Program to find Last non-zero digit in factorial in C

October 13, 2022

Last Non-Zero digit in Factorial in C

Here, in this page we will discuss the program to find the last non-zero digit in factorial in C Programming language. We are given with an integer value and need to return the last non-zero digit in its factorial.

Last Non-Zero digit in Factorial in C

Method Discussed :

  • Method 1 : Recursive way, without handeling overflow condition
  • Method 2 : Recursive way, handeling overflow condition

Method 1 :

Simple approach is to calculate the factorial of the number then, track its unit digit until the non-zero digit found.

  • Create a recursive function say fact(int n) to calculate the factorial,
  • Base condition: if(n<=1) return 1
  • Otherwise return n*fact(n-1).
  • After getting the factorial, and store it in variable say factorial.
  • Run a loop till (factorial % 10 != 0)
  • After that, print(factorial % 10)

Method 1 : Code in C

Run 
#include <stdio.h>
//Recursive function to calculate the factorial
int fact(int n){

   if(n <= 1) //Base Condition
   return 1;

   return n*fact(n-1);
}

//Driver Code
int main(){

   int n=5;
   int factorial = fact(n);

   while(factorial%10==0)
   {
        factorial /= 10;
   }

   printf("%d",factorial%10);
}

Output :

2

Method 2 (Recursive Approach):

In this algorithm we will discuss the recursive approach to find the last non zero digit in factorial of the given number.

  • Create a recursive function say lastDigit(int n, int result[], int countFive).
  • Now, divide each array element into its shortest divisible form by 5 and increase count of such occurrences and store it in variable say countFive.
  • Now divide each array element into its shortest divisible form by 2 and decrease count of such occurrences i.e, countFive–. This way we are not considering the multiplication of 2 and a 5 in our multiplication(number of 2’s present in multiplication result upto n is always more than number 0f 5’s).
  • Multiply each number(after removing pairs of 2’s and 5’s) and store just last digit by taking remainder by 10.
  • Now call recursively for smaller numbers by (currentNumber – 1) as parameter.

Method 2 : Code in C

Run 
#include <stdio.h>

void lastDigit(int n, int result[], int countFive)
{
  int number = n; 

  //Base condition
  if (number == 1)
      return;


  while (number % 5 == 0) {
        number /= 5;
        countFive++;
  }


  while (countFive != 0 && (number & 1) == 0) {
        number /= 2; 
        countFive--;
  }

  result[0] = (result[0] * (number % 10)) % 10;
  lastDigit(n - 1, result, countFive);


}

int lastNon0Digit(int n)
{
   int result[] = { 1 }; // single element array.
   lastDigit(n, result, 0);
   return result[0];
}

int main()
{
   int n=5;
   printf("%d ",lastNon0Digit(n));
   return 0;
}

Output :

2

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