Question

**Please write in Java, in a very simple/beginner coding style/language - thank you!**
Directions: Given a factorial n!. Find the sum of its digits, and the number of trailing zeros (ie: the number of zeros at the end of the factorial).
Input: integer nn, a non-negative integer where n≤20n≤20
Output: integer xyxy, the concatenation of x and y, where x is the sum of the digits in n! and y is the number
of the zeros in n!) Note, 0≤x,y0≤x,y.

Example: Consider the factorial: 5! = 5x4x3x2x1 = 120. The sum of its digits is 1+2+0 = 3, and the number of zeros at the end of 5! = 120 is 1. Then here, x = 3 and y = 1, so the answer/output for 5! is 31.

Extra Credit: allow the user to choose any integer n≥1

Answer 1

import java.util.Scanner;

class Factorial {

    // function to find total zeros

    static int find_zeros(long factorial) {

        int zeros = 0;

        // till factorial is divisible by 10

        while (factorial % 10 == 0) {

            zeros += 1;

            factorial /= 10;

        }

        return zeros;

    }

    // function to find sum of digits

    static int find_sum(long factorial) {

        int sum = 0;

        while (factorial > 0) {

            sum += factorial % 10;

            factorial /= 10;

        }

        return sum;

    }

    public static void main(String[] args) {

        // read n

        Scanner sc = new Scanner(System.in);

        System.out.print("Enter n: ");

        int n = sc.nextInt();

        // compute factorial

        long factorial = 1;

        for (int i = 1; i <= n; i++)

            factorial *= i;

        int sum = find_sum(factorial);

        int zeros = find_zeros(factorial);

        // print

        System.out.printf("%d%d", sum, zeros);

    }

}

.

Output:

.