] Factorial digit : [1]
 [12] Factorial digit : [8]
 [15] Factorial digit : [11]
 [220] Factorial digit : [420]
import math
#  Python 3 program 
#  Count digits in a factorial 
class FactorialNo :
	# Find digits of a factorial number
	def factorial_digits(self, number) :
		digit = 1
		if (number < 0) :
			# When get negative number
			digit = 0
		
		elif(number > 1) :
			# Find digit
			result = ((number * math.log10(int(number / math.e)) + int(math.log10(2 * math.pi * number) / 2.0)))
			digit = (math.floor(result)) + 1
		
		print("\n [{0}] Factorial digit : [{1}]".format(number,digit), end = "")
	

def main() :
	obj = FactorialNo()
	# Test case
	obj.factorial_digits(5)
	obj.factorial_digits(12)
	obj.factorial_digits(15)
	obj.factorial_digits(220)

if __name__ == "__main__": main()

Output

 [5] Factorial digit : [1]
 [12] Factorial digit : [8]
 [15] Factorial digit : [11]
 [220] Factorial digit : [420]
#  Ruby program 
#  Count digits in a factorial 
class FactorialNo 
	# Find digits of a factorial number
	def factorial_digits(number) 
		digit = 1
		if (number < 0) 
			# When get negative number
			digit = 0
		elsif(number > 1) 
			# Find digit
			result = ((number * Math.log10(number / Math::E) + Math.log10(2 * Math::PI * number) / 2.0))
			digit = (result.floor()).to_i + 1
		end

		print("\n [", number ,"] Factorial digit : [", digit ,"]")
	end

end

def main() 
	obj = FactorialNo.new()
	# Test case
	obj.factorial_digits(5)
	obj.factorial_digits(12)
	obj.factorial_digits(15)
	obj.factorial_digits(220)
end

main()

Output

 [5] Factorial digit : [3]
 [12] Factorial digit : [9]
 [15] Factorial digit : [13]
 [220] Factorial digit : [422]
// Scala program 
// Count digits in a factorial 
class FactorialNo
{
	//Find digits of a factorial number
	def factorial_digits(number: Int): Unit = {
		var digit: Int = 1;
		if (number < 0)
		{
			//When get negative number
			digit = 0;
		}
		else if (number > 1)
		{
			//Find digit
			var result: Double = ((number * Math.log10((number / Math.E).toInt) + (Math.log10(2 * Math.PI * number) / 2.0).toInt));
			digit = (Math.floor(result)).toInt + 1;
		}
		print("\n [" + number + "] Factorial digit : [" + digit + "]");
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var obj: FactorialNo = new FactorialNo();
		//Test case
		obj.factorial_digits(5);
		obj.factorial_digits(12);
		obj.factorial_digits(15);
		obj.factorial_digits(220);
	}
}

Output

 [5] Factorial digit : [1]
 [12] Factorial digit : [8]
 [15] Factorial digit : [11]
 [220] Factorial digit : [420]
import Foundation
// Swift 4 program 
// Count digits in a factorial 
class FactorialNo
{
	//Find digits of a factorial number
	func factorial_digits(_ number: Int)
	{
		var digit: Int = 1;
		if (number < 0)
		{
			//When get negative number
			digit = 0;
		}
		else if (number > 1)
		{
			//Find digit
			let result: Double = ((Double(number) * log10(Double(number) / M_E) + log10(2 * Double.pi * Double(number)) / 2.0));
			digit = Int(floor(result)) + 1;
	}
	print("\n [\(number)] Factorial digit : [\(digit)] ", terminator: "");
}
}
func main()
{
	let obj: FactorialNo = FactorialNo();
	//Test case
	obj.factorial_digits(5);
	obj.factorial_digits(12);
	obj.factorial_digits(15);
	obj.factorial_digits(220);
}
main();

Output

 [5] Factorial digit : [3]
 [12] Factorial digit : [9]
 [15] Factorial digit : [13]
 [220] Factorial digit : [422]

Time Complexity

The time complexity of counting the digits in a factorial using the logarithmic property is indeed O(1).

Last updated on by Kalkicode




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