Problem 20: Factorial digit sum

Q)n! means n  (n  1)  ...  3  2  1

For example, 10! = 10  9  ...  3  2  1 = 3628800,
and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.

Find the sum of the digits in the number 100!

Solution: 

now since 100! is too large to handled by long, int or any other primitive data type in java , we again use BigInteger library. The solution is very simple:

 

import java.math.BigInteger;


public class Prob20 {

 
 public static void main(String[] args) {
 
BigInteger sum = BigInteger.ZERO;
BigInteger fact = BigInteger.ONE;
BigInteger rem;

for(int i = 1 ; i <= 100 ; i++ )
{
 fact = fact.multiply(BigInteger.valueOf(i));
 
}

 
 while(!fact.equals(BigInteger.ZERO))
 {  
  rem= fact.mod(BigInteger.TEN);
  fact=fact.divide(BigInteger.TEN);
  sum = sum.add(rem);
 } 
 
 System.out.println(sum);

 }
 }